linear algebra - math.fau.edumath.fau.edu/yiu/linearalgebra2011/2011linearalgebra5a-d.pdf · linear...

Linear Algebra

Paul Yiu

Department of MathematicsFlorida Atlantic University

Fall 2011

Linear Algebra

Paul Yiu


Fall 2011

5A: Permutations

Permutation of n objects

A permutation ofn objects1, 2, . . . ,n isa bijectionθ of the set[n] := {1, 2, . . . , n},and is displayed in two rows listing the set[n] in one row and the images ofits elements in a second row.For example, the permutationθ : [6] // [6] represented by

(

1 2 3 4 5 6

3 1 5 6 2 4

)

has

θ(1) = 3, θ(2) = 1, θ(3) = 5, θ(4) = 6, θ(5) = 2, θ(6) = 4.

1 2 3 4 5 6

1 2 3 4 5 6

Yiu: Linear Algebra 2011

1

Cycle decomposition and transpositions

A transposition of [n] is a permutation which interchanges two elements,and leaves the remaining elements fixed.

Theorem. Every permutation is a composite oftranspositions,and theparity of the number of transposition is invariant.

Therefore, permutations of[n] are classified intoeven permutations andodd permutations,depending on the parity of the numbers of transpositionsthat make up a permutation.


2

Proof of Theorem: decomposition into transpositions

(1) Every permutation decomposes into a number ofdisjoint cycles.Given a permutationθ of [n], define a relationR on [n] as follows:

i R j if and only if θp(i) = j for somep > 0.

This means thatp applications ofθ bringsi into j.

(i) R is an equivalence relation.Reflexivity and transitivity are obvious.For symmetry, note that there is a sufficient large integerN

such thatθN = identity permutation.(For example,N can be taken to ben!).Therefore, ifθp(i) = j, thenθN−p(j) = i, showing thatR is symmetric.

(ii) The equivalences classes ofR are thedisjoint cycles of θ. For exam-ple, the permutation

(

1 2 3 4 5 6

3 1 5 6 2 4

)

consists of two disjoint cycles:(1 3 5 2)(4 6).


3

(iii) A 2-cycle is a transposition.More generally, each cycle of lengthk is a composite ofk−1 transpositions.

(i1 i2 . . . ik) = (i1 ik)(i1 ik−1) · · · (i1 i2).

(iv) If the permutation decomposes intor disjoint cycles, of lengthsk1, . . . ,kr, then it is a composite of

(k1 − 1) + · · · + (kr − 1) = (k1 + · · · + kr) − r = n − r

transpositions.For example, the permutation(1 3 5 2)(4 6)

is a composite of6 − 2 = 4 transpositions.It is an even permutation.On the other hand, the permutation

(

1 2 3 4 5 6

3 1 6 5 2 4

)

consists of one single cycle(1 3 6 4 5 2).It is a composite of5 transpositions, and is an odd permutation.


4

Proof of Theorem: Invariance of parity

(2) Although the number of transpositions expressing a permutation maynot be unique, theparity is. This is justified by consideringthe action of permutations of[n] on the polynomial

P :=∏

1≤i<j≤n

(xi − xj),

as follows:θ(P ) =

∏

1≤i<j≤n

(xθ(i) − xθ(j)).

This action onP permutes the factors, with possible reversals of signs.We writeθ(P ) = sgn(P )P for sgn(P ) = ±1.

(ii) Two key facts:(a) sgn(θ2θ1) = sgn(θ2)sgn(θ1),(b) if θ is a transposition, thenθ(P ) = −P . Equivalently, sgn(θ) = −1.Granting these, it is clear that a permutation cannot be an even permutationand an odd permutation at the same time,since an even permutation preservesP , but an odd permutation changesP

into −P .Yiu: Linear Algebra 2011

5

A permutation as a composite of transpositions

Here is another simple way to calculate sgn(θ),and to writeθ as a composite of transpositions.For i = 1, 2, . . . , n − 1, let ti be the number of entriesgreater than and precedingi in the permutationθ.Then

sgn(θ) = (−1)t1+t2+···+tn−1.

In fact, by interchanging, in succession, eachi with each number greaterthan and preceding it, we obtain a composite of transpositions which bringsθ into the identity permutation.Reversing these transpositions, we obtainθ as a composite of transpositions.

For example, with(

1 2 3 4 5 6

3 1 5 6 2 4

)

we havet1 = 1, t2 = 3, t3 = 0, t4 = 2, t5 = 0, t6 = 0.Therefore, sgn(θ) = (−1)1+3+0+2+0 = 1.The composite(45)(46)(23)(25)(26)(13) bringsθ to the identity.Therefore,θ = (13)(26)(25)(23)(46)(45).


6

Proof of (a): sgn(θ2θ1) = sgn(θ2)sgn(θ1)

Consider∏

1≤i<j≤n

xθ2θ1(i) − xθ2θ1(j)

xi − xj

=∏

1≤i<j≤n

xθ2θ1(i) − xθ2θ1(j)

xθ1(i) − xθ1(j)

∏

1≤i<j≤n

xθ1(i) − xθ1(j)

xi − xj

.

The left hand side and the second product on the right hand side are clearlysgn(θ2θ1) and sgn(θ1) respectively.The first product on the right hand side is sgn(θ2),since each of its factors has in its denominatorε times a factor ofP (ε = ±1),and in its numerator the action ofθ2 onε times the same factor ofP .


7

Proof of (b)

Supposeθ interchangesh andk, and leaves other elements fixed.We may assumeh < k. Among the

(

n2

)

factors ofP only those involvingindicesh andk are affected.• For i = 1, . . . , h − 1, xi − xh becomesxi − xk.• For i = 1, . . . , h − 1, xi − xk becomesxi − xh.• For j = h+1, . . . , k− 1, xh −xj becomesxk −xj = −(xj −xk). Therearek − h − 1 changes in sign.• For (i, j) = (h, k), xh − xk becomesxk − xh = −(xh − xk), one changein sign.• For j = k + 1, . . . , n, xh − xj becomesxk − xj.• For i = h + 1, . . . , k − 1, xi − xk becomesxi − xh = −(xh − xi). Therearek − h − 1 changes in sign.• For j = k + 1, . . . , n, xk − xj becomesxh − xj.Altogether, there are2(k − h − 1) + 1, an odd number of changes in sign.The action of a transposition changesP into −P .


8

Proof of (b)

Supposeθ interchangesh andk, and leaves other elements fixed.We may assumeh < k. Among the

(

n

2

)

factors ofP only those involvingindicesh andk are affected.The factors corresponding to the2(k−h)−1 reddots in the diagram belowbecome those corresponding to thebluedots,while those corresponding to the black dots permute among themselves.The transposition changesP into −P .

h

k

h k1 n

n


9

Linear Algebra

Paul Yiu


Fall 2011

5B: Determinants

Determinant

Let V be ann-dimensional vector space overF ,with basisB = {u1, . . . , un}.To determine if a given set ofn vectorsx1, . . . ,xn ∈ V

is linearly dependent (or independent),it is enough to compute a single scalar inF ,thedeterminant of then×n matrix formed by the given vectors as columns.


1

Alternating n-forms on an n-dimensional vector space

Suppose there is a scalar valued functionD on V n = V × · · · × V // F

satisfying the condition

D(x1, . . . , xn) = 0 if and only if x1, . . . , xn are linearly dependent.

We further assume thatD is linear in each of then variables.We also say thatD is multilinear (n-linear in the present case).

(1) Note that if any two ofx1, . . . ,xn are equal, thenD(x1, . . . , xn) = 0.It follows thatD is skew-symmetric in any two variables.For example, consider the first two variables.

0 = D(x + y, x + y, . . . )

= D(x, x, . . . ) + D(y, x, . . . ) + D(x, y, . . . ) + D(y, x, . . . )

= D(x, y, . . . ) + D(y, x, . . . ).

This givesD(x, y, . . . ) = −D(y, x, . . . ).


2

Consider the givenn vectors as defininga linear transformationf : V // V by

f(uj) = a1,ju1 + a2,ju2 + · · · + an,jun, j = 1, 2, . . . , n.

We compute

D(f(u1), . . . , f(un)) = D

(

n∑

i=1

ai,1ui, . . . ,

n∑

i=1

ai,nui

)

.

By multilinearity, this is a sum ofnn terms,each a scalar multiple ofD(ui1, ui2, . . . , uin),with i1, . . . , in ranging from1 to n.

Note thatD(ui1, ui2, . . . , uin) = 0when any two ofi1, . . . , in are the same.This meansD(ui1, ui2, . . . , uin) is nonzero only ifi1, . . . , in is apermutation of 1, . . . , n.


3

Writing σ for a permutation of1, . . . ,n, we now have a “simpler” expression

D(f(u1), . . . , f(un)) =∑

σ∈Sn

aσ(1),1 · · · aσ(n),nD(uσ(1), . . . , uσ(n))

consisting ofn! terms.Here,Sn denotes the symmetric group of permutations of1, . . . ,n.

Now for a permutationσ,the numberD(uσ(1), . . . , uσ(n)) iseitherD(u1, . . . , un) or −D(u1, . . . , un),depending on whether the permutation

σ(1) σ(2) . . . σ(n)

can be brought into the identity permutation

1 2 . . . n

by an even number or an odd number oftranspositions.This is exactly the notion of sgn(σ).


4

Therefore,

D(uσ(1), . . . , uσ(n)) = sgn(σ)D(u1, . . . , un),

where

sgn(σ) =

{

+1, if σ is an even permutation,

−1, if σ is an odd permutation.

Now, the formula forD(f(u1), . . . , f(un)) becomes

D(f(u1), . . . , f(un)) =

(

∑

σ∈Sn

sgn(σ)aσ(1),1 · · · aσ(n),n

)

D(u1, . . . , un).

We define the scalar∑

σ∈Sn

sgn(σ)aσ(1),1 · · · aσ(n),n

to be thedeterminant of the matrixMB(f) = (ai,j).

det(ai,j) :=∑

σ∈Sn

sgn(σ)aσ(1),1 · · · aσ(n),n


5

We have actually shown thatthe putative multilinear functionD depends only on one value,namely,D(u1, . . . , un) for the (ordered) basisB = {u1, . . . , un}.

For every linear transformationf with matrixMB(f)relative to the basisu1, . . . , un,

D(f(u1), . . . , f(un)) = (detMB(f))D(u1, . . . , un).

This formula has a number of interesting corollaries.


6

det(M1M2) = det(M1) det(M2)

Theorem. The determinant function is multiplicative.

Proof. Define linear transformationsf, g : V // V by M1 = MB(g) andM2 = MB(f) relative to a basisB = {u1, . . . , un} of V .

(i) AssumeM1 andM2 nonsingular, so thatg andf are isomorphisms.

D(g ◦ f(u1), . . . , g ◦ f(un)) = det(MB(g ◦ f))D(u1, . . . , un).

On the other hand,

D(g ◦ f(u1), . . . , g ◦ f(un))

= D(g(f(u1)), . . . , g(f(un))

= det(MB(g))D(f(u1), . . . , f(un))

= det(MB(g)) det(MB(f))D(u1, . . . , un).

Therefore,det(MB(g ◦ f)) = det(MB(g)) det(MB(f)), anddet(M1M2) = det(M1) det(M2).(ii) If one of M1, M2 is singular, then one or both ofg, f ; so isg ◦ f . Thismeansdet(M1M2) = 0 = (detM1)(detM2).


7

Determinant of an endomorphism

Let V be a finite dimensional vector space overF ,andf : V // V a linear transformation.

Proposition. det(MB(f)) is independent of the choice of the basisB.

Proof. For given basesB andB′ of X,MB′(f) = MBB′MB(f)M−1

BB′.From this,

det MB′(f) = det(MBB′MB(f)M−1BB′) = det MB(f).

Therefore, we can speak of the determinant of an endomorphismf : V //V

without reference to a basis:

det(f) := det(MB(f)) for any basisB of V .


8

Determinant of an endomorphism

Let V be a finite dimensional vector space overF ,andf : V // V a linear transformation.

Proposition. det(MB(f)) is independent of the choice of the basisB.

Proof. For given basesB andB′ of V ,MB′(f) = MBB′MB(f)M−1

BB′.From this,

det MB′(f) = det(MBB′MB(f)M−1BB′) = det MB(f).

Therefore, we can speak of the determinant of an endomorphismf : V //V

without reference to a basis:

det(f) := det(MB(f)) for any basisB of V .

Proposition. f : V // V is an isomorphism if and onlydet(f) 6= 0.


9

det M t = det M

Proposition. det M t = det M .

Proof. ForM = (ai,j),

det M :=∑

σ∈Sn

sgn(σ)aσ(1),1 · · · aσ(n),n

Now, for the transposeM t = (aj,i), we have

detM t =∑

σ∈Sn

sgn(σ)a1,σ(1) · · · an,σ(n)

=∑

σ−1∈Sn

sgn(σ−1)aσ−1(1),1 · · · aσ−1(n),n

=∑

σ∈Sn

sgn(σ)aσ(1),1 · · · aσ(n),n

= det M.


10

Linear Algebra

Paul Yiu


Fall 2011

5C: Orientations

Oriented basis

Let F be a field of characteristic not equal2 (so that1 6= −1), andV a finitedimensional vector space overF .Consider theordered bases ofV .We say that two ordered basesB andB′ have thesame orientation if andonly if det(MBB′ > 0.


1

Linear Algebra

Paul Yiu


Fall 2011

5D: Eigenvalues and eigenvectors

Let V be a vector space overF , with dimF V = n,andf : V // V a linear operator.We address the question of finding a basisB relative to whichthe matrix off is simplest possible.


1

Let V be a vector space overF , with dimF V = n,andf : V // V a linear operator.We address the question of finding a basisB relative to whichthe matrix off is simplest possible.

The best scenario is the decomposition ofV intoa direct ofn invariant1-planesV1, . . . ,Vn,each with a single nonzero vectorv1, . . . ,vn:

V = Span(v1) ⊕ · · · ⊕ Span(vn),

with f(vi) = λivi, i = 1, 2 . . . n.Then, relative to the basisB = {v1, . . . , vn},MB(f) is the diagonal matrix

λ1.. .

λn

,

in which the off-diagonal entries are all zeros.In this case, we say thatf is diagonalizable.


2

Eigenvalues and eigenvectors

Let f : V // V be a linear operator onV .A nonzero vectorv ∈ V is aneigenvector of f if

f(v) = λv for someλ ∈ F.

λ is the correspondingeigenvalue of f .


3

Eigenvalues and eigenvectors

Let f : V // V be a linear operator onV .A nonzero vectorv ∈ V is aneigenvector of f if

f(v) = λv for someλ ∈ F.

λ is the correspondingeigenvalue of f .

Such an eigenvector exists as a vector in the kernel of the operatorf − λι,which should be singular.This means that the eigenvalueλ satisfies

det(f − xι) = 0,

or with reference to any basisB of the vector spaceV ,

det(MB(f) − xIn) = 0.


4

Characteristic polynomial

For a linear operatorf on ann-dimensional spaceV overF ,the eigenvalues are the roots of

det(f − xι) = det(MB(f) − xIn),

which is a polynomial inF [x] of degreen.This is called thecharacteristic polynomial of the operatorf .


5





(1) The number of eigenvalues off is not more thann.

(2) If F is algebraically closed, for exampleF = C,there are exactlyn eigenvalues, counting multiplicity.


6





(3) The leading coefficient (ofxn) is (−1)n.

(4) The coefficient ofxn−1 is the sum of the diagonal entries ofM , calledthetrace of the matrix.

(5) The constant term isdet f .

(6) The sum of the eigenvalues is equal to trace(M).

(7) The product of the eigenvalues is equal todet M .In (6) and (7) we interpret the eigenvalues in a suitable extension of the fieldF .


7

Examples

(1) Letf : V // V be the identity operator.Its characteristic polynomial is

det(In − xIn) = det((1 − x)In) = (1 − x)n det In = (1 − x)n.

Therefore, the only eigenvalue isλ = 1 (with multiplicity n).

Every nonzero vectorv ∈ V is an eigenvector (with eigenvalue1):

ι(v) = 1 · v.


8

Examples

(1) Letf : V // V be the identity operator.Its characteristic polynomial is

det(In − xIn) = det((1 − x)In) = (1 − x)n det In = (1 − x)n.

Therefore, the only eigenvalue isλ = 1 (with multiplicity n).

Every nonzero vectorv ∈ V is an eigenvector (with eigenvalue1):

ι(v) = 1 · v.

(2) If M = (aij) is an (upper) triangular matrix,i.e, aij = 0 if i < j,the characteristic polynomial is

(a11 − x)(a22 − x) · · · (ann − x).

Therefore, the eigenvalues are precisely the diagonal entries, counting mul-tiplicity.


9

Example 2

Let the2 × 2 matrix

M =

(

cos θ − sin θ

sin θ cos θ

)

.

The characteristic polynomial is∣

∣

∣

∣

cos θ − x − sin θ

sin θ cos θ − x

∣

∣

∣

∣

= (cos θ − x)2 + sin2 θ = 1 − (2 cos θ)x + x2.

This has real roots if and only if(2 cos θ)2 − 4 ≥ 0.The only possibilities areθ = kπ andM = (−1)kI2.In this case, the only eigenvalue is(−1)k with multiplicity 2.

If θ 6= kπ for an integerk, thenM does not have real eigenvalues.

However, if we regardM as a matrix overC, then the characteristic poly-nomial has rootscos θ ± i sin θ.


10

Example 2 (continued)

Let the2 × 2 matrix

M =

(

cos θ − sin θ

sin θ cos θ

)

.

If we regardM as a matrix overC, then the characteristic polynomial hasrootscos θ ± i sin θ.We may assumesin θ 6= 0.

Forλ = cos θ + i sin θ, the eigenvectors are the solutions of(

−i sin θ − sin θ

sin θ −i sin θ

)

v = 0,

which are clearly multiples of

(

i

1

)

.

Similarly, the eigenvectors corresponding toλ = cos θ−i sin θ are multiples

of

(

−i

1

)

.


11

Example 3

Consider the spaceℓ2 and the linear operatorL which shifts the entries of(xn) one place to the left:

L(x0, x1, x2, . . . ) = (x1, x2, x3, . . . ).

If (xn) is an eigenvector ofL, with corresponding eigenvalueλ, then

(x1, x2, x3, . . . ) = λ(x0, x1, x2, . . . ).

This means

x1 = λx0, x2 = λx1, . . . , xn = λxn−1, . . .

The sequence is a geometric progression withx0 = a 6= 0 and commonratioλ: xn = aλn for n = 0, 1, 2, . . . .Since the series

∑

n x2n is convergent,|λ| < 1.

Everyλ ∈ (−1, 1) is an eigenvalue of the left shift operatorL.


12

Let f : V // V be a linear operator,andv ∈ V an eigenvector with corresponding eigenvalueλ.

f(v) = λv

=⇒ f 2(v) = f(f(v)) = f(λv) = λf(v) = λ(λv) = λ2v

=⇒ f 3(v) = λ3v

...

=⇒ f k(v) = λkv

for every (nonnegative) integerk.

Therefore, for every polynomialP (x) ∈ F [x],the operatorP (f) hasv as an eigenvector,with corresponding eigenvalueP (λ).


13

Theorem. Eigenvectors corresponding to distinct eigenvalues are linearlyindependent.

Proof. Let f : V // V be an operator with distinct eigenvaluesλ1, . . . ,λk,and corresponding eigenectorsv1, . . . ,vk.

Suppose, for a contradiction, that the eigenvectors are linearly dependent,and by relabeling if necessary, that

h∑

i=1

µivi = 0,

with h ≤ k, is the “shortest” linear dependence between them.Now,

0 = f

(

h∑

i=1

µivi

)

=

h∑

i=1

µif(vi) =

h∑

i=1

µiλivi,

and we have

0 =

h∑

i=1

µiλivi − λh

(

h∑

i=1

µivi

)

=

h−1∑

i=1

µi(λi − λh)vi,

a shorter linear dependence between the eigenvector.Yiu: Linear Algebra 2011

14

linear algebra - math.fau.edumath.fau.edu/yiu/linearalgebra2011/2011linearalgebra5a-d.pdf · linear...

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