Canonical Forms��資料來源：�Friedberg, Insel, and Spence, “Linear Algebra”, 2nd ed.,�Prentice-Hall. (Chapter 7)

大葉大學 資訊工程系

黃鈴玲

Linear Algebra

Introduction

• The advantage of a diagonalizable linear operator is the simplicity of its description.
• Not every linear operator is diagonalizable.

Example:�T: P₂→ P₂ with T(f) = f’, the derivative of f.�The matrix of T with respect to the standard basis {1, x, x²} for P₂ is A=���The characteristic polynomial of A is��⇒ A has only one eigenvalue (λ=0)with multiplicity 3. � The eigenspace corresponding to λ=0 is { | r∈R }�⇒ A is not diagonalizable

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• The purpose of this chapter is to consider alternative matrix representations for nondiagonalizable operators.
• These representations are called canonical forms.
• There are different kinds of canonical forms, there advantages and disadvantages depend on how they applied.
• Our focus ⇒ Jordan canonical form
• This form is always available if the underlying field is algebraically closed, that is, if every polynomial with coefficients from the field splits.

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7.1 General Eigenvectors

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Let T: V→V be a linear operator, A be a matrix representation of T.

Suppose A has eigenvalues λ₁, λ₂, …, λ_n, with has n corresponding, linearly independent eigenvectors v₁, v₂, …, v_n. Let B={x₁, x₂, …, x_n} (note that B is a basis for V), then

where [T]_B is the diagonal matrix representation of T.

Since A may be not diagonalizable, we will prove that:

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For any linear operator whose characteristic polynomial splits,� i.e., characteristic polynomial=

there exists an ordered basis B for V such that

for some eigenvalue λ_j of T. Such a matrix J_i is called a Jordan block corresponding to λ_j, and the matrix [T]_B is called the Jordan canonical form of T. The basis B is called a Jordan canonical basis for T.

Example 1

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The 8×8 matrix

is a Jordan canonical form of a linear operator T:C⁸→C⁸; that is, there exists a basis B={x₁, x₂, …, x₈} for C⁸ such that [T]_B=J.

Note that the characteristic polynomial for T and J is det(J−λI) = (λ−2)⁴(λ−3)²λ², and only x₁, x₄, x₅, x₇ of B={x₁, x₂, …, x₈} are eigenvectors of T.

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• It will be proved that every operator whose characteristic polynomial splits has a unique Jordan canonical form (up to the order of the Jordan blocks).
• The Jordan canonical form is not completely determined by the characteristic polynomial of the transformation.�For example:

The characteristic polynomial of J’ is also (λ−2)⁴(λ−3)²λ².

The relationship of vectors in basis B

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Consider the matrix J and the basis B={x₁, x₂, …, x₈} of Example 1.

T(x₂) = x₁+2x₂ T(x₃) = x₂+2x₃ �⇒(T−2I)(x₂)=x₁ ⇒ (T−2I)(x₃)=x₂

• � {x₁, x₂, x₃} = {(T−2I)²(x₃), (T−2I) (x₃), x₃}

Note that in these three vectors, �only x₁=(T−2I)²(x₃) is an eigenvector.

Similarly, {x₅, x₆} = {(T−3I)(x₆), x₆} and {x₇, x₈} = {T(x₈), x₈}

Therefore, if x lies in a Jordan canonical basis of a linear operator T and corresponds to a Jordan block with diagonal entry λ, then (T−λI)^p(x)=0 for a sufficiently large p.

For example, (T−2I)(x₁)=0, (T−2I)²(x₂)=0, (T−2I)³(x₃)=0.

Generalized Eigenvectors

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Definition

Let T be a linear operator on a finite-dimensional vector space V.�A nonzero vector x in V is called a generalized eigenvector of T if there exists a scalar λ such that (T−λI)^p(x)=0 for some positive integer p. We say that x is a generalized eigenvector corresponding to λ.

Cycle of Generalized Eigenvectors

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Definition

Let T be a linear operator on a vector space V, and let x be a generalized eigenvector of T corresponding to the eigenvalue λ.�If p denotes the smallest positive integer such that (T−λΙ)^p(x)=0, then the ordered set� {(T−λI)^p−1(x), (T−λI)^p−2(x), …, (T−λI) (x), x}

is called a cycle of generalized eigenvectors of T corresponding to λ.

(T−λI)^p−1(x) is called the initial vector of the cycle, �x is called the end vector of the cycle, and�the length of the cycle is p.

In Example 1, B₁={x₁, x₂, x₃}, B₂={x₄}, B₃={x₅, x₆}, B₄={x₇, x₈} are the cycles of generalized eigenvectors of T that occur in B.

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Let T be a linear operator on V, and let γ be a cycle of generalized eigenvectors of T corresponding to the eigenvalue λ.

(a) The initial vector of γ is an eigenvector of T corresponding to the eigenvalue λ, and no other member of γ is an eigenvector of T.

(b) γ is linearly independent.

(c) Let B be an ordered basis for V. Then B is a Jordan canonical basis for V if and only if B is a disjoint union of cycles of generalized eigenvectors of T.

Theorem 7.1

Generalized Eigenspace

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Definition

Let λ be an eigenvalue of a linear operator T on a vector space V.

The generalized eigenspace of T corresponding to λ, denoted by K_λ, is the set

K_λ = { x ∈ V : (T−λI)^p(x) = 0 for some positive integer p}.

Theorem 7.2

A subspace W of V is called T-invariant if T(W) ⊆ W.

Let λ be an eigenvalue of a linear operator T on a vector space V.

Then K_λ is a T-invariant subspace of V containing E_λ (the eigenspace of T corresponding to λ).

The connection between the generalized eigenspaces and the characteristic polynomial of an operator

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Theorem 7.6

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Suppose that λ₁, λ₂, ..., λ_κ are distinct eigenvalues of T with corresponding multiplicities m₁, m₂, …, m_k. Then:

1. dim (K_λi) = m_i for all i.
2. If for each i, S_i is a basis for K_λi, � then the union S = S₁∪S₂∪…∪S_k is a basis of V.
3. If B is a Jordan canonical basis for T, then for each i, B_i=B∩ K_λi � is a basis for K_λi.
4. K_λi = Ker((T−λ_iI)^mi) for all i.
5. T is diagonalizable if and only if E_λi = K_λi for all i.

Example 2

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Let T: C ³ → C ³ be defined by T(x) = Ax, where

Solution

Find a basis for each eigenspace and each generalized eigenspace of T.

The characteristic polynomial of T is det(A−λI) = −(t−3)(t−2)².

⇒ λ₁ = 3, λ₂ = 2, with multiplicity m₁=1, m₂=2

⇒ dim(K_λ1) = 1, dim(K_λ2) = 2

• K_λ1 = Ker(T−3I) = E_λ1 = {r(−1,2,1)} ⇒ {r(−1,2,1)} is a basis� K_λ2 = Ker((T−2I)²), E_λ2 = Ker(T−2I)

Example 2 -continued

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Since E_λ2 = {s(1, −3, −1)}, dim(E_λ2)=1< dim(K_λ2)=2

The basis of K_λ2 is a single cycle of length 2.

⇒ Choose (1, −3, −1) be the initial vector of the cycle,

then a vector v is the end vector � if and only if (A−2I)v = (1, −3, −1) .

• Solve (A−2I) v= (1, −3, −1), � we get v= {(−1−r, 2+3r, r)}
• { (1, −3, −1), (−1, 2, 0)} is a basis for K_λ2.

• B={(−1,2,1), (1, −3, −1), (−1, 2, 0)} is a basis for C³,

and

​

is a Jordan canonical form of T.

Example 3

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Let T: P₂ → P₂ be defined by T(f) = −f − f’. Find a basis for each eigenspace and generalized eigenspace of T.

Solution

The characteristic polynomial of T is det(A−λI) = −(t+1)³.

⇒ λ = −1 with multiplicity m=3

⇒ dim(K_λ) = 3 = dim(P₂), � so any basis of P₂, for example, B, is a basis of K_λ

• E_λ = {r} ⇒ {1} is a basis for E_λ

If B={1, x, x²}, then B is an ordered basis for P₂ and

Example 3 -continued

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The basis of K_λ is a single cycle of length 3.

⇒ Choose 1 be the initial vector of the cycle,

then a vector v is the end vector � if and only if (A+I)²v_B = (1, 0, 0) .

• Solve (A+I)² X= (1, 0, 0), � we get X= {(r, s, 0.5)}, let v_B= (0, 0, 0.5), i.e., v = 0.5x²
• (A+I) v_B = (0, −1, 0)
• B={ 1, −x, 0.5x²} is a basis for P₂,� and

is a Jordan canonical form of T.

Homework

• For each of the following linear operators T(x)=Ax, find a basis for each eigenspace and each generalized eigenspace.�(a)

​

​

(b)

​

​

(c) T: P₂→P₂ defined by T(f)=2f − f ’.

• Find the Jordan canonical forms for above linear operators.

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7.2 Jordan Canonical Form

• In this section we will develop a more direct approach for finding the Jordan canonical form and a Jordan canonical basis for a linear operator.
• V: n-dim vector space�T: V → V a linear transformation such that its � characteristic polynomial splits.�Let λ₁, λ₂, ..., λ_κ be distinct eigenvalues of T. �B: a Jordan canonical basis for T.�B_i: the cycle of B that correspond to λ_i form a basis for K_λi.�T_i: the restriction of T to K_λi.�If [T_i]_Bi = A_i for all i, then

​

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How to find A_i and B_i?

• Let Z₁, Z₂, …Z_ki be cycles of B_i with length p₁ ≥ p₂ ≥ … ≥ p_ki. �For example:

​

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k_i = 4 (4 cycles)�p₁ = 3 �p₂ = 3 _�p₃ = 2 �p₄ = 1

Therefore A_i is entirely determined by the number k_i, p₁, …, p_ki.

Dot diagram

• The array consists of k_i columns (one column for each cycle).
• The jth column consists of p_j dots that correspond to the members of Z_j (1 ≤ j ≤ k_i )

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For example: � the dot diagram associated with B_i, where� k_i = 4,p₁ = 3, p₂ = 3, p₃ = 2, p₄ = 1.

Let r_j denote the number of dots in the jth row of a dot diagram for B_i. Then �(a) r₁= dim(V) – rank(T−λ_iI).�(b) r_j= rank((T−λ_iI)^j−1) – rank((T−λ_iI)^j) if j >1.

Theorem 7.9

Example 2

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Let

Find the Jordan canonical form of A and a Jordan canonical basis for the linear transformation T_A.

The characteristic polynomial of A is det(A−λI) = (λ−2)³(λ−3) , Thus A has two eigenvalues, λ₁ =2 and λ₂ = 3 with multiplicities 3 and 1, respectively. Therefore �

Solution

Example 2 -continued

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Since r₁ = dim (R⁴) – rank(A−2I) = 2,� r₂ = rank(A−2I) – rank((A−2I)²) =1� ⇒ dot diagram:

Now we find the Jordan canonical basis for T: � B₁={ x₁, x₂, x₃} = {(T−2I)(x₂), x₂, x₃}, B₂={x₄} �Note that x₁ ∈Ker((T−2I)²), but x₁ ∉Ker(T−2I); and� x₂, x₃ ∈Ker(T−2I), x₄ ∈Ker(T−3I).

• K_λ1= Ker((T−2I)²)�
• { } is a basis for K_λ1.

row 1有r₁個dot�row 2有r₂個dot

⇒

Example 2 -continued

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Then x₁ = (T−2I)(x₂) = (A−2I)(x₂) =

​

• ∉ Ker(T−2I) ⇒ Let x₂=

Now choose x₃ to be an eigenvector which is linearly independent to x₁, so let � x₃ =

For λ=3, it is easily to find an eigenvector. Let x₄ =

Example 2 -continued

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So B={ } is a Jordan canonical basis for T_A.

Notice that if Q = then J = Q ⁻¹AQ.

Example 3

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Let

Find the Jordan canonical form J for A and a matrix Q such that� J = Q⁻¹AQ.

The characteristic polynomial of A is det(A−λI) = (λ−2)²(λ−4)² , Thus A has two eigenvalues, λ₁ =2 and λ₂ = 4 with multiplicities both 2, respectively. Therefore

Solution

Example 3 -continued

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Since r₁ = dim (R⁴) – rank(A−2I) = 2,� ⇒ dot diagram for B₁:

row 1 有r₁個dot

column 1及column 2各有1個dot

⇒

Since r₁ = dim (R⁴) – rank(A−4I) = 4 − 3 = 1,� r₂ = rank(A−4I) – rank((A−4I)²) =1� ⇒ dot diagram for B₂:

row 1 有r₁個dot�row 2 有r₂個dot

column 1有2個dot

⇒

⇒

Example 3 -continued

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Now we find the Jordan canonical basis for T: � B₁={x₁, x₂}, B₂={x₃, x₄} = {(T−4I)(x₄), x₄} �Note that x₁, x₂ ∈Ker(T−2I); � x₃ ∈Ker(T−4I), x₄ ∈Ker((T−4I)²) but x₄ ∉Ker(T−4I).

⇒ { } is a basis for K_λ1.

⇒ { } is a basis for K_λ2.

Example 3 -continued

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Let

⇒

Therefore B = { }.

Notice that if Q = then J = Q ⁻¹AQ.

Homework 1

• Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Let λ₁ = 2, λ₂ = 4, and λ₃ = −3 be the distinct eigenvalues of T, and suppose that the dot diagrams for the restriction of T−λ_iI to K_λi (i=1, 2 ,3) are as follows:������Find the Jordan canonical form of T.

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λ₁ = 2

λ₂ = 4

λ₃ = −3

Homework 2

• Let T be a linear operator on a finite-dimensional vector space V such that the Jordan canonical form of T is

​

​

​

��

(a) Find the characteristic polynomial of T.�(b) Find the dot diagram corresponding to each eigenvalue of T.�(c) For which eigenvalues λ_i, if any, does E_li = K_li?

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Homework 3

• For the following matrix A, find a Jordan canonical form J and a matrix Q such that J = Q⁻¹AQ.

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