1. Diagonalization and Similar Matrices In Section 4.2 we showed how to compute eigenpairs (,p) of a matrix A by determining the roots of the characteristic polynomial c( ) = det(A - I n ) and then using the values of to solve the associated homogeneous linear systems (A - I n )p = 0. For 2  2, 3  3, and other small matrices this two-step procedure could be carried out by hand-calculations. The two-step method is called a direct method since we get the eigenpairs as a result of a specific set of calculations. In this section we show how to obtain an equivalent eigen problem; that is, given a matrix A we show how to find a matrix B which has the same eigenvalues as A. If we design our procedure efficiently, then the eigenvalues of B will be easy to obtain and hence we will have the eigenvalues of the original matrix A. We say that a procedure that yields an equivalent eigen problem preserves eigenvalues. By carefully recording the steps of such procedures we will be able to generate the corresponding eigenvectors.

2. We begin by defining an operation on a matrix that preserves eigenvalues. Definition A similarity transformation of an n  n matrix A is a function f(A) = P -1 AP, where P is an n  n nonsingular matrix. If B = P -1 AP, then we say A is similar to B. We can show that if A is similar to B, then B is similar to A. Hence we say A and B are similar matrices. It follows that A and B are similar matrices if and only if there exists a nonsingular matrix P such that B = P -1 AP. Similar matrices have the same eigenvalues. Proof: We show that similar matrices have the same characteristic polynomial. So similarity transformations preserve eigenvalues.

3. Since similarity transformations preserve eigenvalues, it is natural to ask if A and B are similar with eigenpairs (, p) and (, q), respectively, what is the relationship between the eigenvectors p and q. Suppose we have B = P -1 AP or equivalently A = PBP -1, then we have Ap = p  PBP -1 p = p  P -1 (PBP -1 p) = P -1 ( p)  B(P -1 p) = (P -1 p) This says that if p is an eigenvector of A corresponding to, then q = P -1 p is an eigenvector of B corresponding to. To summarize: If B = P -1 AP and (, p) is an eigenpair of A, then (, P -1 p) is an eigenpair of B. Example: Suppose that A and B are similar with B = P -1 AP where If A has eigen pairand then find the corresponding eigenpair of B. Solution: Since A and B are similar = 2 is an eigenvalue of B and a corresponding eigenvector is

4. From Section 4.2 we know that the eigenvalues of a diagonal matrix, an upper triangular matrix, or a lower triangular matrix are the diagonal entries. With this in mind we pose the following similarity questions for an n  n matrix A. Similarity Questions 1. Given a square matrix A, is A similar to a diagonal matrix? That is, does there exist a nonsingular matrix P so that B = P -1 AP is diagonal? 2. If A is not similar to a diagonal matrix, does there exist a nonsingular matrix P so that B = P -1 AP is upper lower) triangular? That is, is A similar to a triangular matrix? 3. If A is similar to a diagonal or triangular matrix, how do we find the matrix P that performs the similarity transformation? Questions 1 and 2 above are existence questions; that is, is there some property of matrix A that will guarantee that such a transformation is possible. However, question 3 is a construction question; that is, construct the appropriate matrix P. We can answer question 1 using properties developed so far. Questions 2 and 3 are harder.

5. We want to determine conditions under which A similar to a diagonal matrix? We start by deriving information about A if it is similar to a diagonal matrix. Assume that P -1 AP = D, a diagonal matrix. Then the eigenvalues of A are the diagonal entries of D. We have P -1 AP = D  AP = PD. Next express P as a matrix partitioned into its columns; P = [p 1 p 2... p n ]. Then we have A[p 1 p 2... p n ] = [p 1 p 2... p n ]D. Performing the multiplications on both sides we can show that we get [Ap 1 Ap 2... Ap n ] = [d 11 p 1 d 22 p 2... d nn p n ]. Equating corresponding columns gives Ap 1 = d 11 p 1, Ap 2 = d 22 p 2,..., Ap n = d nn p n. These expressions imply that column j of P is an eigenvector of A corresponding to eigenvalue that is the (jj-entry) of diagonal matrix D. Since P is nonsingular its columns are linearly independent and guaranteed not to be 0. The steps here are reversible; that is, if A has n linearly independent eigenvectors, then it is similar to a diagonal matrix. We summarize this important result as: An n  n matrix A is similar to a diagonal matrix if and only if A has n linearly independent eigenvectors.

6. We just provided an argument that showed A is similar to a diagonal matrix if and only if A has n linearly independent eigenvectors. BEWARE: We cannot conclude that if two matrices have the same eigenvalues they are similar. Example: Let A = I 2 and.. Both A and B have eigenvalue 1 with multiplicity 2. (Verify.) A is diagonal so it is certainly similar to a diagonal matrix. However, matrix B is not similar to a diagonal matrix since it has only one linearly independent eigenvector. (Verify.) Thus A and B cannot be similar. (Explain.) From the point of view of using similarity transformations to find eigenvalues the statement, A is similar to a diagonal matrix if and only if A has n linearly independent eigenvectors, says that we need the eigenvectors in order to find the eigenvalues. This is just the reverse of the two step method we developed in Section 4.2. Thus while the statement, A is similar to a diagonal matrix if and only if A has n linearly independent eigenvectors, specifically tells us which matrices are similar to a diagonal matrix, it really isn't very useful for directly computing the eigenvalues of a matrix. However, it does lead us to some criteria that will enable us to recognize that certain matrices are similar to a diagonal matrix. At this point it is convenient to make the following definition.

7. Definition A matrix A is called diagonalizable provided it is similar to a diagonal matrix. For an n × n matrix to be similar to a diagonal matrix it must have n linearly independent eigenvectors. A previous result stated that eigenvectors of a matrix corresponding to distinct eigenvalues are linearly independent. Combining these ideas we have the following result. If an n  n matrix A has n distinct eigenvalues, then A is diagonalizable. Warning: If A has an eigenvalue with multiplicity greater than 1, then A may or may not be diagonalizable. We must check to see if each eigenvalue has as many linearly independent eigenvectors as its multiplicity. That is, in the language of Exercise 6 in Section 4.2, we must check to see if the matrix is defective or not. A defective matrix is not diagonalizable. Example: The matrix has = 1 was an eigenvalue of multiplicity 2 and that the corresponding eigenspace has a basis of two eigenvectors. Since the remaining eigenvalue had a value other than 1, A is diagonalizable.

8. Example: The matrix is lower triangular its eigenvalues are 1, 1, and 3. Since it has an eigenvalue of multiplicity > 1, there is a chance it is not diagonalizable; that is, it may be defective. To check this we determine the number of linearly independent eigenvectors associated with eigenvalue = 1; that is, the dimension of the eigenspace associated with = 1. To do this we find ns(A - 1I 3 ). We compute rref(A - 1I 3 ) and obtain Thus x 1 = x 3 = 0 and x 2 is a free variable. Since there is only one free variable dim(ns(A - 1I 3 )) = 1, and so there is just one linearly independent eigenvector corresponding to = 1. Hence A is defective and is not diagonalizable.

9. Discussion questions. What does it mean to say that two matrices are similar? What eigen feature do similar matrices share? What does it mean to say that a matrix is diagonalizable? If matrix A is diagonalizable, then how are the eigenvectors of A involved? Can a singular matrix be diagonalizable? Explain. Can a defective matrix be diagonalizable? Explain why or why not.