Linear Algebra Math 221 Open Book Exam 3 Open Notes 21 Nov, 2002 Calculators Permitted Show all work (except #4)
1. (30 pts) Given the matrix
−=110
240
001
A
Note throughout this problem that the matrix is block diagonal, with a 1x1 matrix (1) and a 2x2 matrix (4, -2; 1, 1). Each of these problems can be solved separately for the two submatrices, and then combined for the final answer. The characteristic polynomial of matrix A is the product of the char polynomials of the two submatrices, char(A) = char(1)char(4,-2;1,1) = (1-λ)[(1-λ)(1-λ)-(-2)(1)]. The eigenvalues of A are the union of the eigenvalues of the two submatrices. Thus eigenvals(A) = eigenvals(1) U eigenvals(4,-2;1,1) = {1}U{2,3} = {1,2,3}. The eigenvectors for A are the eigenvectors of the submatrices, with suitable padding.
a) (5 pts) Compute the characteristic polynomial of A.
char(A) = det (A-λI) =
−−−
−=
−
−
λλ
λλ
110
240
001
det
100
010
001
110
240
001
det
= (1-λ)(4-λ)(1-λ)+(0)(-2)(0)+(0)(1)(0)–(0)(4-λ)(0)–(0)(0)(1-λ)–(-2)(1)(1-λ) = (1-λ)[(4-λ)(1-λ)-(-2)(1)] = (1-λ)[λ2-5λ+6] = -λ3+6λ2-11λ+6 Answer: char(A) = -λ3+6λ2-11λ+6 Check: The constant term of char(A) is equal to det(A) = 6.
b) (10 pts) Compute the eigenvalues of A.
The eigenvalues of A are the roots of the characteristic polynomial. We note that the characteristic polynomial has the form (1-λ)[λ2-5λ+6], so it has λ=1 as one root. The other two roots can be found as the roots of the quadratic polynomial λ2-5λ+6. Using the quadratic formula we get roots
{ }3,22
15)1(2
)6)(1(4)5()5( 2
=±
=−−±−−
=λ
Answer: λ = {1, 2, 3} Check: The product of the eigenvalues is the determinant.
c) (10 pts) Find the eigenvectors of A.
We compute the eigenvectors for a given eigenvalue λ by solving the matrix equation (A-λI) = 0 and selecting a basis for the solution space. λ = 1:
(A-λI) = (A-I) =
−→
−→
−
000
200
010
000
230
010
010
230
000
So the solution is:
==
0
0
free is
3
2
1
x
x
x
so
=
=
0
0
1
0
0 1
1
3
2
1
x
x
x
x
x
So the eigenvector corresponding to the eigenvalue λ=1 is the vector (1;0;0), or any multiple of it. Check: Confirm that A(1;0;0) = (1;0;0)
λ = 2:
(A-λI) = (A-2I) =
−
−→
−
−→
−−
−
000
110
001
000
220
001
110
220
001
So the solution is:
==
free is
0
3
32
1
x
xx
x
so
=
=
1
1
00
3
3
3
3
2
1
x
x
x
x
x
x
So the eigenvector corresponding to the eigenvalue λ=2 is the vector (0;1;1), or any multiple of it. Check: Confirm that A(0;1;1) = 2(0;1;1) = (0;2;2)
λ = 3:
(A-λI) = (A-I) =
−
−→
−−
−
000
210
002
210
210
002
So the solution is:
==
free is
2
0
3
32
1
x
xx
x
so
=
=
1
2
0
2
0
3
3
3
3
2
1
x
x
x
x
x
x
So the eigenvector corresponding to the eigenvalue λ=3 is the vector (0;2;1), or any multiple of it. Check: Confirm that A(0;2;1) = 3(0;2;1) = (0;6;3)
Answer: λ = 1, 2, 3 have eigenvectors
1
2
0
,
1
1
0
,
0
0
1
respectively
d) (5 pts) Find a matrix P such that P-1AP is diagonal.
The matrix P consists of the linearly independent eigenvectors of A.
Answer: P =
110
210
001
Check: P-1AP =
=
−
−
300
020
001
110
210
001
110
240
001
110
210
0011
= D
Note that if we put the eigenvectors together to form P in a different order we will still get a diagonal matrix D, but the order of the eigenvalues along the diagonal will be different.
2. (25 pts) If
=
233
242
112
121
A
a) (10 pts) Compute rank(A)
We compute rank(A) by row reducing A and counting the number of pivots (equivalently, the number of non-zero rows).
−−
→
−−−−
→
−−
−−→
000
000
130
121
000
130
130
121
130
000
130
121
233
242
112
121
Answer: Thus rank(A) = 2
b) (5 pts) Find a basis for the column space of A.
We compute a basis for the column space by selecting those columns in the original matrix A corresponding to the pivot columns in the reduced form of A. As there are pivots in columns 1 and 2 we select the first and second columns of A as a basis of the column space. (As it happens, for this value of A we could have selected any two columns, but that isn’t always the case.)
Answer: A basis of col(A) is
3
4
1
2
,
3
2
2
1
c) (10 pts) Find a basis for the null space of A.
We compute a basis for the null space by solving the homogeneous matrix equation Ax=0 and finding a basis of the solutions. From (a) above:
→
→
−−
→
000
000
10
01
000
000
10
121
000
000
130
121
233
242
112
121
3
13
1
3
1
so
−=−=
free is
3/
3/
3
32
31
x
xx
xx
and our solutions have the form
−−
=
−−
=
13
13
1
3
3
3
3
3
2
1
3
3
x
xx
x
xx
x
. Thus
−−
13
13
1
, or any
multiple of it, is a basis for the null space.
Answer: A basis of Null(A) is
− 3
1
1
Check: Confirm that Av = 0, for each vector v in the basis of Null(A)
3. (20 pts) Prove that if A is an invertible matrix and {v1, v2, v3, v4} are linearly
independent, then {Av1, Av2, Av3, Av4} are linearly independent. Proof: Assume not – in other words assume that {Av1, Av2, Av3, Av4} is linearly dependent. Thus, for some non-zero coefficients ai we have: a1(Av1)+ a2(Av2)+ a3(Av3)+ a4(Av4)=0 defn of linearly depn A(a1v1)+ A(a2v2)+ A(a3v3)+ A(a4v4)=0 Acx = cAx [Thm 5, Chap 1] A(a1v1+a2v2+a3v3+a4v4) = 0 A(u+v) = Au + Av [Thm 5, Chap 1] A-1A(a1v1+a2v2+a3v3+a4v4) = A-10 A is invertible a1v1+a2v2+a3v3+a4v4 = 0 But this is a contradiction, as we know that {v1, v2, v3, v4} are linearly independent. So our assumption that {Av1, Av2, Av3, Av4} are linearly dependent is false, and they must be linearly independent.
4. (25 pts) Multiple Choice: ALWAYS/SOMETIMES/NEVER (5 pts per question)
a) A 3x3 matrix with 2 distinct eigenvalues is diagonalizable.
SOMETIMES – If the repeated eigenvalue has two linearly independent eigenvectors then yes, if not then no. Examples of the two cases are
200
020
001
diagonalizable, and
200
120
001
non-diagonalizable. [Chap 5,
Suppl Problems, #1o]
b) The rank of a 3x3 invertible matrix is 1.
NEVER – An invertible matrix has full rank (the same as the number of rows or columns). See the continuation of the Invertible Matrix Thm, Sect 4.6, pg 267.
c) If A is row equivalent to B, then A and B have the same eigenvalues.
SOMETIMES – In general no, but there are some (rare) cases for which it is true. Do not confuse row equivalence with similarity (i.e. P-1AP), which does preserve eigenvalues and determinant. Also do not confuse this with the fact that the replacement row operation does preserve determinant (but not eigenvalues).
d) A 3x3 matrix with 3 distinct eigenvalues is diagonalizable.
ALWAYS – Each eigenvalue has at least one eigenvector, and eigenvectors from different eigenvalues are linearly independent. Thus, three distinct eigenvalues gives us three linearly independent eigenvectors, and we can construct P and diagonalize the matrix. [Essentially 1999 Exam 3, problem #4e]
e) If a vector is in the Null space of A then it is also in the Null space of BA.
ALWAYS – If the vector x is in Null(A), then Ax=0. Thus B(Ax) = B(0) = 0 and we see that (BA)x = 0 and x is in Null(BA). [2001 Exam 3, problem #1]