Linear Algebra 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

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 (l=0)with multiplicity 3. The eigenspace corresponding to l=0 is { | rR }  A is not diagonalizable

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.

7.1 General Eigenvectors Let T: VV be a linear operator, A be a matrix representation of T. Suppose A has eigenvalues l1, l2, …, ln, with has n corresponding, linearly independent eigenvectors v1, v2, …, vn. Let B={x1, x2, …, xn} (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:

For any linear operator whose characteristic polynomial splits, i. e For any linear operator whose characteristic polynomial splits, i.e., characteristic polynomial= there exists an ordered basis B for V such that where Ji is a square matrix of the form [lj] or the form for some eigenvalue lj of T. Such a matrix Ji is called a Jordan block corresponding to lj, 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 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-lI) = (l-2)4(l-3)2l2, and only x1, x4, x5, x7 of B={x1, x2, …, x8} are eigenvectors of T.

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 (l-2)4(l-3)2l2.

The relationship of vectors in basis B 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 l, then (T-lI)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 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 l such that (T-lI)p(x)=0 for some positive integer p. We say that x is a generalized eigenvector corresponding to l.

Cycle of Generalized Eigenvectors 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 l. If p denotes the smallest positive integer such that (T-lI)p(x)=0, then the ordered set {(T-lI)p-1(x), (T-lI)p-2(x), …, (T-lI) (x), x} is called a cycle of generalized eigenvectors of T corresponding to l. (T-lI)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.

Theorem 7.1 Let T be a linear operator on V, and let g be a cycle of generalized eigenvectors of T corresponding to the eigenvalue l. (a) The initial vector of g is an eigenvector of T corresponding to the eigenvalue l, and no other member of g is an eigenvector of T. (b) g 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.

Generalized Eigenspace Definition Let l be an eigenvalue of a linear operator T on a vector space V. The generalized eigenspace of T corresponding to l, denoted by Kl, is the set Kl = { x  V : (T-lI)p(x) = 0 for some positive integer p}. A subspace W of V is called T-invariant if T(W)  W. Theorem 7.2 Let l be an eigenvalue of a linear operator T on a vector space V. Then Kl is a T-invariant subspace of V containing El (the eigenspace of T corresponding to l).

The connection between the generalized eigenspaces and the characteristic polynomial of an operator 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 l1, l2, ..., lk are distinct eigenvalues of T with corresponding multiplicities m1, m2, …, mk. Then: dim (Kli) = mi for all i. If for each i, Si is a basis for Kli, then the union S = S1∪S2∪…∪Sk is a basis of V. If B is a Jordan canonical basis for T, then for each i, Bi=B∩ Kli is a basis for Kli. Kli = Ker((T-liI)mi) for all i. T is diagonalizable if and only if Eli = Kli for all i.

Example 2 Let T: C 3  C 3 be defined by T(x) = Ax, where Find a basis for each eigenspace and each generalized eigenspace of T. Solution The characteristic polynomial of T is det(A-lI) = -(t-3)(t-2)2.  l1 = 3, l2 = 2, with multiplicity m1=1, m2=2  dim(Kl1) = 1, dim(Kl2) = 2 Kl1 = Ker(T-3I) = El1 = {r(-1,2,1)}  {r(-1,2,1)} is a basis Kl2 = Ker((T-2I)2), El2 = Ker(T-2I)

Example 2 -continued Since El2 = {s(1, -3, -1)}, dim(El2)=1< dim(Kl2)=2 The basis of Kl2 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 Kl2. 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 Let T: P2  P2 be defined by T(f) = -f - f’. Find a basis for each eigenspace and generalized eigenspace of T. Solution If B={1, x, x2}, then B is an ordered basis for P2 and The characteristic polynomial of T is det(A-lI) = -(t+1)3.  l = -1 with multiplicity m=3  dim(Kl) = 3 = dim(P2), so any basis of P2, for example, B, is a basis of Kl El = {r}  {1} is a basis for El

Example 3 -continued The basis of Kl 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)2vB = (1, 0, 0) . Solve (A+I)2 X= (1, 0, 0), we get X= {(r, s, 0.5)}, let vB= (0, 0, 0.5), i.e., v = 0.5x2 (A+I) vB = (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.

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 l1, l2, ..., lk be distinct eigenvalues of T. B: a Jordan canonical basis for T. Bi: the cycle of B that correspond to li form a basis for Kli. Ti: the restriction of T to Kli. If [Ti]Bi = Ai for all i, then

How to find Ai and Bi? Let Z1, Z2, …Zki be cycles of Bi with length p1  p2  …  pki. For example: ki = 4 (4 cycles) p1 = 3 p2 = 3 p3 = 2 p4 = 1 Therefore Ai is entirely determined by the number ki, p1, …, pki.

Dot diagram The array consists of ki columns (one column for each cycle). The jth column consists of pj dots that correspond to the members of Zj (1  j  ki ) For example: the dot diagram associated with Bi, where ki = 4,p1 = 3, p2 = 3, p3 = 2, p4 = 1. Theorem 7.9 Let rj denote the number of dots in the jth row of a dot diagram for Bi. Then (a) r1= dim(V) – rank(T-liI). (b) rj= rank((T-liI)j-1) – rank((T-liI)j) if j >1.

Example 2 Let Find the Jordan canonical form of A and a Jordan canonical basis for the linear transformation TA. Solution The characteristic polynomial of A is det(A-lI) = (l-2)3(l-3) , Thus A has two eigenvalues, l1 =2 and l2 = 3 with multiplicities 3 and 1, respectively. Therefore

Example 2 -continued Since r1 = dim (R4) – rank(A-2I) = 2, r2 = rank(A-2I) – rank((A-2I)2) =1  dot diagram: column 1有2個dot 故cycle長為2  row 1有r1個dot row 2有r2個dot 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). Kl1= Ker((T-2I)2) { } is a basis for Kl1.

Example 2 -continued  Ker(T-2I)  Let x2= Then x1 = (T-2I)(x2) = (A-2I)(x2) = Now choose x3 to be an eigenvector which is linearly independent to x1, so let x3 = For l=3, it is easily to find an eigenvector. Let x4 =

Example 2 -continued So B={ } is a Jordan canonical basis for TA. Notice that if Q = then J = Q -1AQ.

Example 3 Let Find the Jordan canonical form J for A and a matrix Q such that J = Q-1AQ. Solution The characteristic polynomial of A is det(A-lI) = (l-2)2(l-4)2 , Thus A has two eigenvalues, l1 =2 and l2 = 4 with multiplicities both 2, respectively. Therefore

Example 3 -continued 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 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 Kl1.  { } is a basis for Kl2.

Example 3 -continued 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 l1 = 2, l2 = 4, and l3 = -3 be the distinct eigenvalues of T, and suppose that the dot diagrams for the restriction of T-liI to Kli (i=1, 2 ,3) are as follows: Find the Jordan canonical form of T. l1 = 2 l2 = 4 l3 = -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 li, if any, does Eli = Kli?

Homework 3 For the following matrix A, find a Jordan canonical form J and a matrix Q such that J = Q-1AQ.