7.1 The Jordan Canonical Form I – Linear Algebra, 5th Edition

7.1 The Jordan Canonical Form I

Let T be a linear operator on a finite-dimensional vector space V, and suppose that the characteristic polynomial of T splits. Recall from Section 5.2 that the diagonalizability of T depends on whether the union of ordered bases for the distinct eigenspaces of T is an ordered basis for V. So a lack of diagonalizability means that at least one eigenspace of T is too “small.”

In this section, we extend the definition of eigenspace to generalized eigenspace. From these subspaces, we select ordered bases whose union is an ordered basis β for V such that

[ T ] β = ( A 1 O O O A 2 O O O A k ) ,

where each O is a zero matrix, and each Ai is a square matrix of the form (λ) or

( λ 1 0 0 0 0 λ 1 0 0 0 0 0 λ 1 0 0 0 0 λ )

for some eigenvalue λ of T. Such a matrix Ai is called a Jordan block corresponding to λ, and the matrix [T]β is called a Jordan canonical form of T. We also say that the ordered basis β is a Jordan canonical basis for T. Observe that each Jordan block Ai is “almost” a diagonal matrix—in fact, [T]β is a diagonal matrix if and only if each Ai is of the form (λ).

Example 1

Suppose that T is a linear operator on C8, and β={v1, v2, , v8} is an ordered basis for C8 such that

is a Jordan canonical form of T. Notice that the characteristic polynomial of T is det(JtI)=(t2)4(t3)2t2, and hence the multiplicity of each eigenvalue is the number of times that the eigenvalue appears on the diagonal of J. Also observe that v1, v4, v5, and v7 are the only vectors in β that are eigenvectors of T. These are the vectors corresponding to the columns of J with no 1 above the diagonal entry.

In Sections 7.1 and 7.2, we prove that every linear operator whose characteristic polynomial splits has a Jordan canonical form that is unique up to the order of its Jordan blocks. Nevertheless, it is not the case that the Jordan canonical form is completely determined by the characteristic polynomial of the operator. For example, let T be the linear operator on C8 such that [T]β=J, where β is the ordered basis in Example 1 and

J = ( 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ) .

Then the characteristic polynomial of T is also (t2)4(t3)2t2. But the operator T has the Jordan canonical form J, which is different from J, the Jordan canonical form of the linear operator T of Example 1.

Consider again the matrix J and the ordered basis β of Example 1. Notice that T(v2)=v1+2v2, and so (T2I)(v2)=v1 . Similarly, (T2I)(v3)=v2. Since v1 and v4 are eigenvectors of T corresponding to λ=2, it follows that (T2I)3(vi)=0 for i=1, 2, 3, and 4. Similarly (T3I)2(vi)=0 for i=5, 6, and (T0I)2(vi)=0 for i=7, 8.

Because of the structure of each Jordan block in a Jordan canonical form, we can generalize these observations: If v lies in a Jordan canonical basis for a linear operator T and is associated with a k×k Jordan block with diagonal entry λ, then (TλI)k(v)=0. Eigenvectors satisfy this condition for k=1.

Definition.

Let T be a linear operator on a vector space V, and let λ be a scalar. A nonzero vector x in V is called a generalized eigenvector of T corresponding to λ if (TλI)p(x)=0 for some positive integer p.

Notice that if x is a generalized eigenvector of T corresponding to λ, and p is the smallest positive integer for which (TλI)p(x)=0, then (TλI)p1(x) is an eigenvector of T corresponding to λ. Therefore λ is an eigenvalue of T.

In the context of Example 1, each vector in β is a generalized eigenvector of T. In fact, v1, v2, v3, and v4 correspond to the eigenvalue 2, v5 and v6 correspond to the eigenvalue 3, and v7 and v8 correspond to the eigenvalue 0.

Just as eigenvectors lie in eigenspaces, generalized eigenvectors lie in “generalized eigenspaces.”

Definition.

Let T be a linear operator on a vector space V, and let λ be an eigenvalue of T. The generalized eigenspace of T corresponding to λ, denoted Kλ, is the subset of V defined by

K λ = { x V :   ( T λ I ) p ( x ) = 0 for some positive integer  p } .

Note that Kλ consists of the zero vector and all generalized eigenvectors corresponding to λ.

Recall that a subspace W of V is T-invariant for a linear operator T if T(W)W. In the development that follows, we assume the results of Exercises 3 and 4 of Section 5.4. In particular, for any polynomial g(x), if W is T-invariant, then it is also g(T)-invariant. Furthermore, the range of a linear operator T is T-invariant.

Theorem 7.1.

Let T be a linear operator on a vetor space V, and let λ be an eigenvalue of T. Then

  1. Kλ is a T-invariant subspace of V containing Eλ (the eigenspace of T corresponding to λ).

  2. For any eigenvalue μ of T such that μλ,

    K λ K μ = { 0 } .
  3. For any scalar μλ, the restriction of TμI to Kλ is one-to-one and onto.

Proof.

(a) Clearly, 0Kλ. Suppose that x and y are in Kλ. Then there exist positive integers p and q such that

( T λ I ) p ( x ) = ( T λ I ) q ( y ) = 0.

Therefore

( T λ I ) p + q ( x + y ) = ( T λ I ) p + q ( x ) + ( T λ I ) p + q ( y ) = ( T λ I ) q ( 0 ) + ( T λ I ) p ( 0 ) = 0 ,

and hence x+yKλ. The proof that Kλ is closed under scalar multiplication is straightforward.

To show that Kλ is T-invariant, consider any xKλ. Choose a positive integer p such that (TλI)p(x)=0. Then

( T λ I ) p T ( x ) = T ( T λ I ) p ( x ) = T ( 0 ) = 0.

Therefore T(x)Kλ.

Finally, it is a simple observation that Kλ is contained in Kλ.

(b) Suppose that wKλKμ. Then there exist positive integers p and q such that

( T λ I ) p ( w ) = ( T μ I ) q ( w ) = 0.

As polynomials in t, (tλ)p and (tμ)q are relatively prime, and hence by Theorem E.2 in Appendix E there exist polymomials q1(t) and q2(t) such that

q 1 ( t ) ( t λ ) p + q 2 ( t ) ( t μ ) q = 1.

It follows that

q 1 ( T ) ( T λ I ) p ( w ) + q 2 ( T ) ( T μ I ) q ( w ) = w ,

and hence 0=w.

(c) Let μ be any scalar such that μλ. Since Kλ is T-invariant, it is also (TμI)-invariant. If (TμI)(w)=0 for some wKλ, then wEμKμ. Hence w=0 by (b), and we conclude that the restriction of TμI to Kλis one-to-one.

We now show that the restriction of TμI to Kλ is onto. Let xKλ, p be the smallest positive integer for which (TλI)p(x)=0, and

W = span { x ,   ( T λ I ) ( x ) ,   ,   ( T λ I ) p 1 ( x ) } .

It is easily shown that W is a T-invariant subspace of Kλ, and hence it is also (TμI)-invariant. So TμI maps W to W. Since TμI is one-to-one on Kλ , it must be one-to-one on W. Thus, because W is finite-dimensional, TμI maps W onto W. So there exists a yW such that (TμI)(y)=x. Therefore TμI maps Kλ onto Kλ.

Theorem 7.2.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Suppose that λ is an eigenvalue of T with multiplicity m. Then

  1. dim(Kλ)m.

  2. Kλ=N((TλI)m).

Proof.

(a) Let W=Kλ, and let h(t) be the characteristic polynomial of TW. By Theorem 5.20 (p. 312), h(t) divides the characteristic polynomial of T, and by Theorem 7.1(b), λ is the only eigenvalue of TW. Hence h(t)=(1)d(tλ)d, where d=dim(W), and dm.

(b) Clearly N((TλI)m)Kλ. The characteristic polynomial of T is of the form f(t)=(tλ)mg(t), where g(t) is a product of powers of the form (tμ) for eigenvalues μλ. By Theorem 7.1, g(T) is one-to one on Kλ. Since Kλ is finite-dimensional, it is also onto. Let xKλ. Then there exists yKλ such that g(T)(y)=x. Hence

( T λ I ) m ( x ) = ( T λ I ) m g ( T ) ( y ) = f ( T ) ( y ) = 0

by the Cayley-Hamilton Theorem. Therefore xN((TλI)m), and thus KλN((TλI)m).

Theorem 7.3.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let λ1, λ2, , λk be the distinct eigenvalues of T. Then, for every xV, there exist unique vectors viKλi, for i=1, 2, , k, such that

x = v 1 + v 2 + + v k .

Proof.

Suppose the characteristic polynomial of T is

f ( t ) = ( t λ 1 ) n 1 ( t λ 2 ) n 2     ( t λ k ) n k .

For j=1, 2, , k, let

f j ( t ) = i j i = 1 k ( t λ i ) n i .

First, we prove the existence of the vectors vi. Because the polynomials fj(t) are relatively prime for j=1, 2, , k, by Theorem E.2 there exist polynomials qj(t) for j=1, 2, , k such that

q 1 ( t ) f 1 ( t ) + q 2 ( t ) f 2 ( t ) + + q k ( t ) f k ( t ) = 1.

It follows that

v = q 1 ( T ) f 1 ( T ) ( v ) + q 2 ( T ) f 2 ( T ) ( v ) + + q k ( T ) f k ( T ) ( v ) .

For i=1, 2, , k, let vi=qi(T)fi(T)(v)=fi(T)qi(T)(v). Then for each i,

( T λ i I ) n i ( v i ) = f ( T ) ( q i ( T ) ( v ) ) = 0

by the Cayley-Hamilton theorem. It follows that viKλi for i=1, 2, , k.

Now we prove the uniqueness of the vectors vi. Since each Kλj is T-invariant, each Kλi is invariant under the operator fj(T) for j=1, 2, , k. Furthermore, fj(T) is one-to-one on Kλj by Theorem 7.1(c) and maps Kλi to {0} for ij by Theorem 7.2(b).

Suppose that

v = i = 1 k v i = i = 1 k w i ,

where vi, wiKλi for i=1, 2, , k. Then for j=1, 2, , k,

f j ( T ) ( v ) = i = 1 k f j ( T ) ( v i ) = i = 1 k f j ( T ) ( w i ) ,

and hence fj(T)(vj)=fj(T)(wj) since fj(T)(vi)=fj(T)(wi)=0 for ij. Because fj(T) is one-to-one on Kλj, it follows that vj=wj.

The next result extends Theorem 5.8(b) (p. 267) to all linear operators whose characteristic polynomials split. In this case, the eigenspaces are replaced by generalized eigenspaces.

Theorem 7.4.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial (tλ1)m1(tλ2)m2(tλk)mk splits. For i=1, 2, , k, let βi be an ordered basis for Kλi. Then

  1. βiβj= for ij.

  2. β=β1β2βk is an ordered basis for V.

  3. dim(Kλi)=mi for all i.

Proof.

(a) This is a direct consequence of Theorem 7.1(b).

(b) First, we prove that β is linearly independent. Consider a linear combination of vectors in β equal to 0. For each i, let vi be the sum of the terms of this linear combination involving the vectors in βi. Then

v 1 + v 2 + + v k = 0 = 0 + 0 + + 0 ,

and hence vi=0 for all i by Theorem 7.3. Since each βi is linearly independent, it follows that all of the coefficients in the linear combination are zeros. Hence β is linearly independent.

We now prove that β is a spanning set for V. Consider any vector vV. By Theorem 7.3, for i=1, 2, , k there exist vectors viKλi such that v=v1+v2++vk. Since each vi is a linear combination of the vectors in βi, v is a linear combination of the vectors in β. Hence β spans V.

(c) Since dim(V) is equal to the degree of the characteristic polynomial of T, it follows that dim(V)=i=1kmi. By (b), dim(V)=i=1kdim(Kλi). Hence i=1k(midim(Kλi))=0. By Theorem 7.2(a), midim(Kλi)0 for all i. Consequently, midim(Kλi)=0 for all i, which is the desired result.

Corollary.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Then T is diagonalizable if and only if Eλ=Kλ for every eigenvalue λ of T.

Proof.

Combining Theorems 7.4 and 5.8(a) (p. 267), we see that T is diagonalizable if and only if dim(Eλ)=dim(Kλ) for each eigenvalue λ of T. But EλKλ, and hence these subspaces have the same dimension if and only if they are equal.

We now focus our attention on the problem of selecting suitable bases for the generalized eigenspaces of a linear operator so that we may use Theorem 7.4 to obtain a Jordan canonical basis for the operator. For this purpose, we consider again the basis β of Example 1. We have seen that the first four vectors of β lie in the generalized eigenspace K2. Observe that the vectors in β that determine the first Jordan block of J are of the form

{ v 1 ,   v 2 ,   v 3 } = { ( T 2 I ) 2 ( v 3 ) ,   ( T 2 I ) ( v 3 ) ,   v 3 } .

Furthermore, observe that (T2I)3(v3)=0. The relation between these vectors is the key to finding Jordan canonical bases. This leads to the following definitions.

Definitions.

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. Suppose that p is the smallest positive integer for which (TλI)p(x)=0. Then the ordered set

{ ( T λ I ) p 1 ( x ) ,   ( T λ I ) p 2 ( x ) ,   ,   ( T λ I ) ( x ) ,   x }

is called a cycle of generalized eigenvectors of T corresponding to λ. The vectors (TλI)p1(x) and x are called the initial vector and the end vector of the cycle, respectively. We say that the length of the cycle is p.

Notice that the initial vector of a cycle of generalized eigenvectors of a linear operator T is the only eigenvector of T in the cycle. Also observe that if x is an eigenvector of T corresponding to the eigenvalue λ, then the set {x} is a cycle of generalized eigenvectors of T corresponding to λ of length 1.

In Example 1, the subsets β1={v1, v2, v3}, β2={v4}, β3={v5, v6}, and β4={v7, v8} are the cycles of generalized eigenvectors of T that occur in β . Notice that β is a disjoint union of these cycles. Furthermore, setting Wi=span(βi) for i=1, 2, 3, 4, we see that βi is a basis for Wi and [TWi]βi is the ith Jordan block of the Jordan canonical form of T. This is precisely the condition that is required for a Jordan canonical basis.

Theorem 7.5.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits, and suppose that β is a basis for V such that β is a disjoint union of cycles of generalized eigenvectors of T. Then the following statements are true.

  1. For each cycle γ of generalized eigenvectors contained in β, W=span(γ) is T-invariant, and [TW]γ is a Jordan block.

  2. β is a Jordan canonical basis for V.

Proof.

(a) Suppose that γ corresponds to the eigenvalue λ, γ has length p, and x is the end vector of γ. Then γ={v1, v2, , vp}, where

v i = ( T λ I ) p i ( x )  for  i < p and v p = x .

So

( T λ I ) ( v 1 ) = ( T λ I ) p ( x ) = 0 ,

and hence T(v1)=λv1. For i>1,

( T λ I ) ( v i ) = ( T λ I ) p ( i 1 ) ( x ) = v i 1 .

This shows that TλI maps W into itself, and so T does also. Thus, by the preceding equations, we see that [TW]γ is a Jordan block.

For (b), simply repeat the arguments of (a) for each cycle in β in order to obtain [T]β. We leave the details as an exercise.

In view of this result, we must show that, under appropriate conditions, there exist bases that are disjoint unions of cycles of generalized eigenvectors. Since the characteristic polynomial of a Jordan canonical form splits, this is a necessary condition. We will soon see that it is also sufficient. The next result moves us toward the desired existence theorem.

Theorem 7.6.

Let T be a linear operator on a vector space V, and let λ be an eigenvalue of T. Suppose that γ1, γ2, , γq are cycles of generalized eigenvectors of T corresponding to λ such that the initial vectors of the γi’s are distinct and form a linearly independent set. Then the γi’s are disjoint, and their union γ=i=1qγi is linearly independent.

Proof.

Exercise 5 shows that the γi’s are disjoint.

The proof that γ is linearly independent is by mathematical induction on the number of vectors in γ. If this number is 1, then the result is clear. So assume that, for some integer n>1, the result is true whenever γ has fewer than n vectors, and suppose that γ has exactly n vectors. Let W be the subspace of V generated by γ. Clearly W is (TλI)-invariant, and dim(W)n. Let U denote the restriction of TλI to W.

For each i, let γi denote the cycle obtained from γi by deleting the end vector. Note that if γi has length one, then γi=. In the case that γi, each vector of γi is the image under U of a vector in γi, and conversely, every nonzero image under U of a vector of γi is contained in γi. Let γ=iγi.

Then by the last statement, γ generates R(U). Furthermore, γ consists of nq vectors, and the initial vectors of the γi’s are also initial vectors of the γi’s. Thus we may apply the induction hypothesis to conclude that γ is linearly independent. Therefore γ is a basis for R(U). Hence dim(R(U))=nq. Since the q initial vectors of the γi’s form a linearly independent set and lie in N(U), we have dim(N(U))q. From these inequalities and the dimension theorem, we obtain

n dim ( W ) = dim ( R ( U ) ) + dim ( N ( U ) ) ( n q ) + q = n .

We conclude that dim(W)=n. Since γ generates W and consists of n vectors, it must be a basis for W. Hence γ is linearly independent.

Corollary.

Every cycle of generalized eigenvectors of a linear operator is linearly independent.

Theorem 7.7.

Let T be a linear operator on a finite-dimensional vector space V, and let λ be an eigenvalue of T. Then Kλ has an ordered basis consisting of a union of disjoint cycles of generalized eigenvectors corresponding to λ.

Proof.

The proof is by mathematical induction on n=dim(Kλ). The result is clear for n=1. So suppose that for some integer n>1 the result is true whenever dim(Kλ)<n, and assume that dim(Kλ)=n. Let U denote the restriction of TλI to Kλ. Then R(U) is a subspace of Kλ of lesser dimension, and R(U) is the space of generalized eigenvectors corresponding to λ for the restriction of T to R(U). Therefore, by the induction hypothesis, there exist disjoint cycles γ1, γ2, , γq of generalized eigenvectors of this restriction, and hence of T itself, corresponding to λ for which γ=i=1qγi is a basis for R(U). For i=1, 2, , q, the end vector of γi is the image under U of a vector viKλ, and so we can extend each γi to a larger cycle γ˜i=γi{vi} of generalized eigenvectors of T corresponding to λ. For i=1, 2, , q, let wi be the initial vector of γ˜i (and hence of γi). Since {w1, w2, , wq} is a linearly independent subset of Eλ, this set can be extended to a basis {w1, w2, , wq, u1, u2, , us} for Eλ. Then γ˜1, γ˜2, , γ˜q, {u1}, {u2}, , {us} are disjoint cycles of generalized eigenvectors of T corresponding to λ such that the initial vectors of these cycles are linearly independent. Therefore their union γ˜ is a linearly independent subset of Kλ by Theorem 7.6.

Finally, we show that γ˜ is a basis for Kλ. Suppose that γ consists of r=rank(U) vectors. Then γ˜ consists of r+q+s vectors. Furthermore, since {w1, w2, , wq, u1, u2, , us} is a basis for Eλ=N(U), it follows that nullity(U)=q+s. Therefore

dim ( K λ ) = rank ( U ) + nullity ( U ) = r + q + s .

So γ˜ is a linearly independent subset of Kλ containing dim(Kλ) vectors, and thus γ˜ is a basis for Kλ.

Corollary 1.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits. Then T has a Jordan canonical form.

Proof.

Let λ1, λ2, , λk be the distinct eigenvalues of T. By Theorem 7.7, for each i there is an ordered basis βi consisting of a disjoint union of cycles of generalized eigenvectors corresponding to λi. Let β=β1β2βk. Then, by Theorem 7.5(b), β is a Jordan canonical basis for V.

The Jordan canonical form also can be studied from the viewpoint of matrices.

Definition.

Let AMn×n(F) be such that the characteristic polynomial of A (and hence of LA) splits. Then the Jordan canonical form of A is defined to be the Jordan canonical form of the linear operator LA on Fn.

The next result is an immediate consequence of this definition and Corollary 1.

Corollary 2.

Let A be an n×n matrix whose characteristic polynomial splits. Then A has a Jordan canonical form J, and A is similar to J.

Proof.

Exercise.

We can now compute the Jordan canonical forms of matrices and linear operators in some simple cases, as is illustrated in the next two examples. The tools necessary for computing the Jordan canonical forms in general are developed in the next section.

Example 2

Let

A = ( 3 1 2 1 0 5 1 1 4 ) M 3 × 3 ( R ) .

To find the Jordan canonical form for A, we need to find a Jordan canonical basis for T=LA.

The characteristic polynomial of A is

f ( t ) = det ( A t I ) = ( t 3 ) ( t 2 ) 2 .

Hence λ1=3 and λ2=2 are the eigenvalues of A with multiplicities 1 and 2, respectively. By Theorem 7.4, dim(Kλ1)=1, and dim(Kλ2)=2. By Theorem 7.2, Kλ1=N(T3I), and Kλ2=N((T2I)2). Since Eλ1=N(T3I), we have that Eλ1=Kλ1. Observe that (1, 2, 1) is an eigenvector of T corresponding to λ1=3; therefore

β 1 = { ( 1 2 1 ) }

is a basis for Kλ1.

Since dim(Kλ2)=2 and a generalized eigenspace has a basis consisting of a union of cycles, this basis is either a union of two cycles of length 1 or a single cycle of length 2. Since the rank of A2I is 2 it follows that dim(Eλ2)=1. Hence the former is not the case. It can easily be shown that w=(1, 3, 1) is an eigenvector of A corresponding to λ2. This vector can serve as the initial vector of the cycle. Any solution to the equation (A2I)v=w will serve as the end vector. For example, we can take v=(1, 2, 0). Thus

β 2 = { ( A 2 I ) v ,   v } = { ( 1 3 1 ) ,   ( 1 2 0 ) }

is a cycle of generalized eigenvectors for λ2=2. Finally, we take the union of these two bases to obtain

β = β 1 β 2 = { ( 1 2 1 ) ,   ( 1 3 1 ) ,   ( 1 2 0 ) } ,

which is a Jordan canonical basis for A. Therefore,

J = [ T ] β = ( 3 0 0 0 2 1 0 0 2 )

is a Jordan canonical form for A. Notice that A is similar to J. In fact, J=Q1AQ, where Q is the matrix whose columns are the vectors in β.

Example 3

Let T be the linear operator on P2(R) defined by T(g(x))=g(x)g(x). We find a Jordan canonical form of T and a Jordan canonical basis for T.

Let β be the standard ordered basis for P2(R). Then

[ T ] β = ( 1 1 0 0 1 2 0 0 1 ) ,

which has the characteristic polynomial f(t)=(t+1)3. Thus λ=1 is the only eigenvalue of T, and hence Kλ=P2(R) by Theorem 7.4. So β is a basis for Kλ. Now

dim ( E λ ) = 3 rank ( A + I ) = 3 rank ( 0 1 0 0 0 2 0 0 0 ) = 3 2 = 1.

Therefore a basis for Kλ cannot be a union of two or three cycles because the initial vector of each cycle is an eigenvector, and there do not exist two or more linearly independent eigenvectors. So the desired basis must consist of a single cycle of length 3. If γ is such a cycle, then γ determines a single Jordan block

[ T ] γ = ( 1 1 0 0 1 1 0 0 1 ) ,

which is a Jordan canonical form of T.

The end vector h(x) of such a cycle must satisfy (T+I)2(h(x))0. In any basis for Kλ, there must be a vector that satisfies this condition, or else no vector in Kλ satisfies this condition, contrary to our reasoning. Testing the vectors in β, we see that h(x)=x2 is acceptable. Therefore

γ = { ( T + I ) 2 ( x 2 ) ,   ( T + I ) ( x 2 ) ,   x 2 } = { 2 ,   2 x ,   x 2 }

is a Jordan canonical basis for T.

In the next section, we develop a computational approach for finding a Jordan canonical form and a Jordan canonical basis. In the process, we prove that Jordan canonical forms are unique up to the order of the Jordan blocks.

Let T be a linear operator on a finite-dimensional vector space V, and suppose that the characteristic polynomial of T splits. By Theorem 5.10 (p. 277), T is diagonalizable if and only if V is the direct sum of the eigenspaces of T. If T is diagonalizable, then the eigenspaces and the generalized eigenspaces coincide.

For those familiar with the material on direct sums in Section 5.2, we conclude by stating a generalization of Theorem 5.10 for nondiagonalizable operators.

Theorem 7.8.

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits. Then V is the direct sum of the generalized eigenspaces of T.

Proof.

Exercise.

Exercises

  1. Label the following statements as true or false.

    1. Eigenvectors of a linear operator T are also generalized eigenvectors of T.

    2. It is possible for a generalized eigenvector of a linear operator T to correspond to a scalar that is not an eigenvalue of T.

    3. Any linear operator on a finite-dimensional vector space has a Jordan canonical form.

    4. A cycle of generalized eigenvectors is linearly independent.

    5. There is exactly one cycle of generalized eigenvectors corresponding to each eigenvalue of a linear operator on a finite-dimensional vector space.

    6. Let T be a linear operator on a finite-dimensional vector space whose characteristic polynomial splits, and let λ1, λ2, , λk be the distinct eigenvalues of T. If, for each i, βi is a basis for Kλi, then β1β2βk is a Jordan canonical basis for T.

    7. For any Jordan block J, the operator LJ has Jordan canonical form J.

    8. Let T be a linear operator on an n-dimensional vector space whose characteristic polynomial splits. Then, for any eigenvalue λ of T, Kλ=N((TλI)n).

  2. For each matrix A, find a basis for each generalized eigenspace of LA consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form J of A.

    1. A=(1113)

    2. A=(1232)

    3. A=(114521811310)

    4. A=(2100021000300113)

  3. For each linear operator T, find a basis for each generalized eigenspace of T consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form J of T.

    1. Define T on P2(R) by T(f(x))=2f(x)f(x).

    2. V is the real vector space of functions spanned by the set of real-valued functions {1, t, t2, et, tet}, and T is the linear operator on V defined by T(f)=f.

    3. T is the linear operator on M2×2(R) defined for all AM2×2(R) by T(A)=BA, where B=(1101). 1

    4. T(A)=2A+At for all AM2×2(R).

  4. † Let T be a linear operator on a vector space V, and let γ be a cycle of generalized eigenvectors that corresponds to the eigenvalue λ. Prove that span(γ) is a T-invariant subspace of V. Visit goo.gl/Lw4ahY for a solution.

  5. Let γ1, γ2, , γp be cycles of generalized eigenvectors of a linear operator T corresponding to an eigenvalue λ. Prove that if the initial eigenvectors are distinct, then the cycles are disjoint.

  6. Let T: T: VW be a linear transformation. Prove the following results.

    1. N(T)=N(T).

    2. N(Tk)=N((T)k)..

    3. If V=W (so that T is a linear operator on V) and λ is an eigenvalue of T, then for any positive integer k

      N ( ( T λ I V ) k ) = N ( ( λ I V T ) k ) .
  7. Let U be a linear operator on a finite-dimensional vector space V. Prove the following results.

    1. N(U)N(U2)N(Uk)N(Uk+1).

    2. If rank(Um)=rank(Um+1) for some positive integer m, then rank(Um)=rank(Uk) for any positive integer km.

    3. If rank(Um)=rank(Um+1) for some positive integer m, then N(Um)=N(Uk) for any positive integer km.

    4. Let T be a linear operator on V, and let λ be an eigenvalue of T. Prove that if rank((TλI)m)=rank((TλI)m+1) for some integer m, then Kλ=N((TλI)m).

    5. Second Test for Diagonalizability. Let T be a linear operator on V whose characteristic polynomial splits, and let λ1, λ2, , λk be the distinct eigenvalues of T. Then T is diagonalizable if and only if rank(TλiI)=rank((TλiI)2) for 1ik.

    6. Use (e) to obtain a simpler proof of Exercise 24 of Section 5.4: If T is a diagonalizable linear operator on a finite-dimensional vector space V and W is a T-invariant subspace of V, then TW is diagonalizable.

  8. Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits and has Jordan canonical form J. Prove that for any nonzero scalar c, cJ is a Jordan canonical form for cT.

  9. Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomial splits.

    1. Prove Theorem 7.5(b).

    2. Suppose that β is a Jordan canonical basis for T, and let λ be an eigenvalue of T. Let β=βKλ. Prove that β is a basis for Kλ.

  10. Let T be a linear operator on a finite-dimensional vector space whose characteristic polynomial splits, and let λ be an eigenvalue of T.

    1. Suppose that γ is a basis for Kλ consisting of the union of q disjoint cycles of generalized eigenvectors. Prove that qdim(Eλ).

    2. Let β be a Jordan canonical basis for T, and suppose that J=[T]β has q Jordan blocks with λ in the diagonal positions. Prove that qdim(Eλ).

  11. Prove Corollary 2 to Theorem 7.7.

  12. Let T be a linear operator on a finite-dimensional vector space V, and let λ be an eigenvalue of T with corresponding eigenspace and generalized eigenspace Eλ and Kλ, respectively. Let U be an invertible linear operator on V that commutes with T (i.e., TU=UT). Prove that U(Eλ)=Eλ and U(Kλ)=Kλ.

Exercises 13 and 14 are concerned with direct sums of matrices, defined in Section 5.4 on page 318.

  1. Prove Theorem 7.8.

  2. Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits, and let λ1, λ2, , λk be the distinct eigenvalues of T. For each i, let Ji be the Jordan canonical form of the restriction of T to Kλi. Prove that

    J = J 1 J 2 J k

    is the Jordan canonical form of J.