# 5.2 Diagonalizability

In Section 5.1, we presented the diagonalization problem and observed that not all linear operators or matrices are diagonalizable. Although we are able to diagonalize operators and matrices and even obtain a necessary and sufficient condition for diagonalizability (Theorem 5.1 p. 247), we have not yet solved the diagonalization problem. What is still needed is a simple test to determine whether an operator or a matrix can be diagonalized, as well as a method for actually finding a basis of eigenvectors. In this section, we develop such a test and method.

In Example 6 of Section 5.1, we obtained a basis of eigenvectors by choosing one eigenvector corresponding to each eigenvalue. In general, such a procedure does not yield a basis, but the following theorem shows that any set constructed in this manner is linearly independent.

# Theorem 5.5.

Let T be a linear operator on a vector space, and let  be distinct eigenvalues of T. For each , let  be a finite set of eigenvectors of T corresponding to . If each , is linearly independent, then  is linearly independent.

# Proof.

The proof is by mathematical induction on k. If , there is nothing to prove. So assume that the theorem holds for  distinct eigenvalues, where , and that we have k distinct eigenvalues  of T. For each , let  be a linearly independent set of eigenvectors of T corresponding to . We wish to show that  is linearly independent.

Consider any scalars , where  and , such that



Because  is an eigenvector of T corresponding to , applying  to both sides of (1) yields



But  is linearly independent by the induction hypothesis, so that (2) implies  for  and . Since  are distinct, it follows that  for . Hence  for  and , and therefore (1) reduces to . But  is also linearly independent, and so  for . Consequently  for  and , proving that S is linearly independent.

# Corollary.

Let T be a linear operator on an n-dimensional vector space V. If T has n distinct eigenvalues, then T is diagonalizable.

# Proof.

Suppose that T has n distinct eigenvalues . For each i choose an eigenvector  corresponding to . By Theorem 5.5,  is linearly independent, and since , this set is a basis for V. Thus, by Theorem 5.1 (p. 247), T is diagonalizable.

# Example 1

Let



The characteristic polynomial of A (and hence of ) is



and thus the eigenvalues of  are 0 and 2. Since  is a linear operator on the two-dimensional vector space , we conclude from the preceding corollary that  (and hence A) is diagonalizable.

The converse of the corollary to Theorem 5.5 is false. That is, it is not true that if T is diagonalizable, then it has n distinct eigenvalues. For example, the identity operator is diagonalizable even though it has only one eigenvalue, namely, .

We have seen that diagonalizability requires the existence of eigenvalues. Actually, diagonalizability imposes a stronger condition on the characteristic polynomial.

# Definition.

A polynomial f(t) in P(F) splits over F if there are scalars  (not necessarily distinct) in F such that



For example,  splits over R, but  does not split over R because  cannot be factored into a product of linear factors. However,  does split over C because it factors into the product . If f(t) is the characteristic polynomial of a linear operator or a matrix over a field F, then the statement that f(t) splits is understood to mean that it splits over F.

# Theorem 5.6.

The characteristic polynomial of any diagonalizable linear operator on a vector space V over a field F splits over F.

# Proof.

Let T be a diagonalizable linear operator on the n-dimensional vector space V, and let  be an ordered basis for V such that  is a diagonal matrix. Suppose that



and let f (t) be the characteristic polynomial of T. Then



From this theorem, it is clear that if T is a diagonalizable linear operator on an n-dimensional vector space that fails to have n distinct eigenvalues, then the characteristic polynomial of T must have repeated zeros.

The converse of Theorem 5.6 is false; that is, the characteristic polynomial of T may split, but T need not be diagonalizable. (See Example 3, which follows.) The following concept helps us determine when an operator whose characteristic polynomial splits is diagonalizable.

# Definition.

Let  be an eigenvalue of a linear operator or matrix with characteristic polynomial f(t). The multiplicity (sometimes called the algebraic multiplicity) of  is the largest positive integer k for which  is a factor of f(t).

# Example 2

Let



which has characteristic polynomial . Hence  is an eigenvalue of A with multiplicity 2, and  is an eigenvalue of A with multiplicity 1.

If T is a diagonalizable linear operator on a finite-dimensional vector space V, then there is an ordered basis  for V consisting of eigenvectors of T. We know from Theorem 5.1 (p. 247) that  is a diagonal matrix in which the diagonal entries are the eigenvalues of T. Since the characteristic polynomial of T is , it is easily seen that each eigenvalue of T must occur as a diagonal entry of  exactly as many times as its multiplicity. Hence  contains as many (linearly independent) eigenvectors corresponding to an eigenvalue as the multiplicity of that eigenvalue. So the number of linearly independent eigenvectors corresponding to a given eigenvalue is of interest in determining whether an operator can be diagonalized. Recalling from Theorem 5.4 (p. 250) that the eigenvectors of T corresponding to the eigenvalue  are the nonzero vectors in the null space of , we are led naturally to the study of this set.

# Definition.

Let T be a linear operator on a vector space V, and let  be an eigenvalue of T. Define . The set  is called the eigenspace of T corresponding to the eigenvalue . Analogously, we define the eigenspace of a square matrix A corresponding to the eigenvalue  to be the eigenspace of  corresponding to .

Clearly,  is a subspace of V consisting of the zero vector and the eigenvectors of T corresponding to the eigenvalue . The maximum number of linearly independent eigenvectors of T corresponding to the eigenvalue  is therefore the dimension of . Our next result relates this dimension to the multiplicity of .

# Theorem 5.7.

Let T be a linear operator on a finite-dimensional vector space V, and let  be an eigenvalue of T having multiplicity m. Then .

# Proof.

Choose an ordered basis  for , extend it to an ordered basis  for V, and let . Observe that  is an eigenvector of T corresponding to , and therefore



By Exercise 21 of Section 4.3, the characteristic polynomial of T is



where g(t) is a polynomial. Thus  is a factor of f(t), and hence the multiplicity of  is at least p. But , and so .

# Example 3

Let T be the linear operator on  defined by . The matrix representation of T with respect to the standard ordered basis  for  is



Consequently, the characteristic polynomial of T is



Thus T has only one eigenvalue  with multiplicity 3. Solving  shows that  is the subspace of  consisting of the constant polynomials. So {1} is a basis for , and therefore . Consequently, there is no basis for  consisting of eigenvectors of T, and therefore T is not diagonalizable. Even though T is not diagonalizable, we will see in Chapter 7 that its eigenvalue and eigenvectors are still useful for describing the behavior of T.

# Example 4

Let T be the linear operator on  defined by



We determine the eigenspace of T corresponding to each eigenvalue. Let  be the standard ordered basis for . Then



and hence the characteristic polynomial of T is



So the eigenvalues of T are  and  with multiplicities 1 and 2, respectively.

Since



 is the solution space of the system of linear equations



It is easily seen (using the techniques of Chapter 3) that



is a basis for . Hence .

Similarly,  is the solution space of the system



Since the unknown  does not appear in this system, we assign it a parametric value, say, , and solve the system for  and , introducing another parameter t. The result is the general solution to the system



It follows that



is a basis for , and .

In this case, the multiplicity of each eigenvalue  is equal to the dimension of the corresponding eigenspace . Observe that the union of the two bases just derived, namely,



is linearly independent by Theorem 5.5 and hence is a basis for  consisting of eigenvectors of T. Consequently, T is diagonalizable.

Examples 3 and 4 suggest that an operator on V whose characteristic polynomial splits is diagonalizable if and only if the dimension of each eigenspace is equal to the multiplicity of the corresponding eigenvalue. This is indeed true, as our next theorem shows. Moreover, when the operator is diagonalizable, we can use Theorem 5.5 to construct a basis for V consisting of eigenvectors of the operator by collecting bases for the individual eigenspaces.

# Theorem 5.8.

Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Let  be the distinct eigenvalues of T. Then

1. (a) T is diagonalizable if and only if the multiplicity of  is equal to  for all i.

2. (b) If T is diagonalizable and  is an ordered basis for  for each i, then  is an ordered basis2 for V consisting of eigenvectors of T.

# Proof.

For each i, let  denote the multiplicity of , and .

First, suppose that T is diagonalizable. Let  be a basis for V consisting of eigenvectors of T. For each i, let , the set of vectors in  that are eigenvectors corresponding to , and let  denote the number of vectors in . Then  for each i because  is a linearly independent subset of a subspace of dimension , and  by Theorem 5.7. The ’s sum to n because  contains n vectors. The ’s also sum to n because the degree of the characteristic polynomial of T is equal to the sum of the multiplicities of the eigenvalues. Thus



It follows that



Since  for all i, we conclude that  for all i.

Conversely, suppose that  for all i. We simultaneously show that T is diagonalizable and prove (b). For each i, let  be an ordered basis for , and let . By Theorem 5.5,  is linearly independent. Furthermore, since  for all i,  contains



vectors. Therefore  is an ordered basis for V consisting of eigenvectors of V, and we conclude that T is diagonalizable.

This theorem completes our study of the diagonalization problem. We summarize our results.

# Test for Diagonalizability

Let T be a linear operator on an n-dimensional vector space V. Then T is diagonalizable if and only if both of the following conditions hold.

1. The characteristic polynomial of T splits.

2. For each eigenvalue  of T, the multiplicity of  equals , that is, the multiplicity of  equals .

These same conditions can be used to test if a square matrix A is diagonalizable because diagonalizability of A is equivalent to diagonalizability of the operator .

If T is a diagonalizable operator and  are ordered bases for the eigenspaces of T, then the union  is an ordered basis for V consisting of eigenvectors of T, and hence  is a diagonal matrix.

When testing T for diagonalizability, it is usually easiest to choose a convenient basis  for V and work with . If the characteristic polynomial of B splits, then use condition 2 above to check if the multiplicity of each repeated eigenvalue of B equals . (By Theorem 5.7, condition 2 is automatically satisfied for eigenvalues of multiplicity 1.) If so, then B, and hence T, is diagonalizable.

If T is diagonalizable and a basis  for V consisting of eigenvectors of T is desired, then we first find a basis for each eigenspace of B. The union of these bases is a basis  for  consisting of eigenvectors of B. Each vector in  is the coordinate vector relative to  of an eigenvector of T. The set consisting of these n eigenvectors of T is the desired basis .

Furthermore, if A is an  diagonalizable matrix, we can use the corollary to Theorem 2.23 (p. 115) to find an invertible  matrix Q and a diagonal  matrix D such that . The matrix Q has as its columns the vectors in a basis of eigenvectors of A, and D has as its j th diagonal entry the eigenvalue of A corresponding to the j th column of Q.

We now consider some examples illustrating the preceding ideas.

# Example 5

We test the matrix



for diagonalizability.

The characteristic polynomial of A is , which splits, and so condition 1 of the test for diagonalization is satisfied. Also A has eigenvalues  and  with multiplicities 1 and 2, respectively. Since  has multiplicity 1, condition 2 is satisfied for . Thus we need only test condition 2 for . Because



has rank 2, we see that , which is not the multiplicity of . Thus condition 2 fails for , and A is therefore not diagonalizable.

# Example 6

Let T be the linear operator on  defined by



We first test T for diagonalizability. Let α denote the standard ordered basis for  and . Then



The characteristic polynomial of B, and hence of T, is , which splits. Hence condition 1 of the test for diagonalization is satisfied. Also B has the eigenvalues  and  with multiplicities 2 and 1, respectively. Condition 2 is satisfied for  because it has multiplicity 1. So we need only verify condition 2 for . For this case,



which is equal to the multiplicity of . Therefore T is diagonalizable.

We now find an ordered basis  for  consisting of eigenvectors of B. We consider each eigenvalue separately.

The eigenspace corresponding to  is



which is the solution space for the system



and has



as a basis.

The eigenspace corresponding to  is



which is the solution space for the system



and has



as a basis.

Let



Then  is an ordered basis for  consisting of eigenvectors of B.

Finally, observe that the vectors in  are the coordinate vectors relative to  of the vectors in the set



which is an ordered basis for  consisting of eigenvectors of T. Thus



Our next example is an application of diagonalization that is of interest in Section 5.3.

# Example 7

Let



We show that A is diagonalizable and find a  matrix Q such that  is a diagonal matrix. We then show how to use this result to compute  for any positive integer n.

First observe that the characteristic polynomial of A is , and hence A has two distinct eigenvalues,  and . By applying the corollary to Theorem 5.5 to the operator , we see that A is diagonalizable. Moreover,



are bases for the eigenspaces  and , respectively. Therefore



is an ordered basis for  consisting of eigenvectors of A. Let



the matrix whose columns are the vectors in . Then, by the corollary to Theorem 2.23 (p. 115),



To find  for any positive integer n, observe that . Therefore



We now consider an application that uses diagonalization to solve a system of differential equations.

# Systems of Differential Equations

Consider the system of differential equations



where, for each i,  is a differentiable real-valued function of the real variable t. Clearly, this system has a solution, namely, the solution in which each  is the zero function. We determine all of the solutions to this system.

Let  be the function defined by



The derivative of x, denoted , is defined by



Let



be the coefficient matrix of the given system, so that we can rewrite the system as the matrix equation .

It can be verified that for



we have . Substitute  into  to obtain  or, equivalently, . The function  defined by  can be shown to be differentiable, and  (see Exercise 17). Hence the original system can be written as .

Since D is a diagonal matrix, the system  is easy to solve. Setting



we can rewrite  as



The three equations



are independent of each other, and thus can be solved individually. It is easily seen (as in Example 3 of Section 5.1) that the general solution to these equations is , and , where , and  are arbitrary constants. Finally,



yields the general solution of the original system. Note that this solution can be written as



The expressions in brackets are arbitrary vectors in  and , respectively, where  and . Thus the general solution of the original system is , where  and . This result is generalized in Exercise 16.

# Direct Sums*

Let T be a linear operator on a finite-dimensional vector space V. There is a way of decomposing V into simpler subspaces that offers insight into the behavior of T. This approach is especially useful in Chapter 7, where we study nondiagonalizable linear operators. In the case of diagonalizable operators, the simpler subspaces are the eigenspaces of the operator.

# Definition.

Let  be subspaces of a vector space V. We define the sum of these subspaces to be the set



which we denote by  or .

It is a simple exercise to show that the sum of subspaces of a vector space is also a subspace.

# Example 8

Let , let  denote the xy-plane, and let  denote the yz-plane. Then  because, for any vector , we have



where  and .

Notice that in Example 8 the representation of (a, b, c) as a sum of vectors in  and  is not unique. For example,  is another representation. Because we are often interested in sums for which representations are unique, we introduce a condition that assures this outcome. The definition of direct sum that follows is a generalization of the definition given in the exercises of Section 1.3.

# Definition.

Let  be subspaces of a vector space V such that  for . We call W the direct sum of the subspaces , and write , if



# Example 9

In , and . For any 



Thus



To show that W is the direct sum of , and , we must prove that . But these equalities are obvious, and so .

Our next result contains several conditions that are equivalent to the definition of a direct sum.

# Theorem 5.9.

Let  be subspaces of a finite-dimensional vector space V. The following conditions are equivalent.

1. (a) 

2. (b)  and, for any vectors  such that  , if , then  for all i.

3. (c) Each vector  can be uniquely written as , where .

4. (d) If  is an ordered basis for , then  is an ordered basis for V.

5. (e) For each , there exists an ordered basis  for  such that  is an ordered basis for V.

# Proof.

Assume (a). We prove (b). Clearly



Now suppose that  are vectors such that  for all i and . Then for any j



But  and hence



So , proving (b).

Now assume (b). We prove (c). Let . By (b), there exist vectors  such that  and . We must show that this representation is unique. Suppose also that , where  for all i. Then



But  for all i, and therefore  for all i by (b). Thus  for all i, proving the uniqueness of the representation.

Now assume (c). We prove (d). For each i, let  be an ordered basis for . Since



by (c), it follows that  generates V. To show that this set is linearly independent, consider vectors  and  and scalars  such that



For each i, set



Then for each i,  and



Since  for each i and , (c) implies that  for all i. Thus



for each i. But each  is linearly independent, and hence  for all i and j. Consequently  is linearly independent and therefore is a basis for V.

Clearly (e) follows immediately from (d).

Finally, we assume (e) and prove (a). For each i, let  be an ordered basis for  such that  is an ordered basis for V. Then



by repeated applications of Exercise 14 of Section 1.4. Fix , and suppose that, for some nonzero vector ,



Then



Hence v is a nontrivial linear combination of both  and , so that v can be expressed as a linear combination of  in more than one way. But these representations contradict Theorem 1.8 (p. 44), and so we conclude that



proving (a).

With the aid of Theorem 5.9, we are able to characterize diagonalizability in terms of direct sums.

# Theorem 5.10.

A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if V is the direct sum of the eigenspaces of T.

# Proof.

Let  be the distinct eigenvalues of T.

First suppose that T is diagonalizable, and for each i choose an ordered basis  for the eigenspace . By Theorem 5.8,  is a basis for V, and hence V is a direct sum of the  by Theorem 5.9.

Conversely, suppose that V is a direct sum of the eigenspaces of T. For each i, choose an ordered basis  of . By Theorem 5.9, the union  is a basis for V. Since this basis consists of eigenvectors of T, we conclude that T is diagonalizable.

# Example 10

Let T be the linear operator on  defined by



It is easily seen that T is diagonalizable with eigenvalues , and . Furthermore, the corresponding eigenspaces coincide with the subspaces , and  of Example 9. Thus Theorem 5.10 provides us with another proof that .

# Exercises

1. Label the following statements as true or false.

1. (a) Any linear operator on an n-dimensional vector space that has fewer than n distinct eigenvalues is not diagonalizable.

2. (b) Two distinct eigenvectors corresponding to the same eigenvalue are always linearly dependent.

3. (c) If  is an eigenvalue of a linear operator T, then each vector in  is an eigenvector of T.

4. (d) If  and  are distinct eigenvalues of a linear operator T, then .

5. (e) Let  and  be an ordered basis for  consisting of eigenvectors of A. If Q is the  matrix whose j th column is , then  is a diagonal matrix.

6. (f) A linear operator T on a finite-dimensional vector space is diagonalizable if and only if the multiplicity of each eigenvalue  equals the dimension of .

7. (g) Every diagonalizable linear operator on a nonzero vector space has at least one eigenvalue.

The following two items relate to the optional subsection on direct sums.

8. (h) If a vector space is the direct sum of subspaces , then  for .

9. (i) If



then .

2. For each of the following matrices , test A for diagonalizability, and if A is diagonalizable, find an invertible matrix Q and a diagonal matrix D such that .

1. (a) 

2. (b) 

3. (c) 

4. (d) 

5. (e) 

6. (f) 

7. (g) 

3. For each of the following linear operators T on a vector space V, test T for diagonalizability, and if T is diagonalizable, find a basis  for V such that  is a diagonal matrix.

1. (a)  and T is defined by .

2. (b)  and T is defined by .

3. (c)  and T is defined by


4. (d)  and T is defined by .

5. (e)  and T is defined by .

6. (f)  and T is defined by .

4. Prove the matrix version of the corollary to Theorem 5.5: If  has n distinct eigenvalues, then A is diagonalizable.

5. State and prove the matrix version of Theorem 5.6.

1. (a) Justify the test for diagonalizability and the method for diagonalization stated in this section.

2. (b) Formulate the results in (a) for matrices.

6. For



find an expression for , where n is an arbitrary positive integer.

7. Suppose that  has two distinct eigenvalues,  and , and that . Prove that A is diagonalizable.

8. Let T be a linear operator on a finite-dimensional vector space V, and suppose there exists an ordered basis  for V such that  is an upper triangular matrix.

1. (a) Prove that the characteristic polynomial for T splits.

2. (b) State and prove an analogous result for matrices.

The converse of (a) is treated in Exercise 12(b).

9. Let T be a linear operator on a finite-dimensional vector space V with the distinct eigenvalues  and corresponding multiplicities . Suppose that  is a basis for V such that  is an upper triangular matrix. Prove that the diagonal entries of  are  and that each  occurs  times .

10. Let A be an  matrix that is similar to an upper triangular matrix and has the distinct eigenvalues  with corresponding multiplicities . Prove the following statements.

1. (a) 

2. (b) .

1. (a) Prove that if  and the characteristic polynomial of A splits, then A is similar to an upper triangular matrix. (This proves the converse of Exercise 9(b).) Hint: Use mathematical induction on n. For the general case, let  be an eigenvector of A, and extend  to a basis  for . Let P be the  matrix whose jth column is , and consider . Exercise 13(a) in Section 5.1 and Exercise 21 in Section 4.3 can be helpful.

2. (b) Prove the converse of Exercise 9(a).

Visit goo.gl/gJSjRU for a solution.

11. Let T be an invertible linear operator on a finite-dimensional vector space V.

1. (a) Recall that for any eigenvalue  of T,  is an eigenvalue of  (Exercise 9 of Section 5.1). Prove that the eigenspace of T corresponding to  is the same as the eigenspace of  corresponding to .

2. (b) Prove that if T is diagonalizable, then  is diagonalizable.

12. Let . Recall from Exercise 15 of Section 5.1 that A and  have the same characteristic polynomial and hence share the same eigenvalues with the same multiplicities. For any eigenvalue  of A and , let  and  denote the corresponding eigenspaces for A and , respectively.

1. (a) Show by way of example that for a given common eigenvalue, these two eigenspaces need not be the same.

2. (b) Prove that for any eigenvalue , .

3. (c) Prove that if A is diagonalizable, then  is also diagonalizable.

13. Find the general solution to each system of differential equations.

1. (a) 

2. (b) 

3. (c) 

14. Let



be the coefficient matrix of the system of differential equations



Suppose that A is diagonalizable and that the distinct eigenvalues of A are . Prove that a differentiable function  is a solution to the system if and only if x is of the form



where  for . Use this result to prove that the set of solutions to the system is an n-dimensional real vector space.

15. Let , and let Y be an  matrix of differentiable functions. Prove , where  for all i, j.

Exercises 18 through 20 are concerned with simultaneous diagonalization.

# Definitions.

Two linear operators T and U on a finite-dimensional vector space V are called simultaneously diagonalizable if there exists an ordered basis  for V such that both  and  are diagonal matrices. Similarly,  are called simultaneously diagonalizable if there exists an invertible matrix  such that both  and  are diagonal matrices.

1. (a) Prove that if T and U are simultaneously diagonalizable linear operators on a finite-dimensional vector space V, then the matrices  and  are simultaneously diagonalizable for any ordered basis .

2. (b) Prove that if A and B are simultaneously diagonalizable matrices, then  and  are simultaneously diagonalizable linear operators.

1. (a) Prove that if T and U are simultaneously diagonalizable operators, then T and U commute (i.e., ).

2. (b) Show that if A and B are simultaneously diagonalizable matrices, then A and B commute.

The converses of (a) and (b) are established in Exercise 25 of Section 5.4.

1. Let T be a diagonalizable linear operator on a finite-dimensional vector space, and let m be any positive integer. Prove that T and  are simultaneously diagonalizable.

Exercises 21 through 24 are concerned with direct sums.

1. Let  be subspaces of a finite-dimensional vector space V such that



Prove that V is the direct sum of  if and only if


2. Let V be a finite-dimensional vector space with a basis , and let  be a partition of  (i.e.,  are subsets of  such that  and  if ). Prove that .

3. Let T be a linear operator on a finite-dimensional vector space V, and suppose that the distinct eigenvalues of T are . Prove that


4. Let  be subspaces of a vector space V such that  and . Prove that if , then