final examination in linear algebra: 18.06 ma y professor...

Final Examination in Linear Algebra: 18.06

May 18, 1998 9:00{12:00 Professor Strang

Your name is:

Please circle your recitation:

1) M2 2-132 M. Nevins 2-588 3-4110 monica@math

2) M3 2-131 A. Voronov 2-246 3-3299 voronov@math

3) T10 2-132 A. Edelman 2-380 3-7770 edelman@math

4) T12 2-132 A. Edelman 2-380 3-7770 edelman@math

5) T12 2-131 Z. Spasojevic 2-101 3-4470 zoran@math

6) T1 2-131 Z. Spasojevic 2-101 3-4770 zoran@math

7) T2 2-132 Y. Ma 2-333 3-7826 yanyuan@math

Grading 12345678

Answer all 8 questions on these pages. This is a closed book exam. Calculators are not

needed in any way and therefore not allowed (to be fair to all). Grades are known only to

your recitation instructor. Best wishes for the summer and thank you for taking 18.06.

GS

1 If A is a 5 by 4 matrix with linearly independent columns, �nd each of these explicitly:

(a) (3 points) The nullspace of A.

(b) (3 points) The dimension of the left nullspace N (AT ).

(c) (3 points) One particular solution xp to Axp = column 2 of A.

(d) (3 points) The general (complete) solution to Ax = column 2 of A.

(e) (3 points) The reduced row echelon form R of A.

2

2 (a) (5 points) Find the general (complete) solution to this equation Ax = b:

26664

1 1 2

1 1 2

2 2 2

37775

26664

x1

x2

x3

37775 =

26664

2

2

4

37775 :

(b) (3 points) Find a basis for the column space of the 3 by 9 block matrix [A 2A A2].

3

3 (a) (5 points) The command N = null (A) produces a matrix whose columns are a

basis for the nullspace of A. What matrix (describe its properties) is then produced

by B = null (N 0)?

(b) (3 points) What are the shapes (how many rows and columns) of those matrices

N and B, if A is m by n of rank r?

4

4 Find the determinants of these three matrices:

(a) (2 points)

A =

26666664

0 0 0 1

0 0 2 0

0 3 0 0

1 2 3 4

37777775

(b) (2 points)

B =

24 0 �A

I �I

35 (8 by 8, same A)

(c) (2 points)

C =

24 A �A

I �I

35 (8 by 8, same A)

5

5 If possible construct 3 by 3 matrices A, B, C, D with these properties:

(a) (3 points) A is a symmetric matrix. Its row space is spanned by the vector

(1; 1; 2) and its column space is spanned by the vector (2; 2; 4).

(b) (3 points) All three of these equations have no solution but B 6= 0:

Bx =

26664

1

0

0

37775 Bx =

26664

0

1

0

37775 Bx =

26664

0

0

1

37775 :

(c) (3 points) C is a real square matrix but its eigenvalues are not all real and not all

pure imaginary.

(d) (3 points) The vector (1; 1; 1) is in the row space of D but the vector (1;�1; 0) is

not in the nullspace.

6

6 Suppose u1; u2; u3 is an orthonormal basis for R3 and v1; v2 is an orthonormal basis

for R2.

(a) (5 points) What is the rank, what are all vectors in the column space, and

what is a basis for the nullspace for the matrix B = u1(v1 + v2)T ?

(b) (5 points) Suppose A = u1vT1+u2v

T2. Multiply AAT and simplify. Show that this

is a projection matrix by checking the required properties.

(c) (4 points) Multiply ATA and simplify. This is the identity matrix! Prove this (for

example compute ATAv1 and then �nish the reasoning).

7

7 (a) (4 points) If these three points happen to lie on a line y = C+Dt, what system

Ax = b of three equations in two unknowns would be solvable?

y = 0 at t = �1 ; y = 1 at t = 0 ; y = B at t = 1 :

Which value of B puts the vector b = (0; 1; B) into the column space of A?

(b) (4 points) For every B �nd the numbers C and D that give the best straight line

y = C +Dt (closest to the three points in the least squares sense).

(c) (4 points) Find the projection of b = (1; 0; 0) onto the column space of A.

(d) (2 points) If you apply the Gram-Schmidt procedure to this matrix A, what is

the resulting matrix Q that has orthonormal columns?

8

8 (a) (5 points) Find a complete set of eigenvalues and eigenvectors for the matrix

A =

26664

2 1 1

1 2 1

1 1 2

37775 :

(b) (6 points, 1 each) Circle all the properties of this matrix A:

A is a projection matrix

A is a positive de�nite matrix

A is a Markov matrix

A has determinant larger than trace

A has three orthonormal eigenvectors

A can be factored into A = LU

(c) (4 points) Write the vector u0 =

264

2

0

1

375 as a combination of eigenvectors of A, and

compute the vector u100 = A100u0.

9


May 18, 1998 Solutions Professor Strang

1. (a) zero vector f0g

(b) 5� 4 = 1

(c) xp =

2664

0

1

0

0

3775

(d) x = xp because N (A) = f0g.

(e) R =

266664

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

377775 =

�I

0

�

2. (a) A =

24 1 1 2

1 1 2

2 2 2

35 �! U =

24 1 1 2

0 0 �20 0 0

35 �! R =

24 1 1 0

0 0 1

0 0 0

35 :

The free variable is x2. The complete solution is

x = xp + xn =

24 2

0

0

35+ x2

24 �1

1

0

35 :

(b) A basis is

24 1

1

2

35 and

24 2

2

2

35

3. (a) The columns of B are a basis for the row space of A (because the row space is

the orthogonal complement of the nullspace).

(b) N is n by (n� r); B is n by r.

4. (a) detA = 6

(b) detB = 6

(c) detC = 0

5. (a) A =

24 1 1 2

1 1 2

2 2 4

35

(b) B =

24 1 1 1

1 1 1

1 1 1

35

(c) C =

24 1 1 0

�1 1 0

0 0 0

35

(d) D =

24 1 1 1

1 0 0

1 0 0

35

All four matrices are only examples (many other correct answers exist).

6. (a) rank(B) = 1; all multiples of u1 are in the column space; the vectors v1 � v2 and

v3 are a basis for the nullspace.

(b) AAT = (u1vT1+ u2v

T2)(v1u

T1+ v2u

T2) = u1u

T1+ u2u

T2since vT

1v2 = 0. AAT is

symmetric and it equals (AAT )2:

(u1uT1+ u2u

T2)(u1u

T1+ u2u

T2) = u1u

T1+ u2u

T2

(The eigenvalues of AAT are 1; 1; 0)

(c) ATA = v1vT1+ v2v

T2since uT

1u2 = 0.

ATAv1 = (v1vT1+ v2v

T2)v1 = v1

ATAv2 = (v1vT1+ v2v

T2)v2 = v2 :

Since v1; v2 are a basis for R2, ATAv = v for all v.

2

7. (a) Ax = b is

24 1 �1

1 0

1 1

35�C

D

�=

24 0

1

B

35. This is solvable if B = 2.

(b) ATA�x = AT b is �3 0

0 2

� �C

D

�=

�1 +B

B

�:

Then C = 1+B

3and D = B

2

(c) ATA�x = AT b is

A =

�3 0

0 2

� �x1x2

�=

�1

1

�p = A�x =

24 1 �1

1 0

1 1

35�1=3

1=2

�=

24 �1=6

1=3

5=6

35 :

(d) Q =

24 1=

p3 �1=

p2

1=p3 0

1=p3 1=

p2

35 columns were already orthogonal, now orthonormal

8. (a) � = 1; 1; 4. Eigenvectors can be24 1

�10

3524 1

0

�1

3524 1

1

1

35

(could also be chosen orthonormal because A = AT )

(b) Circle all the properties of this matrix A:

A is a projection matrix

A is a positive de�nite matrix

A is a Markov matrix

A has determinant larger than trace

A has three orthonormal eigenvectors

A can be factored into A = LU

(c) 24 2

0

1

35 =

24 1

�10

35+

24 1

1

1

35

A100

24 2

0

1

35 = 1100

24 1

�10

35 + 4100

24 1

1

1

35 =

24 4100 + 1

4100 � 1

4100

35 :

3

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��z� ;>� ��Z\[^] � nv` DvB©Ó ~�� X�X6X � Ó^Ô jml^n DvodB x6lûbqdt�adhslij F y,¼ odB a DvodB¤Ë n�h D x6lûbqdt�a�lij F ÉÕ`{h2|6~ ~ Ó ~�Ä»|��v~ Ó �ÖÄ×�×�× Ä·| Ô ~ Ó^Ô y�¼ o `bh¤`{h©[�ub`{a B [in�xEl^t%ed`bag[ D `{lâ}lij DpodB xElû{qdt�adh¤l�j F � odB adx B `{h�`{aÀ®¯Z F ]�y�¼ odBhp[�t B n B [ih�lâd`{ar¥�h o l ¢ h Dvo [ DkDporB l DpodB n2x6lûbqdt¯arh2lij F ÉØ[�n B `{a£®¯Z F ]�yZzeg] _gnvl^tÙwg[in D Zz[¦]��Ö®¯Z F É�]¤`{h�[�hvqdedh�wg[ix B lijs®�Z F ]�y�¼ odB n B jml^n B cd`{t�®¯Z F É%] ÚÕcd`{t�®�Z F ]��Ölinnp[�adÎ�Z F É%]2Únv[iadÎ�Z F ]�y

Û 564�ÇÈ9<;>=@? ACBEDkÜ µ¦Ý �sÞ Ý � e B©DvodB u{`{a B [in D np[iarh�jml^n�t�[ D `{lâfhv[ D `bh�jß°8`{ar¥Ü«àáâ IJK PM

TVUW ã6äå G IJK S�OQTVUW [�adc Üæàáâ IJK PP

TVUW ã6äå G IJK S�P�Mç TVUW XZ\[^] _�`badc Üèàáâ IJK M P

TVUW ã6äå yZzeg] � o [ D `bh DpodB t¯[ D n�`±¾ F B ¾-wdn B hvh�`{ad¥ Ü `ba DpB n�t¯hfl�j DvodB h D [iadcg[in�cHed[ihv`bhª B x D l^nvh IJK PMTVUW [iarc IJK M P

TVUW Ñ Zm¼ odB hv[it B ed[ihv`bh�`bh©qdh B cÈjml^n DvodB `{adwdq D [�adcDporB liq D wdq D yé]Zzx�] � o [ D `{h DporB t¯[ D n�`±¾£É B ¾-wdn B hvhv`bad¥ Ü `{a DpB nvt�h�lij DpodB eg[ihv`bh�xElâdhv`bh D `{ad¥�lijB `b¥ B a¦ª B x D l^nvh©l�j F©Ñ Zm¼ odB hp[it B ed[ihv`bh¤`{h�qdh B c1jmlin DporB `{adwdq D [iadcÀliq D wdq D yé]Zz¼ orB n B [in B�D ¢ l�w�l^h�hv`{eru B x6l^n�n B x D [iadh ¢êB nvhE�#c B w B arcd`{ad¥�lâ ¢ko [ D l^nvc B n�°il^qwd`bx�Î DpodB�B `{¥ B a�ª B x D l^n�h6yé]��z� ;>� ��Z\[^] Ü Zv¿�ë ~ Âm] G Ü Z�¿ ~~ Â�] S Ü Zv¿ ~ë Â�] GÙì>í8îï�ð yZzeg] FñGòì�í�óôí8î� ï�ð yZzx�] ¼ odB�B `b¥ B a¦ª�[iubq B hklij F [in B N [iadc S�P �ghvl¯É G ì � ëë í ~ ð y

õ 5648ö:9<;>=@? ACBED�÷ e B©DvodB hvqdedh�wg[ix B lij�Ý � x6lârhv`{h D `bad¥�lij�ª B x D l^n�h IJJJJK� ø ùT UUUUW hv[ D `bh�jß°8`{ar¥

� Ä N ø Sûú ù G M XZ\[^] _�`badc�[ Q�üÀN t¯[ D n�`±¾ F ¢ko l^h B x6liu{qdt�afhvwg[�x B `{h ÷ yZzeg] _�`badc£[ia�l^n Dvo lâdl^n�t�[�u�eg[ih�`{h2jml^n ÷ yZzx�] _�`badc DporB wdnvl@ý B x D `{lâ»t¯[ D n�`±¾(þÙwdn�l@ý B x D `bad¥Àlâ D l DporB u B j D a8qru{u{h�wg[ix B Zzadl DDporB xElû{qrt¯afh�wg[ix B�ÿ ]�lij F yZzcg] _�`badc DpodB u B [ih D h Ð qg[in B h�h�lû{q D `{lia D lF��G IJJJJK

PPPTVUUUUW X

��z� ;>� �� [�n D h Z\[^]��<Zme�]��[iarc(Zmc�] o [�ª B `{a Ë ad` DvB ub°¯t¯[ia¦°�w�l^hvh�`{edu B x6l^n�n B x D [iadh ¢ B n�h6yZ\[^] ¼ odB x6lûbqdt�adh�lij F h o l^qdu{cÈe B�D ¢ lYu{`ba B [�nvub°}`{arc B w B adc B a D h�lû{q D `{lârh6� D l � Ä N ø Sú ù G M y_gl^n B ¾d[�t¯wdu B � FñG�� ïí ~ ëë ~�� yZzeg] ¬�wrwdub°8`{ar¥ DpodB�� np[it�� Ò x o t¯`bc D wdn�l-x B h�h D l DvodB x6lûbqdt�adh2lij F °8` B u{crh ~ ï � �í ~ë � [�adc ~ î � ~� ~ � yZzx�] ACBED� e B�DporB t¯[ D n�`±¾ ¢ko l^h B x6lûbqdt�adh�[in B�DporB ª B x D l^nvh¨`{a:wg[in D Zme�]�y�¼ odB wrnvl@ý B x D `blâ:t�[��D n�`±¾´lâ D l DpodB x6lûbqdt�aûh�wg[ix B lij F `{h � ¸ y Ò `{adx BfDpodB u B j D a8qru{u{h�wg[ix B `bh DpodB l^n Dpo l^¥ilâg[iux6lit¯wdu B t B a D lij DporB xElû{qrt¯a�hvwd[ix B ��` D h2wdn�l@ý B x D `blâ�t�[ D nv` ¾�`bh� S � ¸ G ~� ë � ~ � í ï� ó�í ~ ëí ï í ~ ë � ï � yZzcg] ¼ odB hvliu{q D `{lâ D l F�¸�F��G F�¸ � ~~~ � `{h ��£G � í ~��~ ï�� y

� 564��:9<;>=@? Ò qdwrw�l^h BF��G IJJJJK

PNQTVUUUUW o [ihkadl�hvlûbq D `blâ

edq DF��G IJJJJK

Q N PT UUUUW o [�h�`ba Ë ad` DpB ub°¯t�[ia¦°fh�lû{q D `{liadh X

Z\[^] _�`badc([�u{u,w�l^hvh�`{edu B `barjml^n�t�[ D `{lâ1[�e�l^q D�� À��[iadc��yÈZz¼ odB np[iadÎ1[�adc DporBh o [iw B lij F yé]Zzeg] _�`badcÀ[ia B ¾d[�t¯wdu B lij�h�qdx o [�t�[ D nv` ¾ F ¢ ` Dvo�� À�C[iadc �´[iu{u�[ih¤h�t�[iubu�[�hw�l^hvh�`{edu B yZzx�] Í�l ¢ cdl�°^l^q£Î�adl ¢"Dpo [ D IJJJJK

Q N PTVUUUUW `bh2adl D `{a DporB a�qdubu{h�wg[ix B l�j F ¸ Ñ

��z� ;>� ��Z\[^] � G Q � M�! � !ÆQ ��[iadc � ! ��yZzeg] � ` Dvo"� G P �#� G Q [iadc"� G N µ FHG � � ë� ë~ ë � yZzx�] ¼ odB x6liu{qdt�a�hvwg[ix B [iadc DpodB u B j D a�qdu{ubhvwg[ix B [in B [iu ¢ [�°8hkl^n Dvo l^¥^liag[iuÃy

ö 5648Û:9<;>=@? � a B [ix o x6[ih B ¥^`±ª B [iu{u DpodB `{arjmlinvt¯[ D `blâÆ°^liq x�[�a [ie�l^q D�DvodBÈB `{¥ B a¦ªi[�u{q B h}[�adcB `b¥ B a¦ª B x D l^nvhE� ¢kodB a DpodB t¯[ D n�`±¾ F o [ih DpodB jmlûbu{l ¢ `{ad¥ wdnvl^w B n D °�µZ\[^] ¼ odB w�l ¢ B n�h F%$ [iwdwdn�l¦[ix o�DpodB ¹ B nvl�t¯[ D n�`±¾�yZzeg] ¼ odB t�[ D nv` ¾�`bh2h�°8t�t B�D nv`bx¤w�l^hv` D `bª B c BEË ad` DvB yZzx�] ¼ odB t¯[ D n�`±¾�`{h2adl D cr`�[i¥^liag[iu{`b¹�[ieru B yZzcg] ¼ odB t¯[ D n�`±¾ o [ih DporB jml^nvt F§G'&)( ¸ � ¢korB n B & [iadc ( [in B ª B x D linvh�`{aÀÝ � yZ+*ôliq£t�`{¥ o�Dê¢ [ia D�D l D n °�[ia B ¾r[it�wdu B yé]Z B ] F `{h2h�`{t�`{u{[in D l�[�cd`{[i¥^lâd[iu�t¯[ D n�`±¾ ¢ ` Dvo cr`�[i¥^liag[iu B a D nv` B h P � P �d[iadc N y

��z� ;>� ��Z\[^] ¼ odB�B `b¥ B a¦ª�[iubq B h�[in B e B�D ¢êBEB a S�P [�adc P yZzeg] ¼ odB B `{¥ B a¦ª�[iu{q B hº[in B w�l^h�` D `±ª B y�¼ odB n B [�n B ��u{`ba B [�nvub°¤`{adc B w B adc B a D �^l^n Dpo l^¥ilâg[iu B `b¥ B a¦ª B x D l^nvhEyZzx�] ¼ odB n B `bhº[�n B w B [ DvB c B `b¥ B a¦ª�[iubq B [iarc Dpo [ DºB `b¥ B a¦ª�[iubq Bêo [�h�j BE¢ B nºub`{a B [invu±°©`{arc B w B adc B a D�B `b¥ B a,�ª B x D l^nvh Dpo [�a£` D h2t%qru D `bwdu{`bx6` D °�[ih�[�n�l-l D lij DvodB x o [inv[ix DvB nv`bh D `{x�w�liub°8adl^t�`�[iu\yZzcg] ¼ odB�B `{¥ B a�ª�[iubq B h¤[in B M Z DpodB B `{¥ B a¦ª B x D linvh�[in B [iu{u<ª B x D linvh�l^n Dvo l^¥^liag[iu D l ( ]�[iarc & × ( Z DporBB `b¥ B a¦ª B x D l^nvh�[in B¤DpodB t%qdu D `bwdu B hslij & ]�yZ B ] ¼ odB¤B `{¥ B a¦ªi[�u{q B h2[in B P [iadc N [iarc DpodB n B [in B�Dpo n B6B ub`{a B [invu±°�`{adc B w B adc B a DkB `{¥ B a¦ª B x D linvh�Z D ¢ ljml^n.- G P [�adc}lia B jml^n.- G N ]�y

/ 56487:9<;>=@? 0¤BEË a B [�h B6Ð q B arx B lij,a�qdt%e B n�h©`{a DpodB jmliu{u{l ¢ `bad¥ ¢ [�°�µ21 ë G M �31�~ G P>¶iN ��[�adc1 $54 � G Z�1 $54 ~,Ä61 $ ] ¶^N y�Z\Ì�[ix o a�qdt%e B n�`bh DpodB [�ª B nv[i¥ B lij DporB�D ¢ lÀwdn B ª8`{liqdha�qdt%e B n�h6yé]Z\[^] Ò B�D qrw�[ N�üÀN t�[ D nv` ¾ F D l¯¥ B�D jmnvl^t IJK 1 $54 ~1 $

T UW D l IJK 1 $54 �1 $54 ~T UW yZzeg] _�`badc£[ia B ¾-wdu{`bx6` D jml^nvt qdu�[ jml^n71 $ yZzx�] � o [ D `bh DpodB ub`{t�` D lij81 $ [ih%9 Þ : Ñ��z� ;>� ��Z\[^] FñGòì ~��ô~��~ ë ð yZzeg] F o [ih B `{¥ B a¦ªi[�u{q B h P [�adc S¨P>¶iN ¢ ` Dpo x6linvn B hvw�l^adcd`bad¥ B `b¥ B a¦ª B x D l^nvh�Z Pi²�P ]�[iarc�Z P^²�S�N ]�y � n�` DpBDporB `{ad` D `{[iu,h D [ DpB [ih%[Yu{`{a B [in�x6l^t ed`{ag[ D `{lia(lij DporB�B `b¥ B a¦ª B x D l^nvhEµ ì ~��ë ð G ~� ¿ ~~ Â�Ä ~î ì ~í � ð y¬�wrwdub°8`{ar¥ F $ � ¢ B ¥ BED �<;#=?>A@; = � G ~� Z P ] $ ¿ ~ ~ Â-Ä ~î Z S ~� ] $ ì ~í � ð �gh�lB1 $ G ~� Z P2S Z S ~� ] $ ]�yZzx�] ¬�h%9 Þ : �C1 $ Þ ~� y

D 56487:9<;>=@? Z\[^] Ò qdwrw�l^h B F `bhk[ O¯üYO t¯[ D n�`±¾�lij�np[iadÎ Q ��[iadc�u B�D�fG

IJJJJJJJK®�~ ~®�~z�®�~z�®�~ ó

TVUUUUUUUWe B�DporB x6lijz[�x D linvhêl�j<` D h Ë nvh D nvl ¢ y�Ì�¾-wdu�[�`{a ¢ko ° F��G M y�Z Ò l DpodB xElijz[ix D l^n�h¥^`±ª B [%jml^nvt qdu�[ jml^nk[�a�qdubu{hvwd[ix B ª B x D l^n ÿ ]Í�`ba D µê¼ orB¨Ë n�h D x6l^t�w�lia B a D lij F�� [�adc DvodB h B xElâdcÀx6lit¯w�lâ B a D lij F�� [in Bc B�DpB n�t¯`bag[ia D h2l�jsZmcd`FE B n B a D ]st¯[ D n�`{x B h6y � o [ D [in B�DvodB h B t�[ D nv`bx B hs[iadc ¢ko °cdl DvodB ° o [�ª B ¹ B nvlfc BEDpB nvt�`{ag[�a D h Ñ Zm¼ odB Q n�c1[iadc O Dvo x6l^t�w�lia B a D h�lij F��jmlûbu{l ¢ h�`{t�`{u{[invu±°^��hvl�°^l^q x�[ia�ý�qrh D [�adh ¢ B nÖjml^n DvodB P h D [iadc N arc x6lit¯w�lâ B a D h6yé]Zzeg] Gêl^t�wdq DvB�DvodB c BEDvB n�t¯`bag[ia D lij

É GIJJJJJJJKP P P PP S�P P S�PP P S�P S�PP S�P S�P P

TVUUUUUUUW XÍ�`ba D µH*ôl^q�t¯`b¥ o¦D Ë arcf` D x6lâ¦ª B ad` B a D2D l�qdh B�DpodB jz[ix D Dpo [ D DpodB x6lûbqdt¯arhs[in Bl^n Dpo l^¥ilâg[iu\y

��z� ;>� ��Z\[^] ¼ odB%Ë n�h D x6lit¯w�lâ B a D l�j F�� `bh¤c BED F � ¢ko `{x o `{h�¹ B n�l}e B x6[iqdh B F o [ih©np[iarÎ Q y%¼ orB h B xElâdcx6lit¯w�lâ B a D lij F�� `{h DpodB c B�DpB n�t¯`bag[ia D lij DporB t¯[ D n�`±¾�lê D [i`ba B c}e�°£n B wdu{[ix6`bad¥ DvodB¨Ë nvh D nvl ¢lij F ¢ ` Dvo£DvodB h B xElâdc£nvl ¢ y�¼ o `{hkt¯[ D n�`±¾ o [ihkn B w B [ DpB c}n�l ¢ h6��hvl�` D h2c BEDvB nvt�`{ad[ia D `{h2¹ B nvldyZzeg] c B�D É G P�R yI*ôliq�x�[ia�xEl^t¯wrq DpBsDvo `{h�e�° ednvq DpB jml^n�x B y�¬�adl DvodB n ¢ [�°�`bh D l©arl DpBsDvo [ D É � G O ��hvl DvodB©B `b¥ B a¦ª�[iu{q B h2lij�É§[in B�J N y Ò `{arx B¤DporB¤D np[�x B lij�ÉÕ`{h2¹ B nvld� DpodB�B `b¥ B a¦ª�[iubq B hk[in B - G N �N � S�N ��[iarc S�N �dh�l¯c B�D É G P�R y


May 18, 1999 1:30 { 4:30 Professor Strang

Your name is:

Secret Code (optional):


1) Mon 2{3 2-131 S. Kleiman 5) Tues 12{1 2-131 S. Kleiman

2) Mon 3{4 2-131 S. Hollander 6) Tues 1{2 2-131 S. Kleiman

3) Tues 11{12 2-132 S. Howson 7) Tues 2{3 2-132 S. Howson

4) Tues 12{1 2-132 S. Howson

Grading 123

45

67

89

Answer all 9 questions on these pages. This is a closed book exam. Calculators are not

needed in any way and therefore not allowed (to be fair to all). Grades are known only

to your recitation instructor (who is free to post with secret codes and to return exams in

person). Solutions will be posted on the web in a few days. Best wishes for the summer and

thank you for taking 18.06.

GS

1 Suppose the matrix A is the product

A =

26664

1 0 0

2 1 0

5 4 3

37775

26664

1 0 3 4

0 1 1 0

0 0 0 0

37775 :

(a) (3 points) Give a basis for the nullspace of A.

(b) (3 points) Give a basis for the row space of A.

(c) (2 points) Express row 3 of A as a combination of your basis vectors

in your answer to (b).

(d) (3 points) What is the dimension of the nullspace of AT ?

2 Suppose A is a 5 by 7 matrix, and Ax = b has a solution for every right side b.

(a) (3 points) What do we know about the column space of A?

(b) (3 points) What do we know about the rows of A?

(c) (3 points) What do we know about the nullspace of A?

(d) (3 points) True or false (with reason):

The columns of A are a basis for the column space of A.

Your answers could refer to dimension/basis/linear independence/spanning a space.

3 Assume that A is invertible and permutations are not needed in elimination if possible.

(a) (2 points) Are the pivots of A�1 equal to1

pivots of A?

If yes, give a reason; if no, give an example.

(b) (3 points) Is the product of pivots of A�1 equal to1

product of pivots of A?

If yes, give a reason; if no, give an example.

(c) (3 points) Apply block elimination to the 2n by 2n matrix

M =

24 A I

�I 0

35 :

Multiply block row 1 by a suitable matrix and add to block row 2.

What matrix appears in the (2; 2) block?

(d) (3 points) What is the determinant of M? Explain the �nal plus or minus sign.

4 Suppose A has eigenvalues �1 = 3, �2 = 1, �3 = 0 with corresponding eigenvectors

x1 =

26664

1

1

1

37775 ; x2 =

26664

0

1

1

37775 ; x3 =

26664

0

0

1

37775 :

(a) (3 points) How do you know that the third column of A contains all zeros?

(b) (3 points) Find the matrix A.

(c) (3 points) By transposing S�1AS = ^, �nd the eigenvectors y1; y2; y3 of AT .

(I am looking for speci�c vectors like x1; x2; x3 above.)

5 This problem computes the plane z = Cx +Dy + E that is closest (in the least squares

sense) to these four measurements:

At x = 1; y = 0 measurement gives z = 1




(a) (3 points) Write down the linear systemAx = b with unknown vector x = (C;D;E)

that would give a plane going exactly through the four given points | except that

this particular system has no solution.

(b) (3 points) Show that this system Ax = b has no solution!

(c) (3 points) Find the best least squares solution bx = ( bC; bD; bE).(d) (3 points) The error vector e = (e1; e2; e3; e4) in the underlying projection problem

is perpendicular to which vectors? You don't have to compute e but you do have to

say which speci�c numerical vectors it is perpendicular to.

6 (a) (3 points) The vectors q1 =1p50

26664

3

4

5

37775 and q2 =

1

5

26664

4

�3

0

37775 are orthonormal.

Find one more vector to complete an orthonormal basis for R3.

(b) (3 points) In solving part (a), you might start with a vector like a3 = (0; 0; 1)

and �nd q3. Which vectors a3 would not work as starting vectors to �nd q3 by

Gram-Schmidt? How many di�erent real vectors q3 will give a correct answer to

part (a)?

(c) (3 points) Project the vector a3 = (0; 0; 1) onto the plane spanned by q1 and q2.

Find its projection p.

7 In each part, �nd the required matrix or explain why such a matrix does not exist.

(a) (3 points) The matrices A and AT and A+ AT have ranks 2 and 1 and 3.

(b) (3 points) The solution to Ax = 0 is unique, but the solution to ATx = 0 is

not unique.

(c) (3 points) The powers Ak do not approach the zero matrix as k ! 1, but the

exponential eAt does approach the zero matrix as t!1.

(d) (3 points) The complete solution to

Ax =

26664

1

1

2

37775 is x =

26664

1

1

1

37775+ c1

26664

1

0

1

37775+ c2

26664�1

1

0

37775 :

(e) (3 points) The pivots are �1 and �2 but the eigenvalues are +1 and +2.

(Symmetric matrix not required, row exchanges not required.)

8 (a) (3 points) The \big formula" for a 6 by 6 determinant has 6! = 720 terms. How

many of those terms are sure to be zero if we know that a15 = 0?

(b) (2 points) If U and V are 3 by 3 orthogonal matrices, is their product UV always

orthogonal? Why (give reason) or why not (give example)?

(c) (2 points) If A and B are 3 by 3 symmetric matrices, is their product AB always

symmetric? Why (give reason) or why not (give example)?

(d) (3 points) For which numbers c is the matrix A positive de�nite? For which

numbers c is A2 positive de�nite? Why?

A =

26664

1 2 3

2 c 4

3 4 9

37775

9 Suppose the 3 by 3 matrix A has the following property Z: Along each of its rows, the

entries add up to zero.

(a) (3 points) Find a nonzero vector in the nullspace of A.

(b) (3 points) Prove that A2 also has property Z.

(c) (3 points) What can you say about the dimension of the nullspace of AT and why?

(d) (2 points) Find an eigenvalue of the matrix A+ 4I.


May 18, 1999 Solutions Professor Strang

1. (a)2664�3�11

0

3775 and

2664�40

0

1

3775

(b)2664

1

0

3

4

3775 and

2664

0

1

1

0

3775

(c) 5(row 1) + 4(row 2)

(d) A has rank 2 and AT is 4 by 3 so its nullspace has dimension 3� 2 = 1.

2. (a) C(A) = R5 since every b is in the column space.

(b) The rank is 5 so the �ve rows must be linearly independent.

(c) The nullspace must have dimension 7� 5 = 2.

(d) This is false because the 7 columns cannot be linearly independent.

3. (a) This is generally false, as for A =

�1 1

1 2

�and A�1 =

�2 �1

�1 1

�.

Note that A = LDU gives A�1 = U�1D�1L�1 (upper times lower!).

(b) True because detA�1 = 1=(detA).

(c) Multiply row 1 by A�1 and add to row 2 to obtain

�A I

0 A�1

�.

(d) The determinant is +1. Exchange the �rst n columns with the last n. This pro-

duces a factor (�1)n and leaves

�I A

0 �I

�which is triangular with determinant

(�1)n. Then (�1)n(�1)n = +1.

4. (a) From Ax3 = �3x3 we have A

24 0

0

1

35 =

24 0

0

0

35.

(b) A = S ^ S�1 =

24 1 0 0

1 1 0

1 1 1

3524 3

1

0

3524 1 0 0

�1 1 0

0 �1 1

35 =

24 3 0 0

2 1 0

2 1 0

35.

(c) Transpose S�1AS = ^ to get STAT (S�1)T = ^. Then the columns of (S�1)T are

the eigenvectors of AT , and part (b) gives (S�1)T =

24 1 �1 0

0 1 �10 0 1

35.

5. (a) 1C + 0D + E = 1

1C + 2D + E = 3

0C + 1D + E = 5

0C + 2D + E = 0

is Ax = b.

(b) Subtract equation (1) from equation (2):

2D = 2 gives D = 1

D + E = 5 gives E = 4

2D + E = 0 is now false

(c) Solve ATAx = AT b:

24 1 1 0 0

0 2 1 2

1 1 1 1

352664

1 0 1

1 2 1

0 1 1

0 2 1

377524 C

D

E

35 =

24 1 1 0 0

0 2 1 2

1 1 1 1

352664

1

3

5

0

3775

24 2 2 2

2 9 5

2 5 4

3524 C

D

E

35 =

24 4

11

9

35

24 2 2 2

0 7 3

0 0 5

7

3524 C

D

E

35 =

24 4

7

2

35

Back-substitution gives E = 14

5, D = �1

5, C = �3

5.

(d) The error vector e is perpendicular to the three columns of A.

2

6. (a) One way is to solve for x perpendicular to q1 and q2:

�3 4 5

4 �3 0

�24 x1x2x3

35 =

�0

0

�

Another way is Gram-Schmidt and we might as well start with a3 = (0; 0; 1).

Then Gram-Schmidt subtracts o� projections:

a3 � (aT3q1)q1 � (aT

3q2) =

24 0

0

1

35� 5

50

24 3

4

5

35� 0 =

24 �:3�:4

:5

35 :

Normalizing to a unit vector gives

q3 =1p50

24 �3�4

5

35 :

(b) a3 will not work if it is in the plane of q1 and q2.

The only possible vectors q3 are +(our q3) and �(our q3).

(c) The projection is the vector that was subtracted o� in part (a):

p =5

50

24 3

4

5

35 =

24 0:3

0:4

0:5

35 :

7. (a) Cannot exist because A and AT have the same rank.

(b) A =

�1

2

�or any non-square A with independent columns.

(c) The desired A has an eigenvalue like �2, outside the unit circle and in the left

half-plane. In fact, A = [�2] is a 1 by 1 example.

(d) From the two given nullspace vectors we know that A =�v v �v

�for some

column v. The particular solution (1; 1; 1) determines v:

A

24 1

1

1

35 =

24 1

1

2

35 gives v + v � v =

24 1

1

2

35 so v =

24 1

1

2

35 :

(e) (My favorite this year)

The �rst pivot must be a11 = �1. The the trace 1+ 2 requires a22 = 4. Then the

determinant must be 2, so these matrices will work:

A =

��1 �16 4

�or any A =

��1 �a6=a 4

�:

3

8. (a) 5! = 120 terms are sure to be zero.

(b) Yes, (UV )T (UV ) = V TUTUV = V TV = I.

(c) No, symmetry would need AB = (AB)T = BTAT = BA and we don't normally

have AB = BA.

(d) The 1 by 1, 2 by 2, 3 by 3 determinants are 1, c � 4, and �4 (not depending

on c!). The last is negative so A is not positive de�nite. But detA = �4 so A

has no zero eigenvalues so A2 has all three positive eigenvalues.

9. (a) x0 =

24 1

1

1

35 has Ax0 = 0.

(b) A2x0 = A(Ax0) = 0

(c) The dimension of N(AT ) is at least 1 (because A is square and we know that

(1; 1; 1) is in N(A)).

(d) A is singular so � = 0 is an eigenvalue of A so � = 4 is an eigenvalue of A+ 4I.

4

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18.06 Professor Strang/Ingerman Final Exam December 17, 2002

Your name is:


1) M2 2-131 P.-O. Persson 2-088 2-1194 persson

2) M2 2-132 I. Pavlovsky 2-487 3-4083 igorvp

3) M3 2-131 I. Pavlovsky 2-487 3-4083 igorvp

4) T10 2-132 W. Luo 2-492 3-4093 luowei

5) T10 2-131 C. Boulet 2-333 3-7826 cilanne

6) T11 2-131 C. Boulet 2-333 3-7826 cilanne

7) T11 2-132 X. Wang 2-244 8-8164 xwang

8) T12 2-132 P. Clifford 2-489 3-4086 peter

9) T1 2-132 X. Wang 2-244 8-8164 xwang

10) T1 2-131 P. Clifford 2-489 3-4086 peter

11) T2 2-132 X. Wang 2-244 8-8164 xwang

The ten questions are worth 10 points each.

Thank you for taking 18.06!

1 The 4 by 6 matrix A has all 2’s below the diagonal and elsewhere all 1’s:

A =

1 1 1 1 1 1

2 1 1 1 1 1

2 2 1 1 1 1

2 2 2 1 1 1

(a) By elimination factor A into L (4 by 4) times U (4 by 6).

(b) Find the rank of A and a basis for its nullspace (the special solutions would be

good).

2

2 Suppose you know that the 3 by 4 matrix A has the vector s = (2, 3, 1, 0) as a basis

for its nullspace.

(a) What is the rank of A and the complete solution to Ax = 0?

(b) What is the exact row reduced echelon form R of A?

3

3 The following matrix is a projection matrix :

P =1

21

1 2 −4

2 4 −8

−4 −8 16

.

(a) What subspace does P project onto?

(b) What is the distance from that subspace to b = (1, 1, 1)?

(c) What are the three eigenvalues of P? Is P diagonalizable?

4

4 (a) Suppose the product of A and B is the zero matrix: AB = 0. Then the (1)

space of A contains the (2) space of B. Also the (3) space of B contains

the (4) space of A. Those blank words are

(1) (2) (3) (4)

(b) Suppose that matrix A is 5 by 7 with rank r, and B is 7 by 9 of rank s. What

are the dimensions of spaces (1) and (2) ? From the fact that space (1) contains

space (2) , what do you learn about r + s?

5

5 Suppose the 4 by 2 matrix Q has orthonormal columns.

(a) Find the least squares solution x to Qx = b.

(b) Explain why QQT is not positive definite.

(c) What are the (nonzero) singular values of Q, and why?

6

6 Let S be the subspace of R3 spanned by

1

2

2

and

5

4

−2

.

(a) Find an orthonormal basis q1, q2 for S by Gram-Schmidt.

(b) Write down the 3 by 3 matrix P which projects vectors perpendicularly onto S.

(c) Show how the properties of P (what are they?) lead to the conclusion that Pb

is orthogonal to b − Pb.

7

7 (a) If v 1, v 2, v 3 form a basis for R3 then the matrix with those three columns is

.

(b) If v 1, v 2, v 3, v 4 span R3, give all possible ranks for the matrix with those four

columns. .

(c) If q1, q2, q3 form an orthonormal basis for R3, and T is the transformation that

projects every vector v onto the plane of q1 and q2, what is the matrix for T in

this basis? Explain.

8

8 Suppose the n by n matrix An has 3’s along its main diagonal and 2’s along the

diagonal below and the (1, n) position:

A4 =

3 0 0 2

2 3 0 0

0 2 3 0

0 0 2 3

.

Find by cofactors of row 1 or otherwise the determinant of A4 and then the determi-

nant of An for n > 4.

9

9 There are six 3 by 3 permutation matrices P .

(a) What numbers can be the determinant of P? What numbers can be pivots?

(b) What numbers can be the trace of P? What four numbers can be eigenvalues

of P?

10

10 Suppose A is a 4 by 4 upper triangular matrix with 1, 2, 3, 4 on its main diagonal.

(You could put all 1’s above the diagonal.)

(a) For A − 3I, which columns have pivots? Which components of the eigenvector

x 3 (the special solution in the nullspace) are definitely zero?

(b) Using part (a), show that the eigenvector matrix S is also upper triangular.

11

Course 18.06, Fall 2002: Final Exam, Solutions

1 (a)

A = LU =

1 0 0 02 1 0 02 0 1 02 0 0 1

1 1 1 1 1 10 −1 −1 −1 −1 −10 0 −1 −1 −1 −10 0 0 −1 −1 −1

.

(b) Four pivots ⇒ rank of A = 4. The row reduced echolon form is

R =

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 1 1

.

The two special solutions (0, 0, 0,−1, 1, 0), (0, 0, 0,−1, 0, 1) form a basis for the nullspace.

2 (a) The dimension of the nullspace is 1, so the rank of A is 4−1 = 3. The complete solutionto Ax = 0 is x = c · (2, 3, 1, 0) for any constant c.

(b) The row reduced echelon form has 3 pivots and the special solution x = (2, 3, 1, 0):

R =

1 0 −2 00 1 −3 00 0 0 1

.

3 (a) The projection matrix P projects onto the column space of P which is the line c·(1, 2,−4).

(b) The vector from b to the subspace is

e = b − Pb =

111

− 121

−1−24

=121

222317

and the distance is

‖e‖ =121

√222 + 232 + 172 =

√130221

.

(c) Since P projects onto a line, its three eigenvalues are 0, 0, 1. Since P is symmetric, ithas a full set of (orthogonal) eigenvectors, and is then diagonalizable.

4 (a) When AB = 0, every column of B is in the nullspace of A. So the null space of Acontains the column space of B. Also the left null space of B contains the row spaceof A.

(b) The dimension of the nullspace of A is n−r = 7−r. The dimension of the column spaceof B is s. Since the first contains the second, 7− r ≥ s, or r + s ≤ 7.

5 (a) The least squares solution x to Qx = b is

x = (QTQ)−1QTb = (I)−1QTb = QTb.

(b) One approach: QT is 2 by 4 so there are free variables and many solutions to QTx = 0.Then QQTx = 0 and QQT is singular; not positive definite.Second approach: The rank of Q is 2, so the rank of QQT must be ≤ 2. But QQT isa 4 by 4 matrix, so the dimension of its nullspace is 2. This means that it has 2 zeroeigenvalues, and QQT is not positive definite.

(c) The singular values of Q are the square roots of the eigenvalues of QTQ = I, that is, all1.

6 (a) Let a = (1, 2, 2) and b = (5, 4,−2). The orthonormal vectors are

q1 =a

‖a‖=

13

122

,

q2 =b − qT

1 bqT

1 qq1

‖ · ‖=

1√42 + 22 + (−4)2

42−4

=13

21−2

where ‖ · ‖ means the norm of the numerator.

(b) Let Q =[q1 q2

]. The projection P is then

P = QQT =19

5 4 −24 5 2−2 2 8

(c) The properties of a projection matrix are P 2 = P and PT = P . This gives

(Pb)T(b − Pb) = bTPT(b − Pb) = bT(Pb − P 2b) = 0.

7 (a) If v1, v2, v3 is a basis for R3 then the matrix with those three columns is invertible(non-singular, full rank).

(b) If v1, v2, v3, v4 span R3 then the column space is R3. The only possible rank for thematrix with those four columns is 3.

(c) The transformations of the three basis vectors are

T (q1) = q1

T (q2) = q2

T (q3) = 0

so the transformation matrix T in the basis q1, q2, q3 is

T =

1 0 00 1 00 0 0

.

8 ∣∣∣∣∣∣∣∣3 0 0 22 3 0 00 2 3 00 0 2 3

∣∣∣∣∣∣∣∣ = 3

∣∣∣∣∣∣3 0 02 3 00 2 3

∣∣∣∣∣∣− 2

∣∣∣∣∣∣2 3 00 2 30 0 2

∣∣∣∣∣∣ = 3 · 27− 2 · 8 = 65.

For general n > 4, the determinant is |An| = 3n + (−1)n−12n.

9 (a) The determinant of a permutation matrix P is 1 or −1. The only possible pivot is 1.

(b) For 3 by 3 permutations, the trace of P is 0, 1, or 3. The eigenvalues are 1 and −1 whentwo rows are exchanged. Otherwise∣∣∣∣∣∣

−λ 0 11 −λ 00 1 −λ

∣∣∣∣∣∣ = −λ3 + 1 = 0 ⇒ λ = 1,−12± i

√3

2(or e2πi/3, e4πi/3),

so the four possible eigenvalues are 1,−1,−12 + i

√3

2 , and −12 − i

√3

2 .

10 (a)

A =

1 1 1 10 2 1 10 0 3 10 0 0 4

, A− 3I =

−2 1 1 10 −1 1 10 0 0 10 0 0 1

Columns 1, 2, and 4 have pivots. Because of the nonzero bottom-right element in A−3I,the fourth component of x 3 is definitely zero.

(b) In the same way as above, the special solutions for the matrices A− 1I, A− 2I, A− 3I,and A − 4I must have 3, 2, 1, and 0 zeros as the last components. The eigenvectormatrix S is then upper triangular.

Questions from 18.06 Final, Fall 2003

1. Suppose A = LU where

L =

1 0 02 1 0−2 3 1

, U =

5 0 5 10 3 3 00 0 0 0

.

(a) What are the dimensions of the 4 fundamental subspaces associated with A?

(b) Give a basis for each of the 4 fundamental subspaces.

1

N(A)

R(A)

C(A)

N(AT )

2

2. Let F be the subspace of R4 given by

F = {(x, y, z, w) : x − y + 2z + 3w = 0}.

Let P be the projection matrix for projecting onto F . (Many of the subquestions can be answeredindependently of the others.)

(a) Give an orthonormal basis {v1, · · · , vk} for the orthogonal complement to F .

(b) Find an orthonormal basis {w1, · · · , wl} for F . Explain how you proceed.

3

The following questions refer to the projection matrix P for projecting onto F .

(c) What are the eigenvalues of P ? Give them with their multiplicities.

(d) What is the projection of

1111

onto F ?

4

3. (a) Write down the 2×2 rotation matrix, R(θ), that rotates R2 in the counterclockwise directionby an angle θ (this matrix is a function of θ).

(b) Compute the eigenvalues of R(θ). For which value(s) of θ are the eigenvalues real?

(c) What are the eigenvectors of R(θ).

5

(d) Write down the singular value decomposition of R(θ).

6

4. (a) Give two 3 × 3 matrices A and B such that AB is not equal to BA.

(b) Suppose A and B are n×n matrices with the same set of linearly independent eigenvectorsv1, v2, · · · , vn. However, the eigenvalues might be different: vi is the eigenvector for theeigenvalue λi of A and the eigenvector for the eigenvalue µi of B. Show that AB = BA.

7

5. Consider the differential equation

[

dudtdvdt

]

=

[

0 32 −1

] [

u

v

]

.

(a) Solve the differential equation and express u(t), v(t) as functions of u(0) and v(0).

8

(b) Find a linear transformation

[

p

q

]

= T

[

u

v

]

such that the differential equation simplifies

into two independent differential equations in p and in q (one relating dpdt

and p, the other

relating dqdt

and q)

(c) Are there initial conditions u(0), v(0) that would make u(t) blow up? If yes, give one suchvalue for u(0) and v(0).

(d) Are there initial conditions u(0), v(0) that would make u(t) go to 0? If yes, give one suchvalue for u(0) and u′(0).

9

6. True of False. Circle the appropriate answer. “True” means “always true”, and “false” means“sometimes false”. Justify each answer briefly.

(a) The product of the pivots when performing Gauss-Jordan is equal to the determinant if wedo not have to permute rows.

True or False.

(b) Let A be an m× n matrix whose columns are independent. Then AAT is positive definite.

True or False.

(c) If A is a symmetric matrix then the singular values are the absolute values of the nonzeroeigenvalues.

True or False.

(d) There exists a 5 × 5 unitary matrix with eigenvalues 1, 1 + i, 1 − i, i and −i.

True or False.

(e) Suppose V and W are two vector spaces of dimension n. If T is a linear transformationfrom V to W with only the 0 vector in the kernel, then for any basis of V , there exists anorthonormal basis of W such that the resulting matrix representing T is upper triangular.

True or False.

10

7. Consider the following matrix A:

A =

0.5 0.4 0.20.4 0.5 0.20.1 0.1 0.6

.

(c) Can you immediately tell one of the eigenvalues of A (without computing them)? Explain.

(d) Compute the determinant of A.

(e) Find the eigenvalues of A and the corresponding eigenvectors. (Check your answer.)

11

(f) Two out of the 3 eigenvectors of A should be orthogonal to

111

. How could you have

explained this before computing the eigenvectors?

(g) Write an exact expression for A100.

12

8. Let

A =

1 0 00 3 00 0 5

.

For each of the following matrices, either complete it (find values for the non-diagonal elements)so that it becomes similar to A, or explain why it is impossible to complete it to a matrix similarto A. Circle whether you are able to complete it or not to a matrix similar to A.

(a)

B =

2 . .

. 2 .

. . 4

Able to complete it to similar?: Yes No

If yes, give a completion. If not, why not?

13

(b)

C =

3 . .

. 3 .

. . 3

.

Able to complete it to similar?: Yes No

If yes, give a completion. If not, why not?

14

18.06 Professor Strang Final Exam December 19, 2005

Grading

1

2

3

4

5

6

7

8

Your PRINTED name is:


1) M 2 2-131 P. Lee 2-087 2-1193 lee

2) M 2 2-132 T. Lawson 4-182 8-6895 tlawson

4) T 10 2-132 P-O. Persson 2-363A 3-4989 persson

5) T 11 2-131 P-O. Persson 2-363A 3-4989 persson

6) T 11 2-132 P. Pylyavskyy 2-333 3-7826 pasha

7) T 12 2-132 T. Lawson 4-182 8-6895 tlawson

8) T 12 2-131 P. Pylyavskyy 2-333 3-7826 pasha

9) T 1 2-132 A. Chan 2-588 3-4110 alicec

10) T 1 2-131 D. Chebikin 2-333 3-7826 chebikin

11) T 2 2-132 A. Chan 2-588 3-4110 alicec

12) T 3 2-132 T. Lawson 4-182 8-6895 tlawson

1 (12 pts.) This question is about the matrix A = I +E where E is the all-ones matrix

ones(4, 4):

A =

2 1 1 1

1 2 1 1

1 1 2 1

1 1 1 2

= I +

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

.

(a) By elimination find the pivots of A.

(b) Factor A into LDLT (if that is possible).

(c) The inverse matrix has the form A−1 = I + cE. Figure out E2 and

then choose the number c so that AA−1 = I.

2

2 (12 pts.) Keep the same matrix A as in Problem 1.

(a) Find the matrix P that projects any vector in R4 onto the subspace

spanned by the first column of A.

(b) Describe the nullspace of I − P and the nullspace of PA.

(c) Find all the eigenvalues of P .

3

3 (12 pts.) Now suppose A = I + bE, with the same E = ones(4, 4).

(a) What are the eigenvalues of E ?

(b) If b = 2, what is the determinant of A ?

(c) Suppose you know that xTAx > 0 for every nonzero vector x. (Same

matrix A.) What are the possible values of b ?

4

4 (16 pts.) Suppose A is an 8 by 8 invertible matrix. Throw away any 3 columns of A

to get an 8 by 5 matrix B.

(a) You will correctly think that B has rank 5. Give a mathematical reason

why this is true.

(b) Tell all you know about the nullspace of BT and the reduced row

echelon form rref(B).

(c) Give as much information as possible about the eigenvalues and eigen-

vectors of BTB and BBT (those are separate questions).

5

5 (12 pts.) Suppose Q is an m by n matrix with QTQ = I. Write down the most

important facts about

(a) The columns of Q

(b) m and n and the rank of Q

(c) The least squares solution x to Qx = b

6

6 (12 pts.) (a) The eigenvalues of A =

0 0 0 1

1 0 0 0

0 1 0 0

0 0 1 0

are .

(b) An orthogonal set of 4 eigenvectors is .

(c) CIRCLE every class of matrices to which this matrix A belongs:

diagonalizable permutation nonsingular

Jordan matrix orthogonal projection

skew-symmetric

7

7 (12 pts.) Suppose A is 2 by 3 with this Singular Value Decomposition UΣV T. U and

V are orthogonal matrices:

A =[

u1 u2

]

4 0 0

0 0 0

vT

1

vT

2

vT

3

.

(a) Find a basis for the nullspace of A.

(b) Find all solutions to the equation Ax = u1.

(c) Find the shortest solution to Ax = u1 (minimum length vector) and

prove that it is shortest.

8

8 (12 pts.) Suppose A (3 by 3) has eigenvalues λ1, λ2, λ3 and independent eigenvectors

x1, x2, x3.

(a) What is the general form of the solutions to uk+1 = Auk anddu

dt= Au ?

(Two questions)

(b) Suppose every solution to uk+1 = Auk approaches a multiple c x1 as

k → ∞ (c depends on u0). What does this tell you about λ1, λ2, λ3 ?

(c) For some 3 by 3 matrices, the complete solution todu

dt= Au does not

have the form you gave in part (a). What can go wrong ? Give an

example of such a matrix A.

9

1.(a) 2, 32 ,43 ,

54

(b)A =

112 112

13 1

12

13

14 1

232

43

54

112 112

13 1

12

13

14 1

T

(c) c = −1/5.since E2 = 4E and AA−1 = (I+E)(I+cE) = I+(c+1+4c)Eso 5c+ 1 = 0.

2.(a) P = aaT

aT a= 1

7

4 2 2 22 1 1 12 1 1 12 1 1 1

, Trick: computation easiest using

a = (2, 1, 1, 1)T

.

(b) The nullspace of I − P consists of all multiples of a. (One view is thatx = Px. Another view is that it is the orthogonal complement of the nullspaceof P which are all vectors orthogonal to a.)

(c) The matrix has rank 1 and is a projector, so one eigenvalue is 1 and therest are 0.

3.(a) E is rank one symmetric with trace 4, so the eigenvalues are 0, 0, 0, 4.

(b) The eigenvalues of A are 1, 1, 1, and 1+2 · 4 = 9 so the determinant is 9.

(c) We need 1 + 4b > 0 so b > −1/4.

4.(a) The columns of B are independent since the columns of A are. There-fore the span of B is �ve dimensional. (B has no free columns.)

(b) The nullspace of BT is a three dimensional subspace of R8.It is theorthogonal complement of the column space of B in R8. The rref of B looks likethe �rst �ve columns of I8.

(c) BTB is a 5× 5 matrix with positive eigenvalues that are the squares ofthe singular values of B. BBT has the same �ve positive eigenvalues and threemore 0 eigenvalues as well.

5.(a) The columns of Q are n orthonormal vectors in Rm.

1

(b) n ≤ m and the rank of Q is n.

(c) x = QT b.

6.(a) 1, i,−1,−i

(b) The four columns of the DFT matrix F4 =

1 1 1 11 −i −1 i1 −1 1 −11 i 1 −i

(c) diagonalizable-yes, permutation-yes, nonsingular-yes, Jordan-no, orthogonal-

yes, projection-no, skew-symmetric-no

7.(a) The nullspace of A has basis v2 and v3.

(b) x = v1/4 + c1v2 + c2v3

(c) Shortest is v1/4. It is the projection of x above onto the span of v1.

8.(a) uk = c1λk1x1+c2λ

k2x2+c3λ

k3x3 and u(t) = c1e

λ1tx1+c2eλ2tx2+c3e

λ3tx3

(b)λ1 = 1 and |λi| < 1, for i = 2, 3.

(c) The matrix may not have a complete set of independent eigenveectors.

An example is a three by three Jordan block:

0 1 00 0 10 0 0

.

2

18.06 Professor Strang Final Exam May 16, 2005

Grading

1

2

3

4

5

6

7

8

9

10


Closed book / 10 wonderful problems.Thank you for taking this course !


There is an extra blank page at the end.

1) M 2 2-131 A. Chan 2-588 3-4110 alicec

2) M 3 2-131 A. Chan 2-588 3-4110 alicec

3) M 3 2-132 D. Testa 2-586 3-4102 damiano

4) T 10 2-132 C.I. Kim 2-273 3-4380 ikim

5) T 11 2-132 C.I. Kim 2-273 3-4380 ikim

6) T 12 2-132 W.L. Gan 2-101 3-3299 wlgan

7) T 1 2-131 C.I. Kim 2-273 3-4380 ikim

8) T 1 2-132 W.L. Gan 2-101 3-3299 wlgan

9) T 2 2-132 W.L. Gan 2-101 3-3299 wlgan

1 (10 pts.) Suppose P1, . . . , Pn are points in Rn. The coordinates of Pi are (ai1, ai2, . . . , ain).

We want to find a hyperplane c1x1+· · ·+cnxn = 1 that contains all n points Pi.

(a) What system of equations would you solve to find the c’s for that

hyperplane ?

(b) Give an example in R3 where no such hyperplane exists (of this form),

and an example which allows more than one hyperplane of this form.

(c) Under what conditions on the points or their coordinates is there not a

unique interpolating hyperplane with this equation ?

2

2 (10 pts.) (a) Find a complete set of “special solutions” to Ax = 0 by noticing the

pivot variables and free variables (those have values 1 or 0).

A =

⎡⎢⎢⎢⎣

1 2 3 4 5

1 2 3 4 6

0 0 0 0 0

⎤⎥⎥⎥⎦ .

(b) and (c) Prove that those special solutions are a basis for the nullspace N(A).

What two facts do you have to prove ?? Those are parts (b) and (c)

of this problem.

3

3 (10 pts.) (a) I was looking for an m by n matrix A and vectors b, c such that Ax = b

has no solution and ATy = c has exactly one solution. Why can I not

find A, b, c ?

(b) In Rm, suppose I gave you a vector b and a vector p and n linearly

independent vectors a1, a2, . . . , an. If I claim that p is the projection

of b onto the subspace spanned by the a’s, what tests would you make

to see if this is true ?

4

4 (10 pts.) (a) Find the determinant of

B =

⎡⎢⎢⎢⎢⎢⎢⎣

1 2 1 1

1 1 1 1

1 1 2 1

1 1 1 2

⎤⎥⎥⎥⎥⎥⎥⎦

.

(b) Let A be the 5 by 5 matrix

A =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

2 1 1 1 1

1 2 1 1 1

1 1 2 1 1

1 1 1 2 1

1 1 1 1 2

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

.

Find all five eigenvalues of A by noticing that A − I has rank 1 and

the trace of A is .

(c) Find the (1, 3) and (3, 1) entries of A−1.

5

5 (10 pts.) (a) Complete the matrix A (fill in the two blank entries) so that A has

eigenvectors x1 = (3, 1) and x2 = (2, 1):

A =

⎡⎣ 2 6

⎤⎦

(b) Find a different matrix B with those same eigenvectors x1 and x2, and

with eigenvalues λ1 = 1 and λ2 = 0. What is B10 ?

6

6 (10 pts.) We can find the four coefficients of a polynomial P (z) = c0+c1z+c2z2+c3z

3

if we know the values y1, y2, y3, y4 of P (z) at the four points z = 1, i, i2, i3.

(a) What equations would you solve to find c0, c1, c2, c3 ?

(b) Write down one special property of the coefficient matrix.

(c) Prove that the matrix in those equations is invertible.

7

7 (10 pts.) Suppose S is a 4-dimensional subspace of R7, and P is the projection matrix

onto S.

(a) What are the seven eigenvalues of P ?

(b) What are all the eigenvectors of P ?

(c) If you solvedudt = −Pu (notice minus sign) starting from u(0), the

solution u(t) approaches a steady state as t → ∞. Can you describe

that limit vector u(∞) ?

8

8 (10 pts.) Suppose my favorite −1, 2,−1 matrix swallowed extra zeros to become

A =

⎡⎢⎢⎢⎢⎢⎢⎣

2 0 −1 0

0 2 0 −1

−1 0 2 0

0 −1 0 2

⎤⎥⎥⎥⎥⎥⎥⎦

.

(a) Find a permutation matrix P so that

B = PAPT =

⎡⎢⎢⎢⎢⎢⎢⎣

2 −1 0 0

−1 2 0 0

0 0 2 −1

0 0 −1 2

⎤⎥⎥⎥⎥⎥⎥⎦

.

(b) What are the 4 eigenvalues of B ? Is this matrix diagonalizable or not ?

(c) How do you know that A has the same eigenvalues as B ? Then A is

positive definite—what function of u, v, w, z is therefore positive except

when u = v = w = z = 0 ?

9

9 (10 pts.) (a) Describe all vectors that are orthogonal to the nullspace of this singular

matrix A. You can do this without computing the nullspace.

A =

⎡⎢⎢⎢⎣

1 3 7

2 2 6

2 1 4

⎤⎥⎥⎥⎦ .

(b) If you apply Gram-Schmidt to the columns of this A, what orthonormal

vectors do you get ?

(c) Find a “reduced” LU factorization of A, with only 2 columns in L and

2 rows in U . Can you write A as the sum of two rank 1 matrices ?

10

10 (10 pts.) Suppose the singular value decomposition A = UΣV T has

U =1

3

⎡⎢⎢⎢⎣

−1 2 2

2 −1 2

2 2 −1

⎤⎥⎥⎥⎦ Σ =

⎡⎢⎢⎢⎣

1 0 0 0

0 4 0 0

0 0 0 0

⎤⎥⎥⎥⎦ V =

1

2

⎡⎢⎢⎢⎢⎢⎢⎣

1 1 1 1

1 −1 1 −1

1 1 −1 −1

1 −1 −1 1

⎤⎥⎥⎥⎥⎥⎥⎦

.

(a) Find the eigenvalues of ATA.

(b) Find a basis for the nullspace of A.

(c) Find a basis for the column space of A.

(d) Find a singular value decomposition of −AT.

11

18.06 - Final Exam, Monday May 16th, 2005

solutions

1. (a) We want the coordinates (ai1, . . . , ain) of Pi to satisfy the equation c1x1 + . . .+ cnxn = 1.Thus the system of equations is Ac = ones:

c1a11 + c2a12 + . . . + cna1n = 1

c1a21 + c2a22 + . . . + cna2n = 1

. . .

c1an1 + c2an2 + . . . + cnann = 1

(b) There is no plane of the given form, if one of the points Pi is the origin. More than oneplane contains the Pi’s if the three points are on a line not through the origin.

(c) There is not a unique solution precisely when det A = 0. This means geometrically thatthe points Pi lie in an (n − 1)−dimensional subspace of Rn.

2. (a) Subtracting the first row from the second, we find the matrix

U =

⎡⎣

1 2 3 4 50 0 0 0 10 0 0 0 0

⎤⎦ .

(In the row reduced echelon form R, the 5 changes to 0.) The pivot variables are the firstand the last, while x2, x3, x4 are the free variables. Thus the “special solutions” to Ax = 0are ⎡

⎢⎢⎢⎢⎣

−21000

⎤⎥⎥⎥⎥⎦

,

⎡⎢⎢⎢⎢⎣

−30100

⎤⎥⎥⎥⎥⎦

,

⎡⎢⎢⎢⎢⎣

−40010

⎤⎥⎥⎥⎥⎦

.

(b) and (c) We need to prove that these three vectors are linearly independent and theyspan the nullspace. By considering the second, third and fourth coordinates, a combinationof the vectors adding to zero must have zero coefficients. The vectors span the nullspace,since the dimension of the nullspace is three (note that the rank of the matrix A is 2).

3. (a) If Ax = b has no solution, the column space of A must have dimension less than m. Therank is r < m. Since ATy = c has exactly one solution, the columns of AT are independent.This means that the rank of AT is r = m. This contradiction proves that we cannot find A,b and c.

(b) We need to check two statements: the vector b − p is orthogonal to the space generatedby a1, . . . , an and the vector p lies in that subspace. The first condition we check by seeing if

1

the scalar products a1 · (b− p), . . . , an · (b− p) all equal zero. The second condition we checkby considering the (n + 1)×m matrix whose first n rows are the coordinates of the ai’s andwhose last row consists of the coordinates of p. The vector p is in the span of the ai’s if andonly if the last row becomes zero in elimination.

4. (a) To compute the determinant, subtract the second row from all the other rows:

det B = det

⎡⎢⎢⎣

0 1 0 01 1 1 10 0 1 00 0 0 1

⎤⎥⎥⎦ = − det

⎡⎣

1 1 10 1 00 0 1

⎤⎦ = −1 .

(b) λ = 1, 1, 1, 1, 6. Since A − I has all equal rows, it has rank one. It follows that it hasfour zero eigenvalues. The eigenvalues of A are the eigenvalues of A − I increased by one,so A has the eigenvalue 1 with multiplicity four. The trace of A equals 10 so 10 − 4 = 6 isthe other eigenvalue.

(c) A is symmetric, and thus so is A−1. The cofactor formula gives:

(A−1)13 = (−1)1+3 det B

det A,

and det A = 6 since it equals the product of the eigenvalues of A. We conclude that the(1, 3) and the (3, 1) entries of A−1 are both equal to −1/6.

5. (a) The answer is

A =

[2 6

−1 7

].

Reason: [2 6a b

] [31

]=

[12

3a + b

]= λ1

[31

].

We deduce that λ1 = 4 and 3a + b = 4. Similarly, since x2 is an eigenvector we have[

2 6a b

] [21

]=

[10

2a + b

]= λ2

[21

].

We deduce that λ2 = 5 and therefore that 2a+b = 5. We conclude that a = −1 and b = 7.

(b) B = SΛS−1, where the columns of S are the vectors x1 and x2, and Λ is the diagonalmatrix with entries 1 and 0:

B =

[3 21 1

] [1 00 0

] [1 −2

−1 3

]=

[3 −61 −2

].

Then Λ10 = Λ and therefore B10 = SΛ10S−1 = SΛS−1 = B.

6. (a) We would solve the equations

c0 + c1 + c2 + c3 = y1

c0 + i c1 − c2 − i c3 = y2

c0 − c1 + c2 − c3 = y3

c0 − i c1 − c2 + i c3 = y4 .

2

and the matrix of coefficients is

F =

⎡⎢⎢⎣

1 1 1 11 i −1 −i1 −1 1 −11 −i −1 i

⎤⎥⎥⎦ =

⎡⎢⎢⎣

1 1 1 11 i i2 i3

12 i2 i4 i6

13 i3 i6 i9

⎤⎥⎥⎦ .

(b) F has orthogonal columns and it is symmetric. It is also a Vandermonde matrix: eachcolumn consists of the first four powers of a number (starting from the zero-th power).

(c) Since the columns of F are orthogonal and non-zero, the matrix is invertible. Its in-verse is F/4. The determinant of this Vandermonde matrix is equal to the product of thedifferences of 1, i, i2, i3:

det F = (i − 1)(−1 − 1)(−1 − i)(−i − 1)(−i − i)(−i + 1) = −16i .

7. (a) The seven eigenvalues of P are 1, 1, 1, 1, 0, 0, 0.

(b) The eigenvectors with eigenvalue 1 are the non-zero vectors in S. The eigenvectors witheigenvalue 0 are the non-zero vectors in the orthogonal complement of S.

(c) The solution u(t) to the differential equation has the form

u(t) = v1e−t + v2 ,

where v1 is in S and v2 is in the orthogonal complement of S. Then u(∞) = v2, which is theprojection of u(0) onto the orthogonal complement of S.

8. (a) The required matrix is

P =

⎡⎢⎢⎣

1 0 0 00 0 1 00 1 0 00 0 0 1

⎤⎥⎥⎦ .

(b) Since B is block diagonal, its eigenvalues are the eigenvalues of the diagonal blocks. Inour case, the two blocks are the same and the eigenvalues of each block are 3 and 1. Thusthe eigenvalues of B are 3, 3, 1, 1.

(c) Since P is a permutation matrix, it is orthogonal and therefore P T = P−1. The matrixB is thus similar to the matrix A and we conclude that A and B have the same eigenvalues.

The function of u, v, w, z which is positive except if u = v = w = z = 0 is thus

[u v w z

]A

⎡⎢⎢⎣

uvwz

⎤⎥⎥⎦ = 2

(u2 + v2 + w2 + z2 − uw − vz

).

9. (a) The vectors orthogonal to the nullspace of A are the rows of A. Since we know that thematrix A is singular and it is clearly not rank one, it follows that the rank of A is two. The

3

first two rows are independent and therefore the orthogonal complement of the nullspace ofA is spanned by the two vectors

⎡⎣

137

⎤⎦ and

⎡⎣

226

⎤⎦ .

(b) We get the vectors

1

3

⎡⎣

122

⎤⎦ ,

1√5

⎡⎣

20

−1

⎤⎦ and

⎡⎣

000

⎤⎦ .

(c) The “reduced” LU decomposition, from ignoring the zero row in U , is

Answer A =

⎡⎣

1 02 12 5

4

⎤⎦

[1 3 70 −4 −8

].

Here are the details: Starting elimination we find

⎡⎣

1 0 0−2 1 0−2 0 1

⎤⎦

⎡⎣

1 3 72 2 62 1 4

⎤⎦ =

⎡⎣

1 3 70 −4 −80 −5 −10

⎤⎦ , (1)

and proceeding further we find

⎡⎣

1 0 00 1 00 −5

41

⎤⎦

⎡⎣

1 3 70 −4 −80 −5 −10

⎤⎦ =

⎡⎣

1 3 70 −4 −80 0 0

⎤⎦ .

Collecting all the information together we obtain

⎡⎣

1 3 72 2 62 1 4

⎤⎦ =

⎡⎣

1 0 02 1 02 0 1

⎤⎦

⎡⎣

1 0 00 1 00 5

41

⎤⎦

⎡⎣

1 3 70 −4 −80 0 0

⎤⎦ ,

and multiplying the first two matrices on the right-hand side we deduce that

⎡⎣

1 3 72 2 62 1 4

⎤⎦ =

⎡⎣

1 0 02 1 02 5

41

⎤⎦

⎡⎣

1 3 70 −4 −80 0 0

⎤⎦ .

Since the last row of the last matrix is all zero, we conclude that

⎡⎣

1 3 72 2 62 1 4

⎤⎦ =

⎡⎣

1 02 12 5

4

⎤⎦

[1 3 70 −4 −8

].

4

This is the “reduced” LU factorization of A. Multiplying columns of L by rows of U , this is

A =

⎡⎣

100

⎤⎦ [

1 3 7]+

⎡⎣

0154

⎤⎦ [

0 −4 −8].

10. (a) The eigenvalues of ATA are the same as the eigenvalues of ΣTΣ which is the 4 by 4diagonal matrix with entries 1, 16, 0, 0 along the diagonal.

(b) The nullspace N(A) is spanned by the last two columns of V .

(c) The column space of A is spanned by the first two columns of U .

(d) A singular value decomposition of −AT is −AT = (−V )ΣTUT.

5

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|� ³��~�W��

x = c1q1 + c2q2 + c3q3

|

! �Wq�� W«��³��¤�xTx

�� ©\��ª �¡�xTAx � (��¡« � ��+<� � � c �

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|! �zq�&h ��« �k��+

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v|

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C =

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0 0 1 0

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0 0 0 1

1 0 0 0

0 1 0 0

.

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�� !

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|! ��qr � � �� ¡�� W p � � � $) �� +�³� �©��« �� x��ª��³�� ¤��© ) � � � A

$

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�W ��(� � �&��Wª¬³�« T��+

A�

A =

1 0

3 0

0 −1

0 −3

.

pz�

�� !

p9¥

I ��C �D � D � A�A�� *�� r �� ³��¡� � �W � �� W³ � � � ��« �� ) � � � y � � �� 5� � � « � � © � ��%W�W ��ª �v�� W #� � � © � ��%��W ��ªd�� *3��@p � ��W \� � � © � � %W�W ��ª � ��ª �*) |

A1 =[

3]

A2 =

3 2

1 3

A3 =

3 2 0

1 3 2

0 1 3

An =

3 2 0 0

1 3 2 0

0 1 3 ·

0 0 · ·

! �Wq � � ��P��¡� � � � ©��¡« � �� ¤��+ A2

�� ©A3

$

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a�� ©

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=

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1 0

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Dn−2

.

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|

pu�

�� !

p9�

18.06 Professor Strang Final Exam December 18, 2006

Grading

1

2

3

4

5

6

7

8

9

Your PRINTED name is: SOLUTIONS


1) T 10 2-131 K. Meszaros 2-333 3-7826 karola

2) T 10 2-132 A. Barakat 2-172 3-4470 barakat

3) T 11 2-132 A. Barakat 2-172 3-4470 barakat

4) T 11 2-131 A. Osorno 2-229 3-1589 aosorno

5) T 12 2-132 A. Edelman 2-343 3-7770 edelman

6) T 12 2-131 K. Meszaros 2-333 3-7826 karola

7) T 1 2-132 A. Edelman 2-343 3-7770 edelman

8) T 2 2-132 J. Burns 2-333 3-7826 burns

9) T 3 2-132 A. Osorno 2-229 3-1589 aosorno

1 (4+7=11 pts.) Suppose A is 3 by 4, and Ax = 0 has exactly 2 special solutions:

x1 =

1

1

1

0

and x2 =

−2

−1

0

1

(a) Remembering that A is 3 by 4, find its row reduced echelon form R.

(b) Find the dimensions of all four fundamental subspaces C(A), N(A),

C(AT), N(AT).

You have enough information to find bases for one or more of these

subspaces—find those bases.

2

Solution.

(a) Each special solution tells us the solution to Rx = 0 when we set one free variable = 1

and the others = 0. Here, the third and fourth variables must be the two free variables,

and the other two are the pivots: R =

1 0 ∗ ∗

0 1 ∗ ∗

0 0 0 0

Now multiply out Rx1 = 0 and Rx2 = 0 to find the ∗’s: R =

1 0 −1 2

0 1 −1 1

0 0 0 0

(The ∗’s are just the negatives of the special solutions’ pivot entries.)

(b) We know the nullspace N(A) has n − r = 4 − 2 = 2 dimensions: the special solutions

x1, x2 form a basis.

The row space C(AT) has r = 2 dimensions. It’s orthogonal to N(A), so just pick

two linearly-independent vectors orthogonal to x1 and x2 to form a basis: for example,

x3 =

1

0

−1

2

, x4 =

0

1

−1

1

.

(Or: C(AT) = C(RT) is just the row space of R, so the first two rows are a basis.

Same thing!)

The column space C(A) has r = 2 dimensions (same as C(AT)). We can’t write down

a basis because we don’t know what A is, but we can say that the first two columns of

A are a basis.

The left nullspace N(AT) has m − r = 1 dimension; it’s orthogonal to C(A), so any

vector orthogonal to the first two columns of A (whatever they are) will be a basis.

3

2 (6+3+2=11 pts.) (a) Find the inverse of a 3 by 3 upper triangular matrix U , with

nonzero entries a, b, c, d, e, f . You could use cofactors and the formula

for the inverse. Or possibly Gauss-Jordan elimination.

Find the inverse of U =

a b c

0 d e

0 0 f

.

(b) Suppose the columns of U are eigenvectors of a matrix A. Show that

A is also upper triangular.

(c) Explain why this U cannot be the same matrix as the first factor in

the Singular Value Decomposition A = UΣV T.

4

Solution.

(a) By elimination: (We keep track of the elimination matrix E on one side, and the

product EU on the other. When EU = I, then E = U−1.)

a b c 1 0 0

0 d e 0 1 0

0 0 f 0 0 1

1 b/a c/a 1/a 0 0

0 1 e/d 0 1/d 0

0 0 1 0 0 1/f

1 0 0 1/a −b/ad (be − cd)/adf

0 1 0 0 1/d −e/df

0 0 1 0 0 1/f

=

[I U−1

]

By cofactors: (Take the minor, then “checkerboard” the signs to get the cofactor matrix,

then transpose and divide by det(U) = adf .)

a b c

0 d e

0 0 f

df 0 0

bf af 0

be − cd ae ad

df 0 0

−bf af 0

be − cd −ae ad

df −bf be − cd

0 af −ae

0 0 ad

1/a −b/ad (be − cd)/adf

0 1/d −e/df

0 0 1/f

= U−1

(b) We have a complete set of eigenvectors for A, so we can diagonalize: A = UΛU−1. We

know U is upper-triangular, and so is the diagonal matrix Λ, and we’ve just shown

that U−1 is upper-triangular too. So their product A is also upper-triangular.

(c) The columns aren’t orthogonal! (For example, the product uT1 u2 of the first two

columns is ab + 0d + 0 · 0 = ab, which is nonzero because we’re assuming all the

entries are nonzero.)

5

3 (3+3+5=11 pts.) (a) A and B are any matrices with the same number of rows.

What can you say (and explain why it is true) about the comparison

of

rank of A rank of the block matrix[A B

]

(b) Suppose B = A2. How do those ranks compare ? Explain your reason-

ing.

(c) If A is m by n of rank r, what are the dimensions of these nullspaces ?

Nullspace of A Nullspace of[A A

]

Solution.

(a) All you can say is that rank A ≤ rank [A B]. (A can have any number r of pivot

columns, and these will all be pivot columns for [A B]; but there could be more pivot

columns among the columns of B.)

(b) Now rank A = rank [A A2]. (Every column of A2 is a linear combination of columns

of A. For instance, if we call A’s first column a1, then Aa1 is the first column of A2.

So there are no new pivot columns in the A2 part of [A A2].)

(c) The nullspace N(A) has dimension n − r, as always. Since [A A] only has r pivot

columns — the n columns we added are all duplicates — [A A] is an m-by-2n matrix

of rank r, and its nullspace N ([A A]) has dimension 2n − r.

6

4 (3+4+5=12 pts.) Suppose A is a 5 by 3 matrix and Ax is never zero (except when

x is the zero vector).

(a) What can you say about the columns of A ?

(b) Show that ATAx is also never zero (except when x = 0) by explaining

this key step:

If ATAx = 0 then obviously xTATAx = 0 and then (WHY?) Ax = 0.

(c) We now know that ATA is invertible. Explain why B = (ATA)−1AT

is a one-sided inverse of A (which side of A ?). B is NOT a 2-sided

inverse of A (explain why not).

Solution.

(a) N(A) = 0 so A has full column rank r = n = 3: the columns are linearly independent.

(b) xTATAx = (Ax)T Ax is the squared length of Ax. The only way it can be zero is if

Ax has zero length (meaning Ax = 0).

(c) B is a left inverse of A, since BA = (ATA)−1ATA = I is the (3-by-3) identity matrix.

B is not a right inverse of A, because AB is a 5-by-5 matrix but can only have rank 3.

(In fact, BA = A(ATA)−1AT is the projection onto the (3-dimensional) column space

of A.)

7

5 (5+5=10 pts.) If A is 3 by 3 symmetric positive definite, then Aqi = λiqi with

positive eigenvalues and orthonormal eigenvectors qi.

Suppose x = c1q1 + c2q2 + c3q3.

(a) Compute xTx and also xTAx in terms of the c’s and λ’s.

(b) Looking at the ratio of xTAx in part (a) divided by xTx in part (a),

what c’s would make that ratio as large as possible ? You can assume

λ1 < λ2 < . . . < λn. Conclusion: the ratio xTAx/xTx is a maximum

when x is .

Solution.

(a)

xTx = (c1qT

1 + c2qT

2 + c3qT

3 )(c1q1 + c2q2 + c3q3)

= c2

1qT

1 q1 + c1c2qT

1 q2 + · · ·+ c3c2qT

3 q2 + c2

3qT

3 q3

= c2

1 + c2

2 + c2

3.

xTAx = (c1qT

1 + c2qT

2 + c3qT

3 )(c1Aq1 + c2Aq2 + c3Aq3)

= (c1qT

1 + c2qT

2 + c3qT

3 )(c1λ1q1 + c2λ2q2 + c3λ3q3)

= c2

1λ1qT

1 q1 + c1c2λ2qT

1 q2 + · · · + c3c2λ2qT

3 q2 + c2

3λ3qT

3 q3

= c2

1λ1 + c2

2λ2 + c2

3λ3.

(b) We maximize (c21λ1 + c2

2λ2 + c23λ3)/(c2

1 + c22 + c2

3) when c1 = c2 = 0, so x = c3q3 is a

multiple of the eigenvector q3 with the largest eigenvalue λ3.

(Also notice that the maximum value of this “Rayleigh quotient” xTAx/xTx is the

largest eigenvalue itself. This is another way of finding eigenvectors: maximize xTAx/xTx

numerically.)

8

6 (4+4+4=12 pts.) (a) Find a linear combination w of the linearly independent vec-

tors v and u that is perpendicular to u.

(b) For the 2-column matrix A =[u v

], find Q (orthonormal columns)

and R (2 by 2 upper triangular) so that A = QR.

(c) In terms of Q only, using A = QR, find the projection matrix P onto

the plane spanned by u and v.

Solution.

(a) You could just write down w = 0u+0v = 0 — that’s perpendicular to everything! But

a more useful choice is to subtract off just enough u so that w = v−cu is perpendicular

to u. That means 0 = wTu = vTu − cuTu, so c = (vTu)/(uTu) and

w = v − (vTu

uTu)u.

(b) We already know u and w are orthogonal; just normalize them! Take q1 = u/|u| and

q2 = w/|w|. Then solve for the columns r1, r2 of R: Qr1 = u so r1 =

|u|

0

, and

Qr2 = v so r2 =

c|u|

|w|

. (Where c = (vTu)/(uTu) as before.)

Then Q = [q1 q2] and R = [r1 r2].

(c) P = A(ATA)−1AT = (QR)(RTQTQR)−1(RTQT) = (QR)(RTQT) = QQT.

9

7 (4+3+4=11 pts.) (a) Find the eigenvalues of

C =

0 0 0 1

1 0 0 0

0 1 0 0

0 0 1 0

and C2 =

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

.

(b) Those are both permutation matrices. What are their inverses C−1

and (C2)−1 ?

(c) Find the determinants of C and C + I and C + 2I.

10

Solution.

(a) Take the determinant of C − λI (I expanded by cofactors): λ4 − 1 = 0. The roots of

this “characteristic equation” are the eigenvalues: +1,−1, i,−i.

The eigenvalues of C2 are just λ2 = ±1 (two of each).

(Here’s a “guessing” approach. Since C4 = I, all the eigenvalues λ4 of C4 are 1: so

λ = 1,−1, i,−i are the only possibilities. Just check to see which ones work. Then the

eigenvalues of C2 must be ±1.)

(b) For any permutation matrix, C−1 = CT: so

C−1 =

0 1 0 0

0 0 1 0

0 0 0 1

1 0 0 0

and (C2)−1 = C2 is itself.

(c) The determinant of C is the product of its eigenvalues: 1(−1)i(−i) = −1.

Add 1 to every eigenvalue to get the eigenvalues of C + I (if C = SΛS−1, then C + I =

S(Λ + I)S−1): 2(0)(1 + i)(1 − i) = 0.

(Or let λ = −1 in the characteristic equation det(C − λI).)

Add 2 to get the eigenvalues of C + 2I (or let λ = −2): 3(1)(2 + i)(2 − i) = 15.

11

8 (4+3+4=11 pts.) Suppose a rectangular matrix A has independent columns.

(a) How do you find the best least squares solution x to Ax = b ? By

taking those steps, give me a formula (letters not numbers) for x and

also for p = Ax.

(b) The projection p is in which fundamental subspace associated with A ?

The error vector e = b − p is in which fundamental subspace ?

(c) Find by any method the projection matrix P onto the column space

of A:

A =

1 0

3 0

0 −1

0 −3

.

Solution.

(a)

Ax = b

Least-squares “solution”: ATAx = ATb

ATA is invertible: x = (ATA)−1ATb

and p = Ax is: Ax = A(ATA)−1ATb

(b) p = Ax is a linear combination of columns of A, so it’s in the column space C(A). The

error e = b − p is orthogonal to this space, so it’s in the left nullspace N(AT).

(c) I used P = A(ATA)−1AT. Since ATA =

10 0

0 10

, its inverse is

1/10 0

0 1/10

= 1

10I,

and

P =1

10

1 3 0 0

3 9 0 0

0 0 1 3

0 0 3 9

12

9 (3+4+4=11 pts.) This question is about the matrices with 3’s on the main diagonal,

2’s on the diagonal above, 1’s on the diagonal below.

A1 =[3]

A2 =

3 2

1 3

A3 =

3 2 0

1 3 2

0 1 3

An =

3 2 0 0

1 3 2 0

0 1 3 ·

0 0 · ·

(a) What are the determinants of A2 and A3 ?

(b) The determinant of An is Dn. Use cofactors of row 1 and column 1 to

find the numbers a and b in the recursive formula for Dn:

(∗) Dn = a Dn−1 + bDn−2 .

(c) This equation (∗) is the same as

Dn

Dn−1

=

a b

1 0

Dn−1

Dn−2

.

>From the eigenvalues of that matrix, how fast do the determinants

Dn grow? (If you didn’t find a and b, say how you would answer part

(c) for any a and b ) For 1 point, find D5.

13

Solution.

(a) det(A2) = 3 · 3 − 1 · 2 = 7 and det(A3) = 3 det(A2) − 2 · 1 · 3 = 15.

(b) Dn = 3Dn−1 + (−2)Dn−2. (Show your work.)

(c) The trace of that matrix A is a = 3, and the determinant is −b = 2. So the character-

istic equation of A is λ2 − aλ − b = 0, which has roots (the eigenvalues of A)

λ± =a ±

√a2 − 4(−b)

2=

3 ± 1

2= 1 or 2.

Dn grows at the same rate as the largest eigenvalue of An, λn

+ = 2n.

The final point: D5 = 3D4 + 2D3 = 3(3D3 + 2D2) + 2D3 = 11D3 + 6D2 = 207.

14

18.06 FINAL EXAM May 24, 2007

Grading

1

2

3

4

5

6

7

8

Total:

Your PRINTED name is: SOLUTIONS


(1) M 2 2-131 A. Osorno(2) M 3 2-131 A. Osorno(3) M 3 2-132 A. Pissarra Pires(4) T 11 2-132 K. Meszaros(5) T 12 2-132 K. Meszaros(6) T 1 2-132 Jerin Gu(7) T 2 2-132 Jerin Gu

Problem 1 (10 points)

Let A =

3 2 1 1

6 6 3 3

3 4 2 2

.

(a) Calculate the dimensions of the 4 fundamental subspaces associated with A.

(b) Give a basis for each of the 4 fundamental subspaces.

(c) Find the complete solution of the system Ax =

1

3

2

.

Solution 1

(a) The rank is 2, so the dimensions are:C(A) Ã r = 2

C(AT ) Ã r = 2

N(A) Ã n− r = 4− 2 = 2

N(AT ) Ã m− r = 3− 2 = 1.

(b) We can get these by elimination or by inspection:C(A) Ã

(3 6 3

)T

,(2 6 4

)T

C(AT ) Ã(3 2 1 1

)T

,(6 3 3 1

)T

N(A) Ã(0 −1/2 1 0

)T

,(0 −1/2 0 1

)T

N(AT ) Ã(1 −1 1

)T

(c) x = xparticular + xnullspace =

0

1/2

0

0

+ c1

0

−1/2

1

0

+ c2

0

−1/2

0

1

.


Consider the system of linear equations:

x + y + z = 1

2x + z = 2

−x + y + az = b

In parts (a)�(c) below circle correct answers. Explain your answers.(a) For a = 1, b = −1, the system has:

(1) exactly one solution(2) in�nitely many solutions(3) no solutions

(b) For a = 0, b = 1, the system has:


(c) For a = 0, b = −1, the system has:


(d) Solve the system for a = b = 1.

Solution 2

If we eliminate the augmented matrix we get

1 1 1 1

0 −2 −1 0

0 0 a b + 1

.

(a) Exactly one solution: The matrix is invertible.(b) No solutions: Get a row of zeroes in the matrix with no zero in the augmented column.(c) In�nitely many solutions: Get a row of zeroes with a zero in the augmented column.(d) Using back substitution we get x =

(0 −1 2

)T

.


Let L be the line in R3 spanned by the vector (1, 1, 1)T . Let P be the projection matrix forthe projection onto the line L.(a) What are the eigenvalues of the matrix P? (Indicate their multiplicities.)

(b) Find an orthonormal basis of the orthogonal complement L⊥ to the line L.

(c) Calculate the projection of the vector (1, 2, 3)T onto the line L.

(d) Calculate the projection of the vector (1, 2, 3)T onto the orthogonal complement L⊥.

Solution 3

(a) P is a projection matrix onto a subspace of dimension 1, so the eigenvalues are 1, 0, 0.

(b)

1/√

2

0

−1√

2

,

1/√

6

−2/√

6

1/√

6

.

(c) p =

2

2

2

.

(d) The projection onto L⊥ is b− p =

−1

0

1

.


Let A =

1 2 3

2 2 2

3 2 1

.

In parts (a)�(c) below circle correct answers. Explain your answers.(a) The matrix A is singular: True False(b) The matrix A + 2I is singular: True False(c) The matrix A is positive de�nite: True False(d) Find all eigenvalues of A and the corresponding eigenvectors.(e) Find an orthogonal matrix Q and a diagonal matrix Λ such that A = QΛQT .(f) Solve the system of di�erential equations du(t)

dt= Au(t), u(0) = (1, 0, 0)T .

Solution 4

(a) True ((1 −2 1

)T

is in the nullspace).

(b) True ((1 0 −1

)T

is in the nullspace).(c) False The matrix is singular so has 0 as eigenvalue.(d) A is singular, so 0 is an eigenvalue with eigenvector

(1 −2 1

)T

.

A + 2I is singular, so -2 is an eigenvalue with eigenvector(1 0 −1

)T

.

We can get the last eigenvalue by looking a the trace: 6. The eigenvector is(1 1 1

)T

.(e)

A =

1/√

6 1/√

62 1/√

3

−2/√

6 0 1/√

3

1/√

6 −1/√

2 1/√

3

︸︷︷︸Q

0 0 0

0 −2 0

0 0 6

︸︷︷︸Λ

1/√

6 −2/√

6 1/√

6

1/√

2 0 −1/√

2

1/√

3 1/√

3 1/√

3

︸︷︷︸QT

.

(f) u(t) = eAtu(0) = QeΛtQT u(0) = 16

1

−2

1

+ 1

2e−2t

1

0

−1

+ 1

3e6t

1

1

1

.


Let A =

1 1 1 1

2 2 2 2

3 3 3 3

.

(a) What is the rank of A?(b) Calculate the matrix AT A. Find all its eigenvalues (with multiplicities).(c) Calculate the matrix AAT . Find all its eigenvalues (with multiplicities).(d) Find the matrix Σ in the singular value decomposition A = UΣV T .

Solution 5

(a) 1

(b) AT A =

14 14 14 14

14 14 14 14

14 14 14 14

14 14 14 14

. It was rank 1 so the eigenvalues are 56, 0, 0, 0.

(c) AAT =

4 8 12

8 16 24

12 24 36

. It was rank 1 so the eigenvalues are 56, 0, 0.

(d) Σ =

√56 0 0 0

0 0 0 0

0 0 0 0

.


Let An be the tridiagonal n×n-matrix with 2's on the main diagonal, 1's immediately abovethe main diagonal, 3's immediately below the main diagonal, and 0's everywhere else:

An =

2 1 0 0 · · · 0

3 2 1 0. . . 0

0 3 2 1. . . 0

0 0 3 2. . . 0

... . . . . . . . . . . . . ...0 0 0 0 · · · 2

,

(a) Express the determinant det(An) in terms of det(An−1) and det(An−2).

(b) Explicitly calculate det(An), for n = 1, . . . , 6.

Solution 6

(a) Using cofactors twice we get det(An) = 2 det(An−1)− 3 det(An−2).

(b) det(A1) = det[2] = 2.

det(A2) = det

2 1

3 2

= 1.

det(A3) = 2 · 1− 3 · 2 = −4.det(A4) = 2 · (−4)− 3 · 1 = −11.det(A5) = 2 · (−11)− 3 · (−4) = −10.det(A6) = 2 · (−10)− 3 · (−11) = 13.


Calculate the determinant of the following 6× 6-matrix:

A =

1 1 1 0 1 1

1 1 1 1 0 1

1 1 1 1 1 0

0 1 1 1 1 1

1 0 1 1 1 1

1 1 0 1 1 1

Solution 7

This determinant could be computed using cofactors or doing row operations to simplify andthen cofactors. it could also be computed as follows.

A =

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

︸︷︷︸P

0 1 1 1 1 1

1 0 1 1 1 1

1 1 0 1 1 1

1 1 1 0 1 1

1 1 1 1 0 1

1 1 1 1 1 0

︸︷︷︸B

.

P is a permutation matrix with determinant -1.

B =

1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1 1 1

− I.

The eigenvalues of the matrix with all 1's are 6, 0, 0, 0, 0, 0, 0 so the eigenvalues of B are5, -1, -1, -1, -1, -1, so the determinant of B is -5.Thus the determinant of A is 5.


(a) Calculate A100 for A =

3 4

4 −3

.

(b) Calculate B100 for B =

1 −1

1 1

.

(c) What will happen with the house (shown below) when we apply the linear transformation

T (v) = Bv for B =

1 −1

1 1

? Will it be dilated? Draw the picture of the transformed

house.

Solution 8

(a) A2 =

25 0

0 25

so A100 =

5100 0

0 5100

.

(b) B4 =

−4 0

0 −4

so B100 =

(−4)25 0

0 (−4)25

=

−425 0

0 −425

.

(c) From part (b) we see that B4 is stretching by 4 and rotating by π. Thus B is rotatingby π/4 and stretching by

√2.

Practice 18.06 Final Questions with Solutions

17th December 2007

Notes on the practice questions

The final exam will be on Thursday, Dec. 20, from 9am to 12noon at the Johnson Track, andwill most likely consist of 8–12 questions. The practice problems below mostly concentrate on thematerial from exams 1–2, since you already have practice problems for exam 3. The real final willhave plenty of eigenproblems!

These questions are intended to give you a flavor of how I want you to be able to think aboutthe material, and the flavor of possible questions I might ask. Obviously, these questions do notexhaust all the material that we covered this term, so you should of course still study your lecturenotes and previous exams, and review your homework.

Solutions for these practice problems should be posted on the 18.06 web site by 12/15.

List of potential topics:

Material from exams 1, 2, and 3, and the problem sets (and lectures) up to that point.Definitely not on final: finite-difference approximations, sparse matrices and iterative methods,

non-diagonalizable matrices, generalized eigenvectors, principal components analysis, choosing abasis to convert a linear operator into a matrix, numerical linear algebra and error analysis.

Key ideas:

• The four subspaces of a matrix A and their relationships to one another and the solutions ofAx = b.

• Gaussian elimination A → U → R and backsubstitution. Elimination = invertible rowoperations = multiplying A on the left by an invertible matrix. Multiplying on left by aninvertible matrix preserves N(A), and hence we can use elimination to find the nullspace.Also, it thus preserves C(AH) = N(A)⊥. Also, it thus preserves the linear independence ofthe columns and hence the pivot columns in A are a basis for C(A). Conversely, invertiblecolumn operations = multiplying A on the right by an invertible matrix, hence preservingN(AH) and C(A).

• Solution of Ax = b when A is not invertible: exactly solvable if b ∈ C(A), particularsolutions, not unique if N(A) 6= {0}. Least-squares solution AHAx = AHb if A full columnrank, equivalence to minimizing ‖Ax − b‖2, relationship Ax = PAb to projection matrixPA = A(AHA)−1AH onto C(A). The fact that rank(AHA) = rankA = rankAH , henceAHAx = AHb is always solvable for any A, and AHA is always invertible (and positive-definite) if A has full column rank.

• Vector spaces and subspaces. Dot products, transposes/adjoints, orthogonal complements.Linear independence, bases, and orthonormal bases. Gram-Schmidt and QR factorization.

1

Unitary matrices QH = Q−1, which preserve dot products x · y = (Qx) · (Qy) and lengths‖x‖ = ‖Qx‖.

• Determinants of square matrices: properties, invariance under elimination (row swaps flipsign), = product of eigenvalues. The trace of a matrix = sum of eigenvalues. det AB =(detA)(detB), trace AB = trace BA, trace(A + B) = trace A + trace B.

• For an eigenvector, any complicated matrix or operator acts just like a number λ, and wecan do anything we want (inversion, powers, exponentials...) using that number. To act onan arbitrary vector, we expand that vector in the eigenvectors (in the usual case where theeigenvectors form a basis), and then treat each eigenvector individually. Finding eigenvectorsand eigenvalues is complicated, though, so we try to infer as much as we can about theirproperties from the structure of the matrix/operator (Hermitian, Markov, etcetera). SVD asgeneralization of eigenvectors/eigenvalues.

Problem 1

Suppose that A is some real matrix with full column rank, and we do Gram-Schmidt on it to getan the following orthonormal basis for C(A): q1 = (1, 0, 1, 0,−1)T /

√3, q2 = (1, 2,−1, 0, 0)T /

√6,

q3 = (−2, 1, 0, 0, 2)T /√

9.(a) Suppose we form the matrix B whose columns are

√3q1,

√6q2, and

√9q3. Which of the

four subspaces, if any, are guaranteed to be the same for A and B?(b) Find a basis for the left nullspace N(AT ).(c) Find a basis for the row space C(AT ).

Solution:(a) All four subspaces for A and B are the same. Obviously B has the same column space as

A, i.e. C(A) = C(B). It follows that they have the same left nullspace, N(AT ) = N(BT ), whichare just the orthogonal complement of column space. Since both A and B has full column rank,N(A) = N(B) = {0}. It follows that the row spaces C(AT ) = C(BT ).

(b) As we have observed above, N(AT ) = N(BT ), where B =

1 1 −20 2 11 −1 00 0 0−1 0 2

. To find a

basis for BT , we do Gauss elimination: 1 0 1 0 −11 2 −1 0 0−2 1 0 0 2

1 0 1 0 −1

0 2 −2 0 10 1 2 0 0

1 0 1 0 −1

0 2 −2 0 10 0 3 0 −1/2

1 0 0 0 −5/60 2 0 0 2/30 0 3 0 −1/2

,

So we may take two vectors (0, 0, 0, 1, 0)T and (5,−2, 1, 0, 6)T as a basis of N(AT ).(c) A is a 5 × 3 matrix with full column rank, i.e. rank(A) = 3. So the row space have

dimension dim(C(AT )) = 3. However, C(AT ) lies in R3. So C(AT ) = R3. We may take thevectors (1, 0, 0)T , (0, 1, 0)T , (0, 0, 1)T as a basis.

2

Problem 2

Perverse physicist Pat proposes a permutation: Pat permutes the columns of some matrix A bysome random sequence of column swaps, resulting in a new matrix B.

(a) If you were given B (only), which of the four subspaces of A (if any) could you find? (i.e.which subspaces are preserved by column swaps?)

(b) Suppose B =

1 −1 22 1 3−1 5 0

and b = (0, 1, 2)T . Check whether Ax = b is solvable,

and if so whether the solution x is unique.(c) Suppose you were given B and it had full column rank. You are also given b. Give an

explicit formula (in terms of B and b only) for the minimum possible value of ‖Ax− b‖2.

Solution:(a) Only column swaps are performed, so the column space is not changed. As a consequence,

the left nullspace is also not changed. So given B, we can find C(A) and N(AT ).(b) Since C(A) = C(B), the solvability of Ax = b is equivalent to the solvability of Bx = b.

We do Gauss elimination for the second equation: 1 −1 2 02 1 3 1−1 5 0 2

1 −1 2 0

0 3 −1 10 4 2 2

1 −1 2 0

0 3 −1 10 0 10/3 2/3

.

So the equation Bx = b, and thus the equation Ax = b, is solvable. Moreover, the solution x isunique.

(c) All the vectors Ax are exactly the vectors in the column space C(A) = C(B). Thus theminimum possible value of ‖Ax − b‖2 is exactly the minimum possible value of ‖Bx − b‖2. Tofind the latter one, we need the vector Bx to be the vector in C(B) which is closest to the vectorb. In other words, Bx should be the projection of b on C(B). Since B is of full column rank, theprojection is Bx = B(BT B)−1BT b. So the minimum value we want is ‖B(BT B)−1BT b− b‖2

Problem 3

Which of the following sets are vector spaces (under ordinary addition and multiplication by realnumbers)?

(a) Given a vector x ∈ Rn, the set of all vectors y ∈ Rn with x · y = 3.(b) The set of all functions f(x) whose integral

∫∞−∞ f(x)dx is zero.

(c) Given a subspace V ⊆ Rn and an m× n matrix A, the set of all vectors Ax for all x ∈ V .(d) Given a line L ⊆ Rn and an m× n matrix A, the set of all vectors Ax for all x ∈ L.(e) The set of n× n Markov matrices.(f) The set of eigenvectors with |λ| < 1

2 of an n× n Markov matrix A.

Solution:(a) This is not a vector space, since it doesn’t contain the zero vector.(b) This is a vector space.It is a subset of the vector space consisting all functions. If f and g both has whole integral

zero, af + bg also has whole integral zero. So it is a vector subspace.(c) This is a vector space.It is just the column space of A, which is a vector subspace of Rm.

3

(d) This is not a vector space.A line in Rn doesn’t have to pass the origin, so the set of all vectors Ax need not contain the

zero vector.(e) This is not a vector space.The zero matrix is not a Markov matrix.(f) This is not a vector space in general.If the Markov matrix A has more than one eigenvalue whose absolute value are less than

1/2, then the corresponding set of eigenvectors is not a vector space, since the summation of twoeigenvectors corresponding to different eigenvalues is not a eigenvector.

Problem 4

The rows of an m× n matrix A are linearly independent.(a) Is Ax = b necessarily solvable?(b) If Ax = b is solvable, is the solution necessarily unique?(c) What are N(AH) and C(A)?

Solution:(a) Yes. Since the rows of A are linearly independent, rank(A) = m. So the column space of

A is an m-dimensional subspace of Rm, i.e., is Rm itself. It follows that for any b, the equationAx = b is always solvable.

(b) No, the solution maybe not unique. Since rank(A) = m, the nullspace is n−m dimensional.Thus the solution is not unique if n > m, and is unique if n = m. (It will never have that n < m,otherwise the rows are not linearly independent.)

(c) We have seen in part (a) that C(A) = Rm. So N(AT ) = {0}. Since N(AH) consists thosepoints whose conjugate lies in N(AT ), we see N(AH) = {0}.

Problem 5

Make up your own problem: give an example of a matrix A and a vector b such that the solutionsof Ax = b form a line in R3, b 6= 0, and all the entries of the matrix A are nonzero. Find allsolutions x.Solution:

Such a matrix must be an m× 3 matrix whose nullspace is one dimensional. In other words,the rank is 3− 1 = 2. We may take A to be an 2× 3 matrix whose rows are linearly independent.As an example, we take

A =(

1 1 11 1 2

), b =

(11

).

To find all solutions, we do elimination(1 1 1 11 1 2 1

)

(1 1 1 10 0 1 0

)

(1 1 0 10 0 1 0

),

so the solutions are given by x = (1− t, t, 0)T .

4

Problem 6

Clever Chris the chemist concocts a conundrum: in ordinary least-squares, you minimize ‖Ax−b‖2

by solving AHAx = AHb, but suppose instead that we wanted to minimize (Ax− b)HC(Ax− b)for some Hermitian positive-definite matrix C?

(a) Suppose C = BHB for some matrix B. Which of the following (if any) must be propertiesof B: (i) Hermitian, (ii) Markov, (iii) unitary, (iv) full column rank, (v) full row rank?

(b) In terms of B, A, and b, write down an explicit formula for the x that minimizes (Ax−b)HC(Ax− b).

(c) Rewrite your answer from (b) purely in terms of C, A, and b.(d) Suppose that C was only positive semi-definite. Is there still a minimum value of (Ax −

b)HC(Ax− b)? Still a unique solution x?

Solution:(a) For C = BHB a Hermitian positive-definite matrix, B must be (iv) full column rank. To

show this, we only need to notice that x · Cx = x · BHBx = ‖Bx‖2. Since C is positive definite,B has no nonzero nullspace. It follows that B is of full column rank.

B don’t have to be Hermitian or Markov or unitary or full row rank. For example, we may

take B =

1 00 20 0

. Then C = BHB =(

1 00 4

).

(b) We have

(Ax− b)HC(Ax− b) = (Ax− b)HBHB(Ax− b) = (BAx−Bb)H(BAx−Bb)

= ‖BAx−Bb‖2.

To minimize this, we need to solve the equation (BA)H(BA)x = (BA)HBb. Since B and A areof full column rank, BA is of full column rank. So (BA)H(BA) is invertible. The solution is givenby x = (AHBHBA)−1AHBHV b.

(c) Since BHB = C, we have x = (AHCA)−1AHCb.(d) If C is only positive semi-definite, we still have C = BHB for some B, but B need not to

be of full column rank. As above, we have

(Ax− b)HC(Ax− b) = ‖BAx−Bb‖2.

So the minimum value of (Ax − b)HC(Ax − b) still exists. However, since B may be not of fullcolumn rank, the matrix need not be invertible. The solution x need not be unique. For example,if we take C to be the zero matrix, any x will minimize (Ax− b)HC(Ax− b).

Problem 7

True or false (explain why if true, give a counter-example if false).(a) For n×n real-symmetric matrices A and B, AB and BA always have the same eigenvalues.

[Hint: what is (AB)T ?](b) For n× n matrices A and B with B invertible, AB and BA always have the same eigen-

values. [Hint: you can write det(AB − λI) = det((A − λB−1)B). Alternative hint: think aboutsimilar matrices.]

(c) Two diagonalizable matrices A and B with the same eigenvalues and eigenvectors must bethe same matrix.

5

(d) Two diagonalizable matrices A and B with the same eigenvalues must be the same matrix.(e) For n× n matrices A and B with B invertible, AB and BA always have the same eigen-

vectors.Solution:

(a) True.Since A and B are real-symmetric matrices, (AB)T = BT AT = BA. But (AB)T has the same

eigenvalue as AB. So AB and BA have the same eigenvalues.(b) True.Since B is invertible, we have AB = B−1(BA)B. So BA is similar to AB, and they must

have the same eigenvalues.(c) True.Let Λ be the diagonal matrix consists of eigenvalues, and S be the matrix of eigenvectors

(ordered as the eigenvalue matrix). Then we have A = S−1ΛS and also B = S−1ΛS. So A = B.(d) False.

For example,(

1 00 2

)and

(1 20 2

)are both diagonalizable, with the same eigenvalues, but

they are different.(e) False.

For example, let A =(

0 10 0

)and B =

(1 11 0

). Then AB =

(1 00 0

)and BA =

(0 10 1

).

The eigenvectors of AB are (0, 1)T and (1, 0)T , while the eigenvectors or BA are (0, 1)T and (1, 1)T .

Problem 8

You are given the matrix

A =

0 −1 0 10 1 0 11 0 1 11 0 0 −1

.

(a) What is the sum of the eigenvalues of A?(b) What is the product of the eigenvalues of A?(c) What can you say, without computing them, about the eigenvalues of AAT ?

Solution:(a) The sum of the eigenvalues of A equals trace of A. So the sum of the eigenvalues of A is

0 + 1 + 1− 1 = 1.(b) The product of the eigenvalues of A equals the determinant of A. We do row transforms

0 −1 0 10 1 0 11 0 1 11 0 0 −1

2 0 1 20 1 0 11 0 1 11 0 0 −1

2 0 1 20 1 0 10 0 1/2 00 0 −1/2 −2

2 0 1 20 1 0 10 0 1/2 00 0 0 −2

,

where in the first step we add the other rows to the first row. It follows that the product ofeigenvalues equals det(A) = −2.

(c) AAT is a real symmetric matrix, so its eigenvalues are all real and nonnegative. SinceA is nonsingular, all eigenvalues of AAT are positive. The product of all eigenvalues of AAT are

6

det(AAT ) = det(A) det(AT ) = det(A)2 = 4. The sum of all eigenvalues of AAT is trace(AAT ) =∑i,j a2

ij = 9.

Problem 9

You are given the quadratic polynomial f(x, y, z):

f(x, y, z) = 2x2 − 2xy − 4xz + y2 + 2yz + 3z2 − 2x + 2z.

(a) Write f(x, y, z) in the form f(x, y, z) = xT Ax − bT x where x = (x, y, z)T , A is a real-symmetric matrix, and b is some constant vector.

(b) Find the point (x, y, z) where f(x, y, z) is at an extremum.(c) Is this point a minimum, maximum, or a saddle point of some kind?

Solution:(a) We have

A =

2 −1 −2−1 1 1−2 1 3

, b =

20−2

.

(b) We compute the partial derivatives to find the extremum point:

∂f

∂x= 4x− 2y − 4z − 2 = 0

∂f

∂y= −2x + 2y + 2z = 0

∂f

∂x= −4x + 2y + 6z + 2 = 0

The equation is just 2Ax = b. The solution to the the equations above is x = 1, y = 1, z = 0. Sothe extreme point is (1, 1, 0).

(c) We look for the pivots of A: 2 −1 −2−1 1 1−2 1 3

2 −1 −2

0 1/2 00 0 1

.

Since A is positive definite, the extreme point is a minimum.

Problem 10

Suppose A is some diagonalizable matrix. Consider the vector y(t) = eA2tx for some vector x.(a) If A is 3 × 3 with eigenvalues λ1, λ2, and λ3 with eigenvectors x1, x2, and x3, and

x = x1 + 3x2 + 4x3, what is y(t)?(b) If limt→∞ y(t) = 0 for every vector x, what does that tell you about the eigenvalues of A

and of eA2?

(c) In terms of A, give a system of differential equations that y(t) satisfies, and the initialcondition.

7

(d) In terms of A, give a linear recurrence relation that y(t) satisfies for t = k∆t for integersk and some fixed ∆t.Solution:

(a) A2 has eigenvalues λ21, λ

22 and λ2

3 with eigenvectors x1,x2 and x3, so eA2t has eigenvalueseλ2

1t, eλ22t and eλ2

3t with the same eigenvectors. Thus

y(t) = eA2tx = eA2t(x1 + 3x2 + 4x3) = eλ21tx1 + 3eλ2

2tx2 + 4eλ23tx3.

(b) If limt→∞ y(t) = 0 for every vector x, we must have Re λ21 < 0,Re λ2

2 < 0 and Re λ23 < 0.

The eigenvalues eλ21 , eλ2

2 and eλ23 all sastisfies |eλ2 | < 1.

(c) y(t) satisfies the system y′ = A2y. The initial condition is y(0) = x.

(d) Denote by yk = y(k∆t). Then equation y(t) = eA2tx gives the recurrence relationyk+1 = eA2∆tyk with initial condition y0 = x.

Problem 11

Suppose A is an m × n matrix with full row rank. Which of the following equations always havea solution (possibly non-unique) for any b?

(a) Ax = b(b) AHx = b(c) AHAx = b(d) AAHx = b(e) AHAx = AHb(f) AAHx = Ab

Solution:(a) The equation Ax = b has a solution for any b, as explained in problem 4.(b) The equation AHx = b may has no solution, since the column space of AH is an m-

dimensional subspace in the whole space Rn.(c) The equation AHAx = b may has no solution, since the C(AHA) = C(AH) is an m-

dimensional subspace in Rn. (That C(AHA) = C(AH) comes from the fact N(AHA) = N(A)which we proved in class.)

(d) The equation AAHx = b will always has a unique solution, since AAH is an m×m matrixwhose rank is m, i.e. AAH is invertible.

(e) The equation AHAx = AHb will always has a solution. In fact, any solution to Ax = b(from part (a)) is always a solution to AHAx = AHb.

(f) The equation AAHx = Ab will always has a unique solution, since AAH is invertible, asexplained in part (d).

Additional Practice Problems

Be sure to look at:(i) Exams 1, 2, and 3.(ii) The practice problems for exam 3. (The above problems mostly cover non-eigenvalue

stuff.)(iii) Ideally, also review your homework problems.

8

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18.06 FINAL SOLUTIONS

Problem 1. (10 points) B =

a b a+bb c b+ cx y z

. We know that symmetric matrices have

real eigenvalues and orthogonal eigenvectors. So we set x = a + b and y = b + c. Thisleaves only the singularity of B. For this, we note that setting z = x+y = a+2b+c makesthe third column a sum of the first two, thus ensuring singularity.

Problem 2. (4 points each). A =

0 1 00 0 10 0 p

.

a) To find the eigenvalues of A, we compute det(A−λ I) = det

−λ 1 00 −λ 10 0 p−λ

=

(λ 2)(p−λ ) (because A is upper triangular). So the eigenvalues are 0 and p.

b) If p 6= 0, the we wish to find

abc

so that A

abc

=

papbpc

. But A

abc

=

bcpc

,

so we need b = pa and c = pb; and so

abc

=

1pp2

works.

c) The singular vaues of A are found by first computing the eigenvalues of AT A =0 0 01 0 10 1 p

0 1 00 0 10 0 p

=

0 0 00 1 p0 0 1+ p2

. As this is upper triangular, the eigenval-

ues are 0, 1 and 1+ p2. So the singular values are 0, 1 and√

1+ p2.

d) In general, du/dt = Bu is solved by eBtu(0). For us, B =

2009 1 00 2009 10 0 2009+ p

=

A+2009I. To compute eBt , we note that eBt = exp(t

2009 0 00 2009 00 0 2009+ p

+t

0 1 00 0 10 0 0

)=e2009 0 00 e2009 00 0 epe2009

exp(t

0 1 00 0 10 0 0

). To compute exp(t

0 1 00 0 10 0 0

), we note

that

0 1 00 0 10 0 0

2

=

0 0 10 0 00 0 0

and

0 1 00 0 10 0 0

3

= 0. So exp(t

0 1 00 0 10 0 0

) = I +

t

0 1 00 0 10 0 0

+(t2/2)

0 0 10 0 00 0 0

=

1 t t2/20 1 t0 0 1

, and finally eBt =

e2009 te2009 e2009t2/20 e2009 te2009

0 0 epe2009

.

So our answer is

e2009 te2009 e2009t2/20 e2009 te2009

0 0 epe2009

100

=

e2009

00

.

1

18.06 FINAL SOLUTIONS 2

Problem 3. (8 points) A is 4× 4 and has singular values {3,2,1,0}. As the productof the singular values is (up to sign) the determinant, we get that det(A) = 0, so A hasnontrivial nullspace, and so 0 is an eigenvalue.

Problem 4. (3 points each). A = QR where Q is orthogonal and R is upper triangularwith 1’s on the diagonal.

a) det(AT A) = det(RT QT QR) = det(RT R) = det(R)2 = 1 since det(R) = 1 by the as-sumption on R.

b) The equation AT A = RT R tells us that (R−1)T (AT A) = R; and since (R−1)T is lowertriangular, this is exactly the elimination of AT A. So the pivots are all equal to 1.

c) Yes, since Q−1(QR)Q = RQ.Problem 5. (10 points) C = A−1BX . We know that similar matrices have the same

eigenvalues, so putting X = A forces C and B to have the same eigenvalues.Problem 6. (4 points each) A is 3×3 and has four 0’s and five 1’s.a) A has rank 0 is impossible- it isn’t the zero matrix.

b) A has rank 2: A =

1 1 01 1 01 0 0

.

c) A has rank 3: A =

1 0 01 1 01 0 1

.

Problem 7. (10 points each) A is 100×100.a) A has all even integers as entries. Therefore each column of A has the form 2c where

c is a vector of integers. So we set C = (1/2)A. Then det(A) = det(2C) = 2100det(C); sodet(A) is an even integer (note that det(C) really is an integer because all the entries of Care integers and the det can be computed by the big formula).

b) This time, we use the big formula to compute det(C) = ∑sign(σ)a1σ(1)a2σ(2) · · ·anσ(n). This is a sum containing 100! terms. Now, 100! is an even number, and each termin the sum is odd (as a product of odd integers). Since the sum of two odd numbers is even,the sum of an even number of odd numbers is even; so this sum is an even integer.

Problem 8. (5 points each) We consider the vector space V of functions of the formc1 + c2ex + c3e2x, with basis {1,ex,e2x}.

a) d/dx takes V to the space W spanned by {ex,e2x}. We have that ddx (1) = 0, d

dx ex = ex

, and ddx e2x = 2e2x. So the linear transformation of d/dx in the given bases is

(0 1 00 0 2

).

b) We conisder the transformation φ from V to R defined by f → f (7). This is linear:φ( f +g) = ( f +g)(7) = f (7)+g(7) = φ( f )+φ(g), and φ(c f )→ c f (7) = cφ( f ).

c) No.´ x

0 1 = x is a function not in V .

18.06 Professor Strang Final Exam May 20, 2008

Grading

1

2

3

4

5

6

7

8

9

10



1) M 2 2-131 A. Ritter 2-085 2-1192 afr

2) M 2 4-149 A. Tievsky 2-492 3-4093 tievsky

3) M 3 2-131 A. Ritter 2-085 2-1192 afr

4) M 3 2-132 A. Tievsky 2-492 3-4093 tievsky

5) T 11 2-132 J. Yin 2-333 3-7826 jbyin

6) T 11 8-205 A. Pires 2-251 3-7566 arita

7) T 12 2-132 J. Yin 2-333 3-7826 jbyin

8) T 12 8-205 A. Pires 2-251 3-7566 arita

9) T 12 26-142 P. Buchak 2-093 3-1198 pmb

10) T 1 2-132 B. Lehmann 2-089 3-1195 lehmann

11) T 1 26-142 P. Buchak 2-093 3-1198 pmb

12) T 1 26-168 P. McNamara 2-314 4-1459 petermc

13) T 2 2-132 B. Lehmann 2-089 2-1195 lehmann

14) T 2 26-168 P. McNamara 2-314 4-1459 petermc

Thank you for taking 18.06.

If you liked it, you might enjoy 18.085 this fall.

Have a great summer. GS

1 (10 pts.) The matrix A and the vector b are

A =

1 1 0 2

0 0 1 4

0 0 0 0

b =

3

1

0

(a) The complete solution to Ax = b is x = .

(b) ATy = c can be solved for which column vectors c = (c1, c2, c3, c4) ?

(Asking for conditions on the c’s, not just c in C(AT).)

(c) How do those vectors c relate to the special solutions you found in

part (a) ?

Solution (10 points)

a) The complete solution is a particular solution xp plus any vector in the nullspace xn. Since

the matrix A is already reduced, we can just read the special solutions off: [−1, 1, 0, 0]T and

[−2, 0,−4, 1]T . To find a particular solution to Ax = b, we put any numbers (we may as

well choose 0) in for the free variables. This yields the two equations x1 = 3 and x3 = 1, so

xp = [3, 0, 1, 0]T . In the end we get

xcomp =

3

0

1

0

+ c1

−1

1

0

0

+ c2

−2

0

−4

1

(1)

b) You can do this computation by hand by augmenting AT with the column (c1, c2, c3, c4)

and row reducing. The solution is given by the equations that correspond to 0 rows in the

reduced matrix. A quicker way is to note that ATy = c has a solution whenever c is in the

column space C(AT ), i.e. the row space of A. This is perpendicular to the nullspace. Thus,

we can find the equations by taking a basis for the nullspace and using the components as

coefficients in our equations. We find equations −c1 + c2 = 0 and −2c1 − 4c3 + c4 = 0.

c) Because these c are in the row space, they are perpendicular to vectors in the nullspace

of A, and in particular are perpendicular to the special solutions.

2

2 (8 pts.) (a) Suppose q1 = (1, 1, 1, 1)/2 is the first column of Q. How could you

find three more columns q2, q3, q4 of Q to make an orthonormal basis ?

(Not necessary to compute them.)

(b) Suppose that column vector q1 is an eigenvector of A: Aq1 = 3q1.

(The other columns of Q might not be eigenvectors of A.) Define

T = Q−1AQ so that AQ = QT . Compare the first columns of AQ and

QT to discover what numbers are in the first column of T ?

Solution (8 points)

a) First, we find additional vectors v2, v3 and v4 that (along with q1) make up a basis of R4.

Then we run Gram-Schmidt on q1, v2, v3, v4.

b) Using the column picture of multiplication, we see that the first column of AQ will be

Aq1 = 3q1. Similarly, if we denote the first column of T by (t1, t2, t3, t4), then the first column

of QT will be t1q1 + t2q2 + t3q3 + t4q4. Since these two are equal, we get an equality of vectors

3q1 = t1q1 + t2q2 + t3q3 + t4q4 (2)

Since the qi are linearly independent, we must have t1 = 3 and the other ti = 0, showing

that the first column of T is (3, 0, 0, 0).

We can also note that the first column of T is equal to 3QT q1, which yields the same answer.

3

3 (12 pts.) Two eigenvalues of this matrix A are λ1 = 1 and λ2 = 2. The first two

pivots are d1 = d2 = 1.

A =

1 0 1

0 1 1

1 1 0

.(a) Find the other eigenvalue λ3 and the other pivot d3.

(b) What is the smallest entry a33 in the southeast corner that would

make A positive semidefinite ? What is the smallest c so that A + cI

is positive semidefinite ?

(c) Starting with one of these vectors u0 = (3, 0, 0) or (0, 3, 0) or (0, 0, 3),

and solving uk+1 = 12Auk, describe the limit behavior of uk as k →∞

(with numbers).


a) The sum of the eigenvalues is the trace, so 1 + 2 +λ3 = 2. Thus λ3 = −1. The product of

the pivots is the determinant, which is the product of the eigenvalues as well. So d3 = −2.

Note that this means that A is not positive-definite.

b) We can test positive-definiteness using the determinant method. The two top-left de-

terminants of A are both positive, so we just need to check the third one. We obtain the

relation:

1(c− 1) + 1(−1) ≥ 0 (3)

so the smallest value of c is 2.

For the second part, we test whether the eigenvalues are non-negative. The eigenvalues of

A + cI are just the eigenvalues of A plus c. So when c = 1 all the eigenvalues will be

non-negative.

4

c) The matrix 12A is a Markov matrix. Because it has some 0 entries, we don’t automatically

know that it has a unique steady state vector. However, since the eigenvalues of 12A are 1/2,

−1/2 and 1, it does have a unique steady state vector (only one eigenvalue has absolute value

1). To find it, we calculate the eigenvector of A with eigenvalue 2 by taking the nullspace of

A− 2I:

A− 2I =

−1 0 1

0 −1 1

1 1 −2

(4)

−1 0 1

0 −1 1

0 0 0

(5)

The nullspace is generated by the special solution (1, 1, 1). So, a vector u will have limit

A∞u equal to c(13, 1

3, 1

3), where c is the sum of the components of u. In particular, the vectors

(3, 0, 0), etc., all go to (1, 1, 1).

5

4 (10 pts.) Suppose Ax = b has a solution (maybe many solutions). I want to prove

two facts:

A. There is a solution xrow in the row space C(AT).

B. There is only one solution in the row space.

(a) Suppose Ax = b. I can split that x into xrow + xnull with xnull in the

nullspace. How do I know that Axrow = b ? (Easy question)

(b) Suppose x∗row is in the row space and Ax∗row = b. I want to prove that

x∗row is the same as xrow. Their difference d = x∗row−xrow is in which

subspaces ? How to prove d = 0 ?

(c) Compute the solution xrow in the row space of this matrix A, by solving

for c and d:

1 2 3

1 1 −1

xrow =

14

9

with xrow = c

1

2

3

+ d

1

1

−1

.


a) We have A(xrow + xnull) = A(xrow) + A(xnull) = A(xrow) + 0, so A(xrow) = b.

b) Suppose both A(xrow) = b and A(x∗row) = b. Then x∗row−xrow is in the row space (since it

is a linear combination of vectors in the row space) and is in the nullspace (since multiplying

by A will give us 0). Because the row space and nullspace are orthogonal complements, the

only vector that is in both is the 0 vector: any vector in both will have |x|2 = x · x = 0.

6

c) Substituting in the given expressions for Axrow = b we find

1 2 3

1 1 −1

1 1

2 1

3 −1

c

d

=

14

9

(6)

or 14 0

0 3

c

d

=

14

9

(7)

We find (c, d) = (1, 3), so xrow = (4, 5, 0). Remark: essentially what we are doing here is

projecting onto the row space.

7

5 (10 pts.) The numbers Dn satisfy Dn+1 = 2Dn − 2Dn−1. This produces a first-order

system for un = (Dn+1, Dn) with this 2 by 2 matrix A: Dn+1

Dn

=

2 −2

1 0

Dn

Dn−1

or un = Aun−1 .

(a) Find the eigenvalues λ1, λ2 of A. Find the eigenvectors x1, x2 with

second entry equal to 1 so that x1 = (z1, 1) and x2 = (z2, 1).

(b) What is the inner product of those eigenvectors ? (2 points)

(c) If u0 = c1x1 + c2x2, give a formula for un. For the specific u0 = (2, 2)

find c1 and c2 and a formula for Dn.


a) The eigenvalues of

A =

2 −2

1 0

(8)

satisfy the equation λ2 − 2λ+ 2 = 0, so λ1 = 1 + i and λ2 = 1− i. We find the eigenvectors

by taking the appropriate nullspaces:

A− λ1I =

1− i −2

1 −1− i

(9)

has nullspace generated by x1 = (1 + i, 1), and

A− λ2I =

1 + i −2

1 −1 + i

(10)

has nullspace generated by x2 = (1 − i, 1). If you pick a different vector in the nullspace,

you just rescale so that the bottom entry is 1.

b) The inner product is xH1 x2 = (1−i)2+1 = 1−2i, or its conjugate expression xH

2 x1 = 1+2i.

8

c) If u0 = c1x1 + c2x2, then un = c1λn1x1 + c2λ

n2x2. A matrix always acts on its eigenvectors

in a diagonal way. In particular, (2, 2) = x1 + x2. So we find

un = (1 + i)n

1 + i

1

+ (1− i)n

1− i

1

(11)

with second entry Dn = (1 + i)n + (1− i)n.

9

6 (12 pts.) (a) Suppose q1, q2, a3 are linearly independent, and q1 and q2 are already

orthonormal. Give a formula for a third orthonormal vector q3 as a

linear combination of q1, q2, a3.

(b) Find the vector q3 in part (a) when

q1 =1

2

1

1

1

1

q2 =1

2

1

−1

1

−1

a3 =

1

2

3

4

(c) Find the projection matrix P onto the subspace spanned by the first

two vectors q1 and q2. You can give a formula for P using q1 and q2 or

give a numerical answer.


a) This is the Gram-Schmidt process. We define

w3 = a3 − (q1 · a3)q1 − (q2 · a3)q2 (12)

and then set q3 = w3/‖w3‖. Note that we do not need denominators in the expression for

w3 because the qi are already unit vectors.

b) Substituting in, we find

w3 = a3 − 5q1 − (−1)q2 = (−1,−1, 1, 1) (13)

Renormalizing we get q3 = 12(−1,−1, 1, 1).

c) The projection matrix P is exactly the expression we used for Gram-Schmidt: P =

q1qT1 + q2q

T2 . There are other more complicated expressions which are also correct. We can

start at the most general and simplify to get this one; if A has columns q1 and q2 then

P = A(ATA)−1AT = A(I)AT = q1qT1 + q2q

T2 where we used the column-row picture of

multiplication for the last step.

10

7 (12 pts.) (a) Find the determinant of this N matrix.

N =

1 0 0 4

2 1 0 3

3 0 1 2

4 0 0 1

(b) Using the cofactor formula for N−1, tell me one entry that is zero or

tell me that all entries of N−1 are nonzero.

(c) What is the rank of N − I ? Find all four eigenvalues of N .


a) There are many ways to do this. Perhaps the easiest is cofactors along the top row:

det(N) = 1(1)− 4 det

2 1 0

3 0 1

4 0 0

= 1− 4(4) = −15 (14)

Here I found the determinant of the 3 by 3 by swapping the columns to get an upper

triangular matrix with diagonal entries 1, 1, 4.

b) The cofactor formula is A−1 = CT/ det(A) (we know that A is invertible from part a).

To check for 0 entries we can ignore the det(A) part. We just need to find some cofactors

that are 0, and we can arrange this by crossing out rows and columns that will give us a

smaller matrix with a column of 0s. Some choices are C21, C23, C24, C31, C32, and C34. The

corresponding 0 entries of the inverse are the transposes, so we get the entries (1, 2), (3, 2),

(4, 2), (1, 3), (2, 3), (4, 3).

11

c) The matrix N − I has two columns that are all 0s, and the other two columns are clearly

independent, so it has rank 2. So N − I has eigenvalue 0 with multiplicity 2. This tells us

that N has eigenvalue 1 with multiplicity 2. Calling the other eigenvalues λ1 and λ2, we can

find them solving the trace and determinant equations:

1 + 1 + λ1 + λ2 = 4 (15)

(1)(1)λ1λ2 = −15 (16)

Thus λ1 = 5 and λ2 = −3.

12

8 (8 pts.) Every invertible matrix A is the product A = QH of an orthogonal matrix

Q and a symmetric positive definite matrix H. I will start the proof:

A has a singular value decomposition A = UΣV T.

Then A = (UV T)(V ΣV T).

(a) Show that UV T is an orthogonal matrix Q (what is the test for an

orthogonal matrix ?).

(b) Show that V ΣV T is a symmetric positive definite matrix. What are

its eigenvalues and eigenvectors ? Why did I need to assume that A is

invertible ?

Solution (8 points)

a) To test that Q = UV T is orthogonal, we must show that QTQ = I. But QTQ =

(UV T )TUV T = V UTUV T = V (I)V T = I. We used the fact that U and V are orthogonal

matrices.

b) The matrix H = V ΣV T is definitely symmetric, as HT = V ΣTV T = V ΣV T because Σ

is diagonal. Furthermore, note that the expression H = V ΣV T is a diagonalization of H.

This means that H has eigenvalues given by the entries of Σ and eigenvectors equal to the

columns of V . To show that H is positive-definite, we just need to show that the diagonal

entries of Σ are all positive.

Now, we know that they are all non-negative, because the SVD always gives us non-negative

singular values. We must also show that none of the singular values are zero. Remember

that the singular values are equal to the square roots of the eigenvalues of ATA. However,

because A is invertible, the matrix ATA is also invertible, and so can’t have any eigenvalues

equal to 0. So no singular value is 0 either.

13

9 (7 pts.) (a) Find the inverse L−1 of this real triangular matrix L:

L =

1 0 0

a 1 0

0 a 1

You can use formulas or Gauss-Jordan elimination or any other method.

(b) Suppose D is the real diagonal matrix D = diag(d, d2, d3). What are

the conditions on a and d so that the matrix A = LDLT is (three

separate questions, one point each)

(i) invertible ? (ii) symmetric ? (iii) positive definite ?

Solution (7 points)

a) I’ll do Gauss-Jordan elimination.1 0 0 1 0 0

a 1 0 0 1 0

0 a 1 0 0 1

1 0 0 1 0 0

0 1 0 −a 1 0

0 a 1 0 0 1

(17)

1 0 0 1 0 0

0 1 0 −a 1 0

0 0 1 a2 −a 1

(18)

b) Note that L is invertible no matter what a is, and D is invertible so long as d 6= 0. So

A = LDLT will be invertible whenever d 6= 0. If d = 0, then of course A can’t be invertible.

The matrix A is always symmetric, since AT = (LDLT )T = LDTLT = LDLT .

Because A is always symmetric, to check positive-definiteness we just need to check that the

pivots are all positive. But A = LDLT is the “pivot” decomposition for A. So the pivots of

A are d, d2, d3, and we need d > 0.

14

10 (11 pts.) This problem uses least squares to find the plane C + Dx + Ey = b that

best fits these 4 points:

x= 0 y = 0 b= 2

x= 1 y = 1 b= 1

x= 1 y = −1 b= 0

x= −2 y = 0 b= 1

(a) Write down 4 equations Ax = b with unknown x = (C,D,E) that

would hold if the plane went through the 4 points. Then write down the

equations to solve for the best (least squares) solution x = (C, D, E).

(b) Find the best x and the error vector e (is the vector e in R3 or R4 ?).

(c) If you change this b = (2, 1, 0, 1) to the vector p = Ax, what will be

the best plane to fit these four new points (p1, p2, p3, p4) ? What will

be the new error vector ?


a) The equations are of the form C + 0D + 0E = 2, etc., or in matrix form1 0 0

1 1 1

1 1 −1

1 −2 0

C

D

E

=

2

1

0

1

(19)

Of course this system is not solvable. The best solution is given by ATAx = AT b.

15

b) We have

ATA =

4 0 0

0 6 0

0 0 2

(20)

and

AT b =

4

−1

1

(21)

It is a diagonal system, so we immediately find (C,D,E) = (1,−1/6, 1/2). The error vector

is the difference of the real b and the approximate values we get for our plane: e = b− Ax.

Since Ax = [1, 4/3, 1/3, 4/3]T , we get e = (1,−1/3,−1/3,−1/3).

c) We know p = Ax is the projection of b onto the column space of A. So the system Ax = p

is solvable exactly; we don’t need any approximations. The best fit plane will be the same

plane as in part b: 1 − x/6 + y/2 = b (we changed the b-coordinates of the points so that

they lie on this plane, so of course it is the best fit). The error vector will become 0 because

it is an exact fit.

16

18.06 Final SolutionHold on Tuesday, 19 May 2009 at 9am in Walker Gym.

Total: 100 points.

Problem 1:A sequence of numbers f0, f1, f2, . . . is defined by the recurrence

fk+2 = 3fk+1 − fk,

with starting values f0 = 1, f1 = 1. (Thus, the first few terms in the sequence are1, 1, 2, 5, 13, 34, 89, . . ..)

(a) Defining uk =

(fk+1

fk

), re-express the above recurrence as uk+1 = Auk, and

give the matrix A.

(b) Find the eigenvalues of A, and use these to predict what the ratio fk+1/fk ofsuccessive terms in the sequence will approach for large k.

(c) The sequence above starts with f0 = f1 = 1, and |fk| grows rapidly with k.Keep f0 = 1, but give a different value of f1 that will make the sequence (withthe same recurrence fk+2 = 3fk+1 − fk) approach zero (fk → 0) for large k.

Solution (18 points = 6+6+6)(a) We have(

fk+2

fk+1

)=

(3 −11 0

)(fk+1

fk

)⇒ A =

(3 −11 0

).

(b) Eigenvalues of A are roots of det(A − λI) = λ2 − 3λ + 1 = 0. They are

λ1 =3 +√

5

2and λ2 =

3−√

5

2. Note that λ1 > λ2, so the ratio fk+1/fk will

approach λ1 =3 +√

5

2for large k.

(c) Let v1,v2 be the eigenvectors with eigenvalues λ1 and λ2, respectively. So,we can write u0 = c1v1 + c2v2 and then uk = c1λ

k1v1 + c2λ

k2v2. If we need fk → 0,

1

we have to make c1 = 0. In other words, u0 must be proportional to the eigenvectorv2.

A− λ2I =

3 +√

5

2−1

1 −3−√

5

2

⇒ v2 =

3−√

5

21

.

Hence, we need to take f1 =3−√

5

2so that fk will approach zero for large k.

2

Problem 2: For the matrix A =

1 0 −11 1 12 1 0

with rank 2, consider the system of

equations Ax = b.

(i) Ax = b has a solution whenever b is orthogonal to some nonzero vector c.Explicitly compute such a vector c. Your answer can be multiplied by anyoverall constant, because c is any basis for the spaceof A.

(ii) Find the orthogonal projection p of the vector b =

999

onto C(A). (Note:

The matrix ATA is singular, so you cannot use your formula P = A(ATA)−1AT

to obtain the projection matrix P onto the column space of A. But I haverepeatedly discouraged you from computing P explicitly, so you don’t need tobe reminded anyway, right?)

(iii) If p is your answer from (ii), then a solution y of Ay = p minimizes what?[You need not answer (ii) or compute y for this part.]

Solution (18 points = 7+7+4)(i) The system of equations Ax = b has a solution if and only if b lies in the

column space of A, which is orthogonal to the left nullspace of A. We solve for a(nonzero) vector c in the left nullspace using Gaussian elimination, as follows.

AT =

1 1 20 1 1−1 1 0

;

1 1 20 1 10 2 2

;

1 0 10 1 10 0 0

⇒ c =

−1−11

.

The answer can by any nonzero multiple of c, which will be a basis for the leftnullspace of A.

(ii) Method 1: Since c is a basis of the orthogonal complement of the columnspace C(A), the projection of b onto C(A) can be computed as

p = b− cTbc

‖c‖2=

999

− −9

3

−1−11

=

6612

.

3

Method 2: (not recommended) We know that p is the best linear approximationof b. So we solve

ATA

y1

y2

y3

= AT

999

,

6 3 03 2 10 1 2

y1

y2

y3

=

36180

We can get a particular solution y = (6, 0, 0)T. (There are other solutions too.)Hence,

p = A

y1

y2

y3

=

1 0 −11 1 12 1 0

600

=

6612

.

(iii) Since p is the orthogonal projection of b onto C(A), A solution y of Ay = pminimizes the distance ‖Ay − b‖.

4

Problem 3: True or false. Give a counter-example if false. (You need not providea reason if true.)

(a) If Q is an orthogonal matrix, then detQ = 1.

(b) If A is a Markov matrix, then du/dt = Au approaches some finite constantvector (a “steady state”) for any initial condition u(0).

(c) If S and T are subspaces of R2, then their intersection (points in both S andT ) is also a subspace.

(d) If S and T are subspaces of R2, then their union (points in either S or T ) isalso a subspace.

(e) The rank of AB is less than or equal to the ranks of A and B for any A andB.

(f) The rank of A + B is less than or equal to the ranks of A and B for any Aand B.

Solution (12 points = 2+2+2+2+2+2)(a) False. For example, Q = (−1) is an orthogonal matrix: QTQ = (−1)(−1) =

(1).REMARK: In general, for a real orthogonal matrix Q, detQ = ±1. This is

because det(QTQ) = det(I) = 1 ⇒ det(Q)2 = det(QT) det(Q) = 1.

(b) False. Be careful here that we are discussing differential equations but notthe powers of A. For example, A = (1), the differential equation has solution u = cet

for some constant c, which does not approach to any finite constant vector.REMARK: It is true that for the Markov process uk+1 = Auk, uk approaches

some finite constant vector (a “steady state”) for any initial condition u0.

(c) True. Intersections of subspaces are always subspaces.

(d) False. For example, S and T are the x- and y-axes. Then (1, 1) = (1, 0) +(0, 1) is a linear combination of points in the union of S and T , but does not lie inthe union itself. So the union of S and T is not a subspace.

(e) Ture. One may see this by arguing as follows. Since the column space ofAB is a subspace of the column space of A, the rank of AB is less than or equal to

5

the rank of A. Similarly, since the row space of AB is a subspace of the row spaceof B, the rank of AB is less than or equal to the rank of B.

(f) False. A =

(0 00 1

), B =

(1 00 0

)both have rank 1. But A + B =

(1 00 1

)has rank 2.

REMARK: It is true that rank(A+B) ≤ rankA+ rankB.

6

Problem 4: Consider the matrix

A =

1 1 11 −1 −11 0 −31 0 −1

(a) Find an orthonormal basis for C(A) using Gram-Schmidt, forming the columns

of a matrix Q.

(b) Write each step of your Gram-Schmidt process from (a) as a multiplication ofA on the (left or right) by some invertible matrix. Explain howthe product of these invertible matrices relates to the matrix R from the QRfactorization A = QR of A.

(c) Gram-Schmidt on another matrix B (of the same size as A) gives the sameorthonormal basis (the same Q) as in part (a). Which of the four subspaces,if any, must be the same for the matrices AAT and BBT? [You can do thispart without doing (a) or (b).]

Solution (18 points = 6+6+6)(a) From u1 = (1, 1, 1, 1)T, we get q1 = u1/‖u1‖ = 1

2(1, 1, 1, 1)T.

v2 = (1,−1, 0, 0)T,

u2 = v2 − qT1 v2q1 = v2 = (1,−1, 0, 0)T,

q2 = v2/‖v2‖ =1√2

(1,−1, 0, 0)T;

v3 = (1,−1,−3,−1)T,

u3 = v3 − qT1 v3q1 − qT

1 v3q1 = v3 + u1 − u2 = (1, 1,−2, 0)T,

q3 = v3/‖v3‖ =1√6

(1, 1,−2, 0)T.

Hence, we have

Q =

1/2 1/

√2 1/

√6

1/2 −1/√

2 1/√

6

1/2 0 −2/√

61/2 0 0

.

7

(b) Each step of the Gram Schmidt process from (a) is a multiplication of A onthe right as follows.

A ; A

1/2 0 00 1 00 0 1

; A

1/2 0 00 1 00 0 1

1 0 0

0 1/√

2 00 0 1

; A

1/2 0 00 1 00 0 1

1 0 0

0 1/√

2 00 0 1

1 0 10 1 −10 0 1

; A

1/2 0 00 1 00 0 1

1 0 0

0 1/√

2 00 0 1

1 0 10 1 −10 0 1

1 0 00 1 0

0 0 1/√

6

= Q.

The product of these invertible 3× 3 matrices is exactly R−1.

(c) Since the Gram-Schmidt of A and B gives the same outcome, the columnspace of A and B are the same. We know that A and AAT have the same columnspace, and B and BBT have the same column space. Hence AAT and BBT havethe same column space. Moreover, since left nullspace is always orthogonal to thecolumn space, AAT and BBT have the same left nullspace too. Also, notice thatAAT and BBT are symmetric matrices, their row spaces are the same as the columnspaces, and their nullspaces are the same as the left nullspaces. Therefore, all foursubspaces of AAT are the same as BBT.

8

Problem 5: The complete solution to Ax = b is

x =

10−1

+ c

110

+ d

−201

for any arbitrary constants c and d.

(i) If A is an m×n matrix with rank r, give as much true information as possibleabout the integers m, n, and r.

(ii) Construct an explicit example of a possible matrix A and a possible right-handside b with the solution x above. (There are many acceptable answers; youonly have to provide one.)

Solution (16 points = 8+8)(i) Since we can multiply A with x, n = 3. Also, since the nullspace of A is

2-dimensional, r = n− 2 = 1. There is no restriction on m except that m ≥ r = 1.

(ii) We construct a minimal one, namely, A = (a1 a2 a2) is 1 × 3. For this, we

need A

110

= 0 and A

−201

= 0. That is

(1 1 0−2 0 1

)a1

a2

a3

=

(00

).

A special solution is A = (1 −1 2). In this case, b = Ax = (1 −1 2)

10−1

= (−1).

So, an example is (1 −1 2

)x = (−1).

9

Problem 6: Consider the matrix

A =

1 −1 −1−1 1 −1−1 −1 1

(i) A has one eigenvalue λ = −1, and the other eigenvalue is a double root of

det(A− λI). What is the other eigenvalue? (Very little calculation required.)

(ii) Is A defective? Why or why not?

(iii) Using the above A, suppose we want to solve the equation

du

dt= Au + cu

where c is some real number, for some initial condition u(0).

(a) For what values of c will the solutions u(t) always to go zero as t→∞?

(b) For what values of c will the solutions u(t) typically diverge (‖u(t)‖ → ∞)as t→∞?

(c) For what values of c will the solutions u(t) approach a constant vector(possibly zero) as t→∞?

Solution (18 points = 6+6+6 (2+2+2))(i) Let λ1 = −1 and let λ2 = λ3 denote the double roots. Then from the trace

of A, we have λ1 + 2λ2 = trace(A) = 3. Hence, λ2 = 2.

(ii) A is not defective. There are two ways to see it. For one way, since Ais symmetric, it is always non-defective; for another way, we compute A − λ2I =−1 −1 −1−1 −1 −1−1 −1 −1

, which has rank 1 and hence its nullspace is 2-dimentional.

(iii) The key point here is that A+ cI would have eigenvalues λ1 + c and λ2 + c(with multiplicity 2). An alternative point of view is as follows. If we write theinitial condition u(t) = c1(t)v1 + c2(t)v2 + c3(t)v3, then the differential equationbecomes

dc1(t)

dtv1 +

dc2(t)

dtv2 +

dc3(t)

dtv3 = c1λ1v1 + c2λ2v2 + c3λ3v3 + cc1v1 + cc2v2 + cc3v3.

10

We have dc1(t)

dt= c1λ1 + cc1, ⇒ c1 = e(λ1+c)t;

dc2(t)

dt= c2λ2 + cc2, ⇒ c2 = e(λ2+c)t;

dc3(t)

dt= c3λ3 + cc3, ⇒ c3 = e(λ3+c)t;

(a) If we require u(t) always go zero as t→∞, λ1 + c < 0, λ2 + c = λ3 + c < 0.Hence, we require c < −2.

(b) If the solution u(t) typically diverge, we need either λ1 + c > 0 or λ2 + c =λ3 + c > 0. Hence, we require c > −2.

(c) If we allow the solution to approach to some constant vector, we allow tohave the extreme case of (a), that is to say c ≤ −2.

11

final examination in linear algebra: 18.06 ma y professor...

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