Linear Algebra ITutorial 18th Tune 2020

Afd is notdiagonalitable

Linear Algebra II • Eigenvalues and eigenvectors

Definition 4.15. For � 2 R with � 6= 0 and 1 i, j n we define the elementary matricesR�,j

i , R�i , Ri,j 2 Rn⇥n

by

R�,ji =

0

BBBBBB@

1

...1

...� 1

...1

1

CCCCCCA, R�

i =

0

BBBB@

1

...1�

1

...1

1

CCCCA, Ri,j =

0

BBBBBB@

1

...0 1

...1 0

...1

1

CCCCCCA.

Here the � in R�,ji is in the i-th row and j-th column, in R�

i it is in the i-th row, and in Ri,j the 0 are

on the diagonal in the i-th row and j-th column.

Proposition 4.16. Let A 2 Rn⇥n.

i) R�,ji A is the matrix obtained from A by row operation (R1). (Add �-times the j-th row to the i-th row)

ii) R�i A is the matrix obtained from A by row operation (R2). (Multiply the i-th row with �)

iii) Ri,jA is the matrix obtained from A by row operation (R3). (Swap the j-th row with the i-th row)

Corollary 4.17. Let A 2 Rn⇥n.

i) The matrix A is invertible if and only if it is a product of elementary matrices.

ii) If C is an elementary matrix then det(CA) = det(C) det(A).

5 Eigenvalues and eigenvectors

In this section V always denotes a vector space.

Definition 5.1. Let F : V ! V be a linear map.

i) A � 2 R is called an eigenvalue of F , if there exist a vector v 2 V with v 6= 0, such that

F (v) = �v . (5.1)

ii) A vector v 2 V with v 6= 0, satisfying (5.1), is called an eigenvector of F with eigenvalue �.

Notice that v = 0 always satisfies (5.1) for any � 2 R, since F is a linear map. This is one of manyreasons why v = 0 is not called an eigenvector of F .

In the following we always assume that V is a finitely generated vector space.

Definition 5.2. Let F : V ! V be a linear map let idV : V ! V be the identity map on V .

i) The polynomial fF (�) = det(F � � idV ) is called the characteristic polynomial of F .

ii) Let � 2 R be an eigenvalue of F . Then the space

E�(F ) = ker(F � � idV )

= {v 2 V | F (v) = �v}

is called the eigenspace of F with respect to the eigenvalue �.

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Linear Algebra II • Eigenvalues and eigenvectors

The eigenspace E�(F ) contains therefore all eigenvectors of F with eigenvalue � and the zero vector.

Definition 5.3. i) Let dimV = n. A linear map F : V ! V is called diagonalizable if there exist

a basis B of V , such that

[F ]B =

0

BB@

d1 0. . .

0 dn

1

CCA

for some d1, . . . , dn 2 R.

ii) A matrix A 2 Rn⇥nis called diagonalizable if there exists an invertible matrix S 2 Rn⇥n

with

S�1AS =

0

BB@

d1 0. . .

0 dn

1

CCA

for some d1, . . . , dn 2 R.

Lemma 5.4. Let B be a basis of V and let F : V ! V be a linear map. Then the following two

statements are equivalent

i) The linear map F is diagonalizable.

ii) The matrix [F ]B is diagonalizable.

Lemma 5.5. Let F : V ! V be a linear map and B = (b1, . . . , bn) be a basis of V , such that all bi areeigenvectors of F , i.e. F (bi) = dibi for some di 2 R and i = 1, . . . , n. Then F is diagonalizable and

[F ]B =

0

BB@

d1 0. . .

0 dn

1

CCA

Conversely, if F is diagonalizable then there exists a basis of eigenvectors.

Theorem 5.6. Let v1, . . . , vm 2 V be eigenvectors of a linear map F : V ! V with di↵erent eigen-

values �1, . . . ,�m 2 R. Then v1, . . . , vm are linearly independent.

Corollary 5.7. Let F : V ! V be a linear map with eigenvalues �1, . . . ,�m 2 R and dimV = n.

i) If F has n distinct eigenvalues, i.e. m = n, then F is diagonalizable.

ii) If B1, . . . , Bm are bases of E�1(F ), . . . , E�m(F ), then B1 [ · · · [Bm are linearly independent.

iii) The map F is diagonalizable if and only if

mX

j=1

dimE�j (F ) = n .

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basisofeigenvectorsOFF

A E l

Linear Algebra II • Eigenvalues and eigenvectors

Definition 5.8. Let F : V ! V be a linear map and let � 2 R be an eigenvalue of F .

i) The algebraic multiplicity of �, denoted by algmuF (�), is the multiplicity of � in the charac-

teristic polynomial fF .

ii) The geometric multiplicity of � is given by geomuF (�) = dimE�(F ).

Theorem 5.9. Let F : V ! V be a linear map and � 2 R be an eigenvalue of F . Then

geomuF (�) algmuF (�) .

Corollary 5.10. If F is diagonalizable then geomuF (�) = algmuF (�) for all eigenvalues � of F .

Version 8 (June 16, 2020)- 11 -

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Linear Algebra II & Mathematics Tutorial 2bNagoya University, G30 Program

Spring 2020

Instructor: Henrik Bachmann

Homework 7: Eigenvalues & Eigenvectors II

•

Deadline: 28th June, 2020

Exercise 1. (10 Points) Let F : V ! V be a linear map with eigenvalues �1, . . . ,�m 2 R and dimV = n.Show the following statements by using Theorem 5.6:

i) If F has n distinct eigenvalues, i.e. m = n, then F is diagonalizable.

ii) If B1, . . . , Bm are bases of E�1(F ), . . . , E�m(F ), then B1 [ · · · [Bm are linearly independent.

(Here we mean by B1 [ · · · [Bm the collection of all vectors in the bases B1, . . . , Bm.)

iii) The map F is diagonalizable if and only if

mX

j=1

dimE�j (F ) = n .

Exercise 2. (6 Points) Consider the matrix

A =

0

@1 0 1

0 2 0

1 0 1

1

A .

Diagonalize the matrix A, i.e. find a matrix S, such that S�1AS is a diagonal matrix.

Exercise 3. (6 Points)

i) Let U ⇢ Rnbe a subspace and let PU : Rn ! Rn

be the orthogonal projection to U . Show that PU

is diagonalizable. What are the eigenvalues of PU?

ii) Let rot↵ : R2 ! R2be the rotation by an angle ↵ 2 [0, 2⇡]. For which ↵ is rot↵ diagonalizable?

Version: June 16, 2020- 1 -

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