5.II. Similarity

5.II.1. Definition and Examples5.II.2. Diagonalizability5.II.2. Eigenvalues and Eigenvectors

5.II.1. Definition and Examples

H and H are matrix-equivalent if nonsingular P and Q s.t. H = PHQ.

1

h

id id

h

V W

V W

H

Q P

H

B D

B D

For the special case W = V, we have H = PHP1.

h

id id

h

V V

V V

H

P P

H

B B

B B

1h id h id

B B B B B B B B

Definition 1.1: Similar MatricesMatrices T and S are similar if nonsingular P s.t. T = PSP1.

Hv u → PHv Pu 1 PHP Pv Pu H v u

Example 1.2:

2 1

1 1

P2 3

1 1

S0 1

1 1

T

1

1 2 1 2 3 2 1

1 1 1 1 1 1

P SP

1 1 1 1

1 2 1 0

0 1

1 1

T

Example 1.3:The only matrix similar to the zero matrix Z is itself: PZP1 = PZ = Z. The only matrix similar to the identity matrix I is itself: PIP1 = PP1 = I.

Similarity is a special case of matrix equivalent.

Similarity → matrix equivalent (but not vice versa).

Matrix equivalence classes can be subdivided into similarity classes.

Canonical form of similarity class is the Jordan form.

Exercises 5.II.1.

1. Show that these matrices are not similar.

1 0 4

1 1 3

2 1 7

1 0 1

0 1 1

3 1 2

2. Are there two matrices A and B that are similar while A2 and B2 are not similar?

3. Show that if TλI and N are similar, then T and N+λI are also similar.

5.II.2. Diagonalizability

Definition 2.1: Diagonalizable MatricesA transformation is diagonalizable if it has a diagonal representation wrt the same basis for the codomain as for the domain.A diagonalizable matrix is one that is similar to a diagonal matrix: T is diagonalizable if there is a nonsingular P such that PTP1 is diagonal.

Example 2.2: 4 2

1 1

is diagonalizable.

11 2 4 2 1 2 2 4 1 2

1 1 1 1 1 1 3 3 1 1

2 0

0 3

Example 2.3: 0 0

1 0

N is not diagonalizable.

Proof: Since N2 = 0, the corresponding map n n is the zero map.

→ There is no s.t. N = n→ is diagonal because the square of a nonzero diagonal matrix cannot be the zero matrix.

Thus, the diagonal matrix cannot be a canonical form.

Corollary 2.4:A transformation t is diagonalizable iff a basis = β1 , …, βn and scalars λ1, …, λn s.t. t (βi ) = λi βi i.

Proof:By definition, t is diagonalizable iff a basis = β1 , …, βn s.t.

1 nt t t β βB B B B

1 0

0 n

QED

1 1 n n β βB

Example 2.5:To diagonalize

3 2

0 1

T2 2

t E E is to find s.t.1

2

0

0t

B B

i.e., 1 1 1t β β 2 2 2t β βand

2 2

1 1 1

3 2

0 1

β βE E

2 2

2 2 2

3 2

0 1

β βE E

1 1

2 2

3 2

0 1

b b

b b

→ 1

2

3 2 0

0 1 0

b

b

2nd eq. → 21 0or b

λ1 = 1 → 1 22 2 0b b 1 2b b→

b2 = 0 → 2 13 0 discardedb

→ 1

1

1a

β

→ 2

1

0b

β

3 2

0 1

T

λ1 = 1 1

1

1a

β

2 3 2

1

0b

β

11 1 3 2 1 1 0 1 1 3

1 0 0 1 1 0 1 1 1 0

1 0

0 3

Exercises 5.II.2.

1. Find a formula for the powers of this matrix.3 1

4 2

Hint: What form do the powers of a diagonal matrix have?

2. Show that matrices of this form are not diagonalizable.

1

0 1

c

0c

5.II.3. Eigenvalues and Eigenvectors

Definition 3.1: Eigenvalues & Eigenvectors of a MapA transformation t : V → V has a scalar eigenvalue λ if a nonzero eigenvector ζ s.t. t (ζ) =λζ.

Example 3.2: Projection map

0

x x

y y

z

x, y, z

has an eigenvalue of 1 associated with any eigenvector of the form x, y 0. 0

x

y

Example 3.3: Trivial Space The only transformation on the trivial space {0} is 0 → 0. This map has no eigenvalues because there are no non-zero vectors v mapped to a scalar multiple λv of themselves.

Example 3.4: 1

Consider the homomorphism t : 1 → 1 given by c0 + c1x (c0 +c1) + (c0 +c1) x

The range of t is one-dimensional. Thus an application of t to a vector in the range will simply rescale that vector:

c + c x (2c) + (2c) x i.e., t has an eigenvalue of 2 associated with eigenvectors c + cx, where c 0.It also has an eigenvalue of 0 associated with eigenvectors c cx , where c 0.

Definition 3.5: Eigenvalues & Eigenvectors of a MatrixA square matrix T has a scalar eigenvalue λ associated with the non-zero eigenvector ζ if T ζ= λ ζ.

Remark 3.6: Eigenvalues of a map are also the eigenvalues of matrices representing that map.So similar matrices have the same eigenvalues.But the eigenvectors are different:

similar matrices need not have the same eigenvectors.Example 3.4a: 1

Consider the homomorphism t : 1 → 1 given by c0 + c1x (c0 +c1) + (c0 +c1) x

It has an eigenvalue of 2 associated with eigenvectors c + cx, where c 0.Wrt basis = 1+x, 1x , we have

2 0

0 0t

T B B

2 is an eigenvalue of T associated with eigenvector 0

c

where c 0.→

Wrt basis = 2+x, 1, we have

3 1

3 1t

S D D

2 is an eigenvalue of S associated with eigenvector

c

c

where c 0.→

Example 3.7:What are the eigenvalues and eigenvectors of this matrix?

1 2 1

2 0 2

1 2 3

T

Solution:

0 Tζ ζ1

2

3

1 2 1 0

2 2 0

1 2 3 0

z

z

z

T I ζ

There is a non-zero solution iff TλI is singular, i.e. | TλI | = 0.

1 2 1

0 2 2

1 2 3

3 24 4 22

For λ1 = 0, 1

2

3

1 2 1 0

2 0 2 0

1 2 3 0

z

z

z

→1

2

3

1

1

1

z

z a

z

λ1 = 0, λ2 = 2, λ3 = 2.

~

1

2

3

1 0 1 0

0 1 1 0

0 0 0 0

z

z

z

For λ2 = 2, or λ3 = 2,

1

2

3

1 2 1 0

2 2 2 0

1 2 1 0

z

z

z

→1

2

3

1

0

1

z

z b

z

~

1

2

3

1 0 1 0

0 1 0 0

0 0 0 0

z

z

z

Example 3.8: 1

0 3

S

1

0 3

S I 3 0 → λ1 = π, λ2 = 3.

For λ1 = π, 1

2

0 1 0

0 3 0

z

z

→

1

2

1

0

za

z

For λ2 = 3, 1

2

3 1 0

0 0 0

z

z

→

1

2

1

31

zb

z

Definition 3.9: Characteristic Polynomial The characteristic polynomial of a square matrix T is the determinant of the matrix T λI, where λ is a variable. The characteristic (secular) equation is | T λI | = 0. The characteristic polynomial of a transformation t is the polynomial of any t→.

The characteristic polynomial of a transformation is well-defined, that is, any choice of basis yields the same polynomial.

Lemma 3.10:A linear transformation on a nontrivial vector space has at least one eigenvalue.Proof: Any root of the characteristic polynomial is an eigenvalue. Over the complex numbers, any polynomial of degree one or greater has a root. (This is the reason that in this chapter we’ve gone to scalars that are complex.) QED

Definition 3.11: EigenspaceThe eigenspace of a transformation t associated with the eigenvalue λ is

V t ζ ζ ζ 0

The eigenspace of a matrix is defined analogously.

Lemma 3.12: An eigenspace is a subspace.

Proof: Only closure need be checked (Follows directly from linearity of t).

Example 3.13:In Example 3.8 the eigenspace associated with the eigenvalue π and the eigenspace associated with the eigenvalue 3 are these.

0

aV a

R 3 3

bV b

b

R

Example 3.14:In Example 3.7, these are the eigenspaces associated with the eigenvalues 0 and 2.

0

1

1

1

V a a

R 2

1

0

1

V b b

R

Remark 3.15:The characteristic equation is 0 = x(x2)2 so 2 is an eigenvalue “twice”. However V2 is 1-D so there is a deficient of eigenvectors.

Example 3.16:With respect to the standard bases, this matrix

1 0 0

0 1 0

0 0 0

represents projection:

0

x x

y y

z

, ,x y zC

The eigenspaces are0

0

0

1

V a a

C 1 ,

0

b

V c b c

C

Note that V1 is 2-D.

Any linear combination of eigenvectors belonging to the same eigenvalue is an eigenvector belonging to that eigenvalue.

Theorem 3.17:For any set of distinct eigenvalues of a map or matrix, a set of associated eigenvectors, one per eigenvalue, is linearly independent.

Proof by induction: see Hefferon p.364.

Example 3.18:The eigenvalues of

2 2 2

0 1 1

4 8 3

are distinct: λ1 = 1, λ2 = 2, λ3 = 3.

A set of associated eigenvectors like

2 9 2

1 , 4 , 1

0 4 2

is L.I.

Corollary 3.19:An nn matrix with n distinct eigenvalues is diagonalizable.

Proof: The eigenvectors form a basis of the column space. By corollary 2.4, QED.

Exercises 5.II.3.

1. Find the characteristic equation, and the eigenvalues and associated eigenvectors for

2 1

5 2

2. Find the eigenvalues and associated eigenvectors of the differentiation operatord/dx : 2 → 2 .

3. Prove that if T is nonsingular and has eigenvalues λ1 , …, λn then T1 has eigenvalues 1/λ1 , …, 1/λn . Is the converse true?

4. Show that if A is an n square matrix and each row (column) sums to c then c is a characteristic root of A.