Linear Algebra II Lecture 14 - ualberta.caxichen/math22514f/20141119_printable.pdf · Linear Algebra II Lecture 14 Xi Chen 1 ... Solution. Let C be the ... Let V be a vector space

Characteristic Polynomial, Eigenvalue, Eigenvector and Diagonalization

Linear Algebra II Lecture 14

Xi Chen 1

1University of Alberta

November 19, 2014

Xi Chen Linear Algebra II Lecture 14


Outline

1 Characteristic Polynomial, Eigenvalue, Eigenvector andDiagonalization



Eigenvalue, Eigenvector and Diagonalization

Let T : V → V be a linear endomorphism. We call λ aneigenvalue of T and v 6= 0 an eigenvector of Tcorresponding to λ if T (v) = λv. All eigenvectors v of Tcorresponding to λ lie in the eigenspace

{v : T (v) = λv} = {v : (λI − T )v} = K (λI − T )

of T corresponding to λ.If dim V <∞, we call

det(xI − [T ]B←B) = xn + a1xn−1 + ...+ an

the characteristic polynomial of T , which is independent ofthe choice of B.T (v) = λv⇒ f (T )(v) = f (λ)v for all polynomials f (x).

(Cayley-Hamilton) f (T ) = 0 for the characteristicpolynomial f (x) of T .



Characteristic Polynomial, Eigenvalue, Eigenvectorand Diagonalization

We call T : V → V diagonalizable if there exists anordered basis B of V such that [T ]B←B is diagonal.Let V be a vector space of dimension n. Then a lineartransformation T : V → V is diagonalizable if and only if Thas n linearly independent eigenvectors v1,v2, ...,vn. Inaddition,

[T ]B←B =

λ1

λ2. . .

λn

for B = {v1,v2, ...,vn} if v1,v2, ...,vn are linearlyindependent eigenvectors of T corresponding toλ1, λ2, ..., λn.



Eigenvalues, Eigenvectors and Diagonalization

Let λ1, λ2, ..., λm be all the distinct eigenvalues ofT : V → V . Then T is diagonalizable if and only if

V = K (λ1I − T ) + K (λ2I − T ) + ...+ K (λmI − T )

or equivalently,

dim V = dim K (λ1I−T )+dim K (λ2I−T )+...+dim K (λmI−T )

or equivalently,

dim V ≤ dim K (λ1I−T )+dim K (λ2I−T )+...+dim K (λmI−T )

T is diagonalizable⇒ f (T ) is diagonalizable for allpolynomials f (x).



Examples of Eigenvalues, Eigenvectors andDiagonalization

Let T : R3 → R3 be a linear transformation with twoeigenvalues 1 and 2. If

dim K (I − T ) = 1 and dim K (2I − T ) = 1,

is T diagonalizable?No since dim K (I − T ) + dim K (2I − T ) < 3. There exists anordered basis B such that

[T ]B←B =

1 1 00 1 00 0 2

or

1 0 00 2 10 0 2

.




Let T : R3 → R3 be a linear transformation with twoeigenvalues 1 and 2. If

dim K (I − T ) = 1 and dim K (2I − T ) = 2,

is T diagonalizable?Yes since dim K (I − T ) + dim K (2I − T ) = 3. LetB = {v1,v2,v3} with {v1} a basis of K (I − T ) and {v2,v3} abasis of K (2I − T ). Then

[T ]B←B =

12

2

.




Let T : R3 → R3 be the linear transformation

T (x , y , z) = (x + y , y + z,2z)

Find the characteristic polynomial, eigenvalues andeigenvectors of T and find a basis B such that [T ]B←B isdiagonal if such B exists.Solution. Let C be the standard basis. Then

[T ]C←C =

1 1 00 1 10 0 2

.So the characteristic polynomial of T is

det(xI − [T ]C←C) = (x − 1)2(x − 2).




Solution (CONT). T has eigenvalues 1 and 2 with eigenvectors

{v : T (v) = v} = K (I − T )

=

x

yz

:

0 −1 00 0 −10 0 −1

xyz

=

000

= Span

1

00

{v : T (v) = 2v} = K (2I − T )

=

x

yz

:

1 −1 00 1 −10 0 0

xyz

=

000

= Span

1

11

T is not diagonalizable since Span{(1,0,0), (1,1,1)} 6= R3.




Let T : R3 → R3 be the linear transformation

T (x , y , z) = (x , y + z,2z)

Find the characteristic polynomial, eigenvalues andeigenvectors of T and find a basis B such that [T ]B←B isdiagonal if such B exists.Solution. Let C be the standard basis. Then

[T ]C←C =

1 0 00 1 10 0 2

.So the characteristic polynomial of T is

det(xI − [T ]C←C) = (x − 1)2(x − 2).




Solution (CONT). T has eigenvalues 1 and 2 with eigenvectors

{v : T (v) = v} = K (I − T )

=

x

yz

:

0 0 00 0 −10 0 −1

xyz

=

000

= Span

1

00

,0

10

{v : T (v) = 2v} = K (2I − T )

=

x

yz

:

1 0 00 1 −10 0 0

xyz

=

000

= Span

0

11

T is diagonalizable since

Span{(1,0,0), (0,1,0), (0,1,1)} = R3.




Solution (CONT). Let B = {(1,0,0), (0,1,0), (0,1,1)}. Then

[T ]B←B =

11

2

= P−1C←B[T ]C←CPC←B

=

1 0 00 1 10 0 1

−1 1 0 00 1 10 0 2

1 0 00 1 10 0 1

1 1 0

0 1 10 0 2

and

1 0 00 1 10 0 2

have the same characteristic polynomials and but they are notsimilar.




Let V be a vector space of finite dimension and T : V → V be alinear transformation satisfying T 2 = I. Show that T isdiagonalizable.

Proof.

Since T 2 = I, I − T 2 = (I − T )(I + T ) = 0⇒

R(I − T ) ⊂ K (I + T )⇒ rank(I − T ) ≤ dim K (I + T ).

By Rank Theorem, rank(I − T ) = dim V − dim K (I − T ).Therefore, dim K (I + T ) + dim K (I − T ) ≥ dim V

⇒ dim K (I + T ) + dim K (I − T ) = dim V

So T is diagonalizable.




Let T : M2×2(R)→ M2×2(R) be the linear transformation

T (A) =[3 14 3

]︸︷︷︸

D

A

Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Let λ be an eigenvalue of T with correspondingeigenvector A. Then

T (A) = λA⇔ DA = λA⇔ (λI − D)A = 0

⇔ (λI − D)[v1 v2

]= 0

⇔ (λI − D)v1 = (λI − D)v2 = 0⇔ v1,v2 ∈ Nul(λI − D).




Solution (CONT). It suffices to find the eigenvalues andeigenspaces of D: D has characteristic polynomial x2 − 6x + 5and eigenvalues 1 and 5 with eigenspaces

Nul(I − D) = Span{[

1−2

]}and Nul(5I − D) = Span

{[12

]}.

Correspondingly, T has eigenvalues 1 and 5 with eigenspaces

K (I − T ) = Span{[

1 0−2 0

],

[0 10 −2

]}K (5I − T ) = Span

{[1 02 0

],

[0 10 2

]}.




Solution (CONT). So T diagonalizable and

[T ]B←B =

1

15

5

where

B =

{[1 0−2 0

],

[0 10 −2

],

[1 02 0

],

[0 10 2

]}.

We also see that the characteristic polynomial of T is(x − 1)2(x − 5)2.




Let T : Pn → Pn be the linear transformation

T (f (x)) = f (x + 1)

Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Let D = {1, x , ..., xn} be the standard basis. Then

[T ]D←D︸︷︷︸A

=[(j−1

i−1

)](n+1)×(n+1)

=

1 1 ... 1

1 .... n. . .

...1

.Then the characteristic polynomial of T or equivalently A is(x − 1)n and A 6= I.




Solution (CONT). T is diagonalizable if and only if A is.Suppose that A is diagonalizable. Then there exists aninvertible matrix P such that PAP−1 is diagonal. And since thecharacteristic polynomial of A is (x − 1)n,

PAP−1 =

1

1. . .

1

= I ⇒ A = P−1IP = I.

Contradiction. So T is not diagonalizable.





T (f (x)) = f (1− x)

Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Note that T 2 = I. Such T is always diagonalizable.The eigenvectors of T are (1

2 − x)k with correspondingeigenvalue (−1)k . So

[T ]B←B =

1−1

. . .(−1)n

(n+1)×(n+1)

where B = {1, 12 − x , (1

2 − x)2, ..., (12 − x)n}.





T (f (x)) = xf ′(x) + f (1)

Is T diagonalizable? If it is, find a basis B such that [T ]B←B isdiagonal.Solution. Let D be the standard basis. Then

[T ]D←D︸︷︷︸A

=

1 1 1 ... 1

12

. . .n

.




Solution (CONT). So the characteristic polynomial of T is(x − 1)2(x − 2)...(x − n).Note that

dim K (I − T ) = dim Nul(I − A) = 1dim K (2I − T ) = dim Nul(2I − A) = 1

... =...

dim K (nI − T ) = dim Nul(nI − A) = 1.

Therefore,

dim K (I−T )+dim K (2I−T )+...+dim K (nI−T ) = n < n+1 = dim Pn

and hence T is not diagonalizable.




Let T : R2 → R2 be the linear transformation given by

T (x , y) = (3x + y ,4x + 3y).

Find T n and T 2014(1,1).Solution 1. We try to diagonalize T . Let C be the standardbasis. Then

[T ]C←C =

[3 14 3

]and det(xI − [T ]C←C) = x2 − 6x + 5.

So T has eigenvalues 1 and 5 with eigenvectors

K (I − T ) = Span{(1,−2)} and K (5I − T ) = Span{(1,2)}.Let B = {(1,−2), (1,2)}. Then

[T ]B←B =

[1

5

]Xi Chen Linear Algebra II Lecture 14



Solution 1 (CONT). So

[T ]C←C = PC←B[T ]B←BP−1C←B

=

[1 1−2 2

] [1

5

] [1 1−2 2

]−1

Therefore,

[T n]C←C =

[1 1−2 2

] [1

5

]n [ 1 1−2 2

]−1

=14

[1 1−2 2

] [1

5n

] [2 −12 1

]=

14

[2(5n + 1) (5n − 1)4(5n − 1) 2(5n + 1)

]




Solution 1 (CONT). So

T n(x , y) =14(2(5n + 1)x + (5n − 1)y ,4(5n − 1)x + 2(5n + 1)y)

andT 2014(1,1) =

14(3 · 52014 + 1,6 · 52014 − 2).

Solution 2. By Cayley-Hamilton, f (T ) = 0 for the characteristicpolynomial of T . Therefore, T 2 − 6T + 5I = 0. By long division,

T n = (T 2 − 6T + 5I)g(T ) + aT + bI ⇒ T n = aT + bI.

It suffices to find a and b.




Solution 2 (CONT). Suppose that

xn ≡ (x2 − 6x + 5)g(x) + ax + b = (x − 1)(x − 5)g(x) + ax + b.

Let x = 1. Then a + b = 1.Let x = 5. Then 5a + b = 5n.Solve the system of linear equations:

{a + b = 1

5a + b = 5n ⇒

a =

14(5n − 1)

b =14(5− 5n)

T n = aT + bI ⇒ T n(x , y) = aT (x , y) + b(x , y)= ((3a + b)x + ay ,4ax + (3a + b)y).


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Linear Algebra II Lecture 14 - ualberta.caxichen/math22514f/20141119_printable.pdf · Linear Algebra II Lecture 14 Xi Chen 1 ... Solution. Let C be the ... Let V be a vector space