# 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  for V such that



where each O is a zero matrix, and each  is a square matrix of the form  or



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

# Example 1

Suppose that T is a linear operator on , and  is an ordered basis for  such that

is a Jordan canonical form of T. Notice that the characteristic polynomial of T is , and hence the multiplicity of each eigenvalue is the number of times that the eigenvalue appears on the diagonal of J. Also observe that , and  are the only vectors in  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  be the linear operator on  such that , where  is the ordered basis in Example 1 and



Then the characteristic polynomial of  is also . But the operator  has the Jordan canonical form , 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  of Example 1. Notice that , and so  . Similarly, . Since  and  are eigenvectors of T corresponding to , it follows that  for  and 4. Similarly  for , and  for .

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  Jordan block with diagonal entry , then . Eigenvectors satisfy this condition for .

# Definition.

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

Notice that if x is a generalized eigenvector of T corresponding to , and p is the smallest positive integer for which , then  is an eigenvector of T corresponding to . Therefore  is an eigenvalue of T.

In the context of Example 1, each vector in  is a generalized eigenvector of T. In fact, , and  correspond to the eigenvalue 2,  and  correspond to the eigenvalue 3, and  and  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  be an eigenvalue of T. The generalized eigenspace of T corresponding to , denoted , is the subset of V defined by



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

Recall that a subspace W of V is T-invariant for a linear operator T if . 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  be an eigenvalue of T. Then

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

2. For any eigenvalue  of T such that ,


3. For any scalar , the restriction of  to  is one-to-one and onto.

# Proof.

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



Therefore



and hence . The proof that  is closed under scalar multiplication is straightforward.

To show that  is T-invariant, consider any . Choose a positive integer p such that . Then



Therefore .

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

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



As polynomials in  and  are relatively prime, and hence by Theorem E.2 in Appendix E there exist polymomials  and  such that



It follows that



and hence .

(c) Let  be any scalar such that . Since  is T-invariant, it is also -invariant. If  for some , then . Hence  by (b), and we conclude that the restriction of  to is one-to-one.

We now show that the restriction of  to  is onto. Let , p be the smallest positive integer for which , and



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

# 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  is an eigenvalue of T with multiplicity m. Then

1. 

2. 

# Proof.

(a) Let , and let h(t) be the characteristic polynomial of . By Theorem 5.20 (p. 312), h(t) divides the characteristic polynomial of T, and by Theorem 7.1(b),  is the only eigenvalue of . Hence , where , and .

(b) Clearly . The characteristic polynomial of T is of the form , where g(t) is a product of powers of the form  for eigenvalues . By Theorem 7.1, g(T) is one-to one on . Since  is finite-dimensional, it is also onto. Let . Then there exists  such that . Hence



by the Cayley-Hamilton Theorem. Therefore , and thus .

# 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  be the distinct eigenvalues of T. Then, for every , there exist unique vectors , for , such that



# Proof.

Suppose the characteristic polynomial of T is



For , let



First, we prove the existence of the vectors . Because the polynomials  are relatively prime for , by Theorem E.2 there exist polynomials  for  such that



It follows that



For , let . Then for each i,



by the Cayley-Hamilton theorem. It follows that  for .

Now we prove the uniqueness of the vectors . Since each  is T-invariant, each  is invariant under the operator  for . Furthermore,  is one-to-one on  by Theorem 7.1(c) and maps  to {0} for  by Theorem 7.2(b).

Suppose that



where  for . Then for ,



and hence  since  for . Because  is one-to-one on , it follows that .

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  splits. For , let  be an ordered basis for . Then

1.  for .

2.  is an ordered basis for V.

3.  for all i.

# Proof.

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

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



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

We now prove that  is a spanning set for V. Consider any vector . By Theorem 7.3, for  there exist vectors  such that . Since each  is a linear combination of the vectors in , v is a linear combination of the vectors in . Hence  spans V.

(c) Since dim(V) is equal to the degree of the characteristic polynomial of T, it follows that . By (b), . Hence . By Theorem 7.2(a),  for all i. Consequently,  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  for every eigenvalue  of T.

# Proof.

Combining Theorems 7.4 and 5.8(a) (p. 267), we see that T is diagonalizable if and only if  for each eigenvalue  of T. But , 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  of Example 1. We have seen that the first four vectors of  lie in the generalized eigenspace . Observe that the vectors in  that determine the first Jordan block of J are of the form



Furthermore, observe that . 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 . Suppose that p is the smallest positive integer for which . Then the ordered set



is called a cycle of generalized eigenvectors of T corresponding to . The vectors  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 , then the set  is a cycle of generalized eigenvectors of T corresponding to  of length 1.

In Example 1, the subsets , and  are the cycles of generalized eigenvectors of T that occur in  . Notice that  is a disjoint union of these cycles. Furthermore, setting  for , we see that  is a basis for  and  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  is a basis for V such that  is a disjoint union of cycles of generalized eigenvectors of T. Then the following statements are true.

1. For each cycle  of generalized eigenvectors contained in  is T-invariant, and  is a Jordan block.

2.  is a Jordan canonical basis for V.

# Proof.

(a) Suppose that  corresponds to the eigenvalue  has length p, and x is the end vector of . Then , where



So



and hence . For ,



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

For (b), simply repeat the arguments of (a) for each cycle in  in order to obtain . 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  be an eigenvalue of T. Suppose that  are cycles of generalized eigenvectors of T corresponding to  such that the initial vectors of the ’s are distinct and form a linearly independent set. Then the ’s are disjoint, and their union  is linearly independent.

# Proof.

Exercise 5 shows that the ’s are disjoint.

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

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

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



We conclude that . Since  generates W and consists of n vectors, it must be a basis for W. Hence  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  be an eigenvalue of T. Then  has an ordered basis consisting of a union of disjoint cycles of generalized eigenvectors corresponding to .

# Proof.

The proof is by mathematical induction on . The result is clear for . So suppose that for some integer  the result is true whenever , and assume that . Let U denote the restriction of  to . Then R(U) is a subspace of  of lesser dimension, and R(U) is the space of generalized eigenvectors corresponding to  for the restriction of T to R(U). Therefore, by the induction hypothesis, there exist disjoint cycles  of generalized eigenvectors of this restriction, and hence of T itself, corresponding to  for which  is a basis for R(U). For , the end vector of  is the image under U of a vector , and so we can extend each  to a larger cycle  of generalized eigenvectors of T corresponding to . For , let  be the initial vector of  (and hence of ). Since  is a linearly independent subset of , this set can be extended to a basis  for . Then  are disjoint cycles of generalized eigenvectors of T corresponding to  such that the initial vectors of these cycles are linearly independent. Therefore their union  is a linearly independent subset of  by Theorem 7.6.

Finally, we show that  is a basis for . Suppose that  consists of  vectors. Then  consists of  vectors. Furthermore, since  is a basis for , it follows that . Therefore



So  is a linearly independent subset of  containing  vectors, and thus  is a basis for .

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

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

# Definition.

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

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

# Corollary 2.

Let A be an  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



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

The characteristic polynomial of A is



Hence  and  are the eigenvalues of A with multiplicities 1 and 2, respectively. By Theorem 7.4, , and . By Theorem 7.2, , and . Since , we have that . Observe that  is an eigenvector of T corresponding to ; therefore



is a basis for .

Since  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  is 2 it follows that . Hence the former is not the case. It can easily be shown that  is an eigenvector of A corresponding to . This vector can serve as the initial vector of the cycle. Any solution to the equation  will serve as the end vector. For example, we can take . Thus



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



which is a Jordan canonical basis for A. Therefore,



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

# Example 3

Let T be the linear operator on  defined by . We find a Jordan canonical form of T and a Jordan canonical basis for T.

Let  be the standard ordered basis for . Then



which has the characteristic polynomial . Thus  is the only eigenvalue of T, and hence  by Theorem 7.4. So  is a basis for . Now



Therefore a basis for  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  is such a cycle, then  determines a single Jordan block



which is a Jordan canonical form of T.

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



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.

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  be the distinct eigenvalues of T. If, for each  is a basis for , then  is a Jordan canonical basis for T.

7. For any Jordan block J, the operator  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  of .

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

1. 

2. 

3. 

4. 

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  by .

2. V is the real vector space of functions spanned by the set of real-valued functions , and T is the linear operator on V defined by .

3. T is the linear operator on  defined for all  by , where . 1

4.  for all .

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

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

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

1. 

2. 

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


7. Let U be a linear operator on a finite-dimensional vector space V. Prove the following results.

1. 

2. If  for some positive integer m, then  for any positive integer .

3. If  for some positive integer m, then  for any positive integer .

4. Let T be a linear operator on V, and let  be an eigenvalue of T. Prove that if  for some integer m, then .

5. Second Test for Diagonalizability. Let T be a linear operator on V whose characteristic polynomial splits, and let  be the distinct eigenvalues of T. Then T is diagonalizable if and only if  for .

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  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  is a Jordan canonical basis for T, and let  be an eigenvalue of T. Let . Prove that  is a basis for .

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

1. Suppose that  is a basis for  consisting of the union of q disjoint cycles of generalized eigenvectors. Prove that .

2. Let  be a Jordan canonical basis for T, and suppose that  has q Jordan blocks with  in the diagonal positions. Prove that .

11. Prove Corollary 2 to Theorem 7.7.

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

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  be the distinct eigenvalues of T. For each i, let  be the Jordan canonical form of the restriction of T to . Prove that



is the Jordan canonical form of J.