Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Linear Algebra II Lecture 12

Xi Chen 1

1University of Alberta

March 17, 2017

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Outline

1 Characteristic Polynomials, Eigenvalues and Eigenvectorsof Linear Endomorphisms

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Linear Endomorphism

We call a linear transformation T : V → V from a vector spaceV to itself a linear endomorphism or simply endomorphism.Given two ordered bases B and B′ of V , [T ]B←B and [T ]B′←B′

satisfy

[T ]B′←B′ = PB′←B[T ]B←BP−1B′←B = P[T ]B←BP−1.

Therefore, [T ]B←B and [T ]B′←B′ are similar to each other.For example, let T : R2 → R2 be given byT (x , y) = (x + y , x − y), B = {e1,e2} and B′ = {(1,2), (2,3)}.Then [

−11 −177 11

]︸ ︷︷ ︸

[T ]B′←B′

=

[−3 22 −1

]︸ ︷︷ ︸

P−1B←B′

[1 11 −1

]︸ ︷︷ ︸[T ]B←B

[1 22 3

]︸ ︷︷ ︸

PB←B′

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Linear Endomorphism

We call a linear transformation T : V → V from a vector spaceV to itself a linear endomorphism or simply endomorphism.Given two ordered bases B and B′ of V , [T ]B←B and [T ]B′←B′

satisfy

[T ]B′←B′ = PB′←B[T ]B←BP−1B′←B = P[T ]B←BP−1.

Therefore, [T ]B←B and [T ]B′←B′ are similar to each other.For example, let T : R2 → R2 be given byT (x , y) = (x + y , x − y), B = {e1,e2} and B′ = {(1,2), (2,3)}.Then [

−11 −177 11

]︸ ︷︷ ︸

[T ]B′←B′

=

[−3 22 −1

]︸ ︷︷ ︸

P−1B←B′

[1 11 −1

]︸ ︷︷ ︸[T ]B←B

[1 22 3

]︸ ︷︷ ︸

PB←B′

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Linear Endomorphism

We call a linear transformation T : V → V from a vector spaceV to itself a linear endomorphism or simply endomorphism.Given two ordered bases B and B′ of V , [T ]B←B and [T ]B′←B′

satisfy

[T ]B′←B′ = PB′←B[T ]B←BP−1B′←B = P[T ]B←BP−1.

Therefore, [T ]B←B and [T ]B′←B′ are similar to each other.For example, let T : R2 → R2 be given byT (x , y) = (x + y , x − y), B = {e1,e2} and B′ = {(1,2), (2,3)}.Then [

−11 −177 11

]︸ ︷︷ ︸

[T ]B′←B′

=

[−3 22 −1

]︸ ︷︷ ︸

P−1B←B′

[1 11 −1

]︸ ︷︷ ︸[T ]B←B

[1 22 3

]︸ ︷︷ ︸

PB←B′

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Linear Endomorphism

We call a linear transformation T : V → V from a vector spaceV to itself a linear endomorphism or simply endomorphism.Given two ordered bases B and B′ of V , [T ]B←B and [T ]B′←B′

satisfy

[T ]B′←B′ = PB′←B[T ]B←BP−1B′←B = P[T ]B←BP−1.

Therefore, [T ]B←B and [T ]B′←B′ are similar to each other.For example, let T : R2 → R2 be given byT (x , y) = (x + y , x − y), B = {e1,e2} and B′ = {(1,2), (2,3)}.Then [

−11 −177 11

]︸ ︷︷ ︸

[T ]B′←B′

=

[−3 22 −1

]︸ ︷︷ ︸

P−1B←B′

[1 11 −1

]︸ ︷︷ ︸[T ]B←B

[1 22 3

]︸ ︷︷ ︸

PB←B′

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Characteristic Polynomial, Eigenvalue and Eigenvector

Let T : V → V be a linear endomorphism. We call λ aneigenvalue of T and v 6= 0 an eigenvector of T correspondingto λ if T (v) = λv. The space K (λI − T ) spanned by theeigenvectors of T corresponding to λ is the eigenspace of Tassociated to λ.

If dim V = n <∞, the eigenvalues and eigenvectors of T arethe same as those of [T ]B←B for all ordered bases B of V .

λ is an eigenvalue of T ⇔ K (λI − T ) 6= {0} ⇔ λI − T is notbijective⇔ λ is a zero of the characteristic polynomial of T :

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

which is independent of the choice of B. Note thatan = det(−[T ]B←B) = (−1)n det([T ]B←B) anda1 = −Tr([T ]B←B) is the trace of [T ]B←B.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Characteristic Polynomial, Eigenvalue and Eigenvector

Let T : V → V be a linear endomorphism. We call λ aneigenvalue of T and v 6= 0 an eigenvector of T correspondingto λ if T (v) = λv. The space K (λI − T ) spanned by theeigenvectors of T corresponding to λ is the eigenspace of Tassociated to λ.

If dim V = n <∞, the eigenvalues and eigenvectors of T arethe same as those of [T ]B←B for all ordered bases B of V .

λ is an eigenvalue of T ⇔ K (λI − T ) 6= {0} ⇔ λI − T is notbijective⇔ λ is a zero of the characteristic polynomial of T :

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

which is independent of the choice of B. Note thatan = det(−[T ]B←B) = (−1)n det([T ]B←B) anda1 = −Tr([T ]B←B) is the trace of [T ]B←B.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Characteristic Polynomial, Eigenvalue and Eigenvector

Let T : V → V be a linear endomorphism. We call λ aneigenvalue of T and v 6= 0 an eigenvector of T correspondingto λ if T (v) = λv. The space K (λI − T ) spanned by theeigenvectors of T corresponding to λ is the eigenspace of Tassociated to λ.

If dim V = n <∞, the eigenvalues and eigenvectors of T arethe same as those of [T ]B←B for all ordered bases B of V .

λ is an eigenvalue of T ⇔ K (λI − T ) 6= {0} ⇔ λI − T is notbijective⇔ λ is a zero of the characteristic polynomial of T :

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

which is independent of the choice of B. Note thatan = det(−[T ]B←B) = (−1)n det([T ]B←B) anda1 = −Tr([T ]B←B) is the trace of [T ]B←B.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Characteristic Polynomial, Eigenvalue and Eigenvector

Let T : V → V be a linear endomorphism. We call λ aneigenvalue of T and v 6= 0 an eigenvector of T correspondingto λ if T (v) = λv. The space K (λI − T ) spanned by theeigenvectors of T corresponding to λ is the eigenspace of Tassociated to λ.

If dim V = n <∞, the eigenvalues and eigenvectors of T arethe same as those of [T ]B←B for all ordered bases B of V .

λ is an eigenvalue of T ⇔ K (λI − T ) 6= {0} ⇔ λI − T is notbijective⇔ λ is a zero of the characteristic polynomial of T :

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

which is independent of the choice of B. Note thatan = det(−[T ]B←B) = (−1)n det([T ]B←B) anda1 = −Tr([T ]B←B) is the trace of [T ]B←B.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Dictionary Between Matrices and LinearTransformations

A ∈ Mn×n(R) T : V → V∼ A ∼ A′ if A′ = P−1AP [T ]B′←B′ = P−1

B←B′ [T ]B←BPB←B′

C.P. det(xI − A) det([xI − T ]B←B)

E.V. det(λI − A) = 0 det([λI − T ]B←B) = 0E.V.T Av = λv T (v) = λvE.S. Nul(λI − A) K (λI − T )

D.G. P−1AP diagonal [T ]B←B diagonal

C.P. = Characteristic PolynomialE.V. = Eigenvalue

E.V.T. = EigenvectorE.S. = EigenspaceD.G. = Diagonalizable

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Criterion for Diagonalization

TheoremA linear endomorphism T : V → V be on a vector space V ofdimension n is diagonalizable if and only if one of the followingholds:

T has n linearly independent eigenvectors.The sum of the dimensions of the eigenspaces of T is n:∑

λ

dim K (T − λI) = n.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Algorithm for Diagonalization

If T has n linearly independent eigenvectors v1,v2, ...,vn, then

[T ]B←B =

λ1

λ2. . .

λn

for the ordered basis B = {v1,v2, ...,vn}, where λi is theeigenvalue associated to vi , i.e.,

T (vi) = λivi

for i = 1,2, ...,n.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : R2 → R2 given by T (x , y) = (x + 3y , x − y).Is T diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.

Solution. Let B be the standard basis. Then [T ]B←B =

[1 31 −1

]and the characteristic polynomial of T isdet(xI − [T ]B←B) = x2 − 4. So the eigenvalues of T are 2 and−2 and the eigenspaces are

K (2I − T ) = Nul[

1 −3−1 3

]= Span

{[31

]}K (−2I − T ) = Nul

[−3 −3−1 −1

]= Span

{[1−1

]}

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : R2 → R2 given by T (x , y) = (x + 3y , x − y).Is T diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.

Solution. Let B be the standard basis. Then [T ]B←B =

[1 31 −1

]and the characteristic polynomial of T isdet(xI − [T ]B←B) = x2 − 4. So the eigenvalues of T are 2 and−2 and the eigenspaces are

K (2I − T ) = Nul[

1 −3−1 3

]= Span

{[31

]}K (−2I − T ) = Nul

[−3 −3−1 −1

]= Span

{[1−1

]}

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : R2 → R2 given by T (x , y) = (x + 3y , x − y).Is T diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.

Solution. Let B be the standard basis. Then [T ]B←B =

[1 31 −1

]and the characteristic polynomial of T isdet(xI − [T ]B←B) = x2 − 4. So the eigenvalues of T are 2 and−2 and the eigenspaces are

K (2I − T ) = Nul[

1 −3−1 3

]= Span

{[31

]}K (−2I − T ) = Nul

[−3 −3−1 −1

]= Span

{[1−1

]}

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : R2 → R2 given by T (x , y) = (x + 3y , x − y).Is T diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.

Solution. Let B be the standard basis. Then [T ]B←B =

[1 31 −1

]and the characteristic polynomial of T isdet(xI − [T ]B←B) = x2 − 4. So the eigenvalues of T are 2 and−2 and the eigenspaces are

K (2I − T ) = Nul[

1 −3−1 3

]= Span

{[31

]}K (−2I − T ) = Nul

[−3 −3−1 −1

]= Span

{[1−1

]}

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Since

dim K (2I − T ) + dim K (−2I − T ) = 2 = dimR2,

T is diagonalizable and

[T ]D←D =

[2−2

]

for D =

{[31

],

[1−1

]}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : Pn → Pn given by T (f (x)) = f (x) + f (1). IsT diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.Solution. Let B = {1, x , ..., xn} be the standard basis. Then

[T ]B←B =

2 1 1 ... 1

11

. . .1

(n+1)×(n+1)

and the characteristic polynomial of T is (x − 1)n(x − 2). So theeigenvalues of T are 1 and 2 with eigenspaces

K (I − T ) = {f (x) : f (1) = 0} = Span{x − 1, x2 − 1, ..., xn − 1}K (2I − T ) = {f (x) : f (x) = f (1)} = Span{1}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : Pn → Pn given by T (f (x)) = f (x) + f (1). IsT diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.Solution. Let B = {1, x , ..., xn} be the standard basis. Then

[T ]B←B =

2 1 1 ... 1

11

. . .1

(n+1)×(n+1)

and the characteristic polynomial of T is (x − 1)n(x − 2). So theeigenvalues of T are 1 and 2 with eigenspaces

K (I − T ) = {f (x) : f (1) = 0} = Span{x − 1, x2 − 1, ..., xn − 1}K (2I − T ) = {f (x) : f (x) = f (1)} = Span{1}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : Pn → Pn given by T (f (x)) = f (x) + f (1). IsT diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.Solution. Let B = {1, x , ..., xn} be the standard basis. Then

[T ]B←B =

2 1 1 ... 1

11

. . .1

(n+1)×(n+1)

and the characteristic polynomial of T is (x − 1)n(x − 2). So theeigenvalues of T are 1 and 2 with eigenspaces

K (I − T ) = {f (x) : f (1) = 0} = Span{x − 1, x2 − 1, ..., xn − 1}K (2I − T ) = {f (x) : f (x) = f (1)} = Span{1}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : Pn → Pn given by T (f (x)) = f (x) + f (1). IsT diagonalizable? If it is, find a basis D such that [T ]D←D isdiagonal.Solution. Let B = {1, x , ..., xn} be the standard basis. Then

[T ]B←B =

2 1 1 ... 1

11

. . .1

(n+1)×(n+1)

and the characteristic polynomial of T is (x − 1)n(x − 2). So theeigenvalues of T are 1 and 2 with eigenspaces

K (I − T ) = {f (x) : f (1) = 0} = Span{x − 1, x2 − 1, ..., xn − 1}K (2I − T ) = {f (x) : f (x) = f (1)} = Span{1}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Since

dim K (I − T ) + dim K (2I − T ) = n + 1 = dim Pn,

T is diagonalizable and

[T ]D←D =

2

11

. . .1

(n+1)×(n+1)

for D = {1, x − 1, x2 − 1, ..., xn − 1}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : M2×2(R)→ M2×2(R) given by

T (A) = A[1 31 −1

].

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

[T ]B←B =

1 13 −1

1 13 −1

and the characteristic polynomial of T is (x − 2)2(x + 2)2.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : M2×2(R)→ M2×2(R) given by

T (A) = A[1 31 −1

].

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

[T ]B←B =

1 13 −1

1 13 −1

and the characteristic polynomial of T is (x − 2)2(x + 2)2.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

Find the characteristic polynomial, eigenvalues andeigenvectors of T : M2×2(R)→ M2×2(R) given by

T (A) = A[1 31 −1

].

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

[T ]B←B =

1 13 −1

1 13 −1

and the characteristic polynomial of T is (x − 2)2(x + 2)2.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Example

So the eigenvalues of T are 2 and −2 and the eigenspaces are

K (2I − T ) =

{A : A

[1 −3−1 3

]=

[0 00 0

]}= Span

{[1 10 0

],

[0 01 1

]}K (−2I − T ) =

{A : A

[−3 −3−1 −1

]=

[0 00 0

]}= Span

{[1 −30 0

],

[0 01 −3

]}.

Xi Chen Linear Algebra II Lecture 12

Characteristic Polynomials, Eigenvalues and Eigenvectors of Linear Endomorphisms

Since

dim K (2I − T ) + dim K (−2I − T ) = dim M2×2(R) = 4

T is diagonalizable and

[T ]D←D =

2

2−2

−2

for D =

{[1 10 0

],

[0 01 1

],

[1 −30 0

],

[0 01 −3

]}.

Xi Chen Linear Algebra II Lecture 12