Given an matrix A, we may ask how many linearly independent eigenvectors the matrix A has. In Section 6.1, we saw several examples (with and ) in which the matrix A has n linearly independent eigenvectors—the largest possible number. By contrast, in Example 5 of Section 6.1, we saw that the matrix
has the single eigenvalue corresponding to the single eigenvector
Something very nice happens when the matrix A does have n linearly independent eigenvectors. Suppose that the eigenvalues (not necessarily distinct) of A correspond to the n linearly independent eigenvectors , respectively. Let
be the matrix having these eigenvectors as its column vectors. Then
and hence
because for each . Thus the product matrix AP has column vectors .
Now consider the diagonal matrix
whose diagonal elements are the eigenvalues corresponding (in the same order) to the eigenvectors forming the columns of P. Then
because the product of the ith row of P and the jth column of D is simply the product of and the ith component of .
Finally, upon comparing the results in (2) and (4), we see that
But the matrix P is invertible, because its n column vectors are linearly independent. So we may multiply on the right by to obtain
Equation (6) expresses the matrix A having n linearly independent eigenvectors in terms of the eigenvector matrix P and the diagonal eigenvalue matrix D. It can be rewritten as , but the form in (6) is the one that should be memorized.
In Example 1 of Section 6.1 we saw that the matrix
has eigenvalues and corresponding to the linearly independent eigenvectors and respectively. Then
So
in accord with Eq. (6).
The following definition embodies the precise relationship in (6) between the original matrix A and the diagonal matrix D.
Note that this relationship between A and B is symmetric, for if , then for some invertible matrix Q—just take .
An matrix A is called diagonalizable if it is similar to a diagonal matrix D; that is, there exist a diagonal matrix D and an invertible matrix P such that , and so
The process of finding the diagonalizing matrix P and the diagonal matrix D in (8) is called diagonalization of the matrix A. In Example 1 we showed that the matrices
are similar, and hence that the matrix A is diagonalizable.
Now we ask under what conditions a given square matrix is diagonalizable. In deriving Eq. (6), we showed that if the matrix A has n linearly independent eigenvectors, then A is diagonalizable. The converse of this statement is also true.
It is important to remember not only the fact that an matrix A having n linearly independent eigenvectors is diagonalizable, but also the specific diagonalization in Eq. (6), where the matrix P has the n eigenvectors as its columns, and the corresponding eigenvalues are the diagonal elements of the diagonal matrix D.
In Example 5 of Section 6.1 we saw that the matrix
has only one eigenvalue, and that (to within a constant multiple) only the single eigenvector is associated with this eigenvalue. Thus the matrix A does not have linearly independent eigenvectors. Hence Theorem 1 implies that A is not diagonalizable.
In Example 6 of Section 6.1 we saw that the matrix
has the following eigenvalues and associated eigenvectors:
It is obvious (why?) that the three eigenvectors are linearly independent, so Theorem 1 implies that the matrix A is diagonalizable. In particular, the inverse of the eigenvector matrix
is
and the diagonal eigenvalue matrix is
Therefore, Eq. (6) in the form yields the diagonalization
of the matrix A.
The following theorem tells us that any set of eigenvectors associated with distinct eigenvalues (as in Example 3) is automatically linearly independent.
If the matrix A has n distinct eigenvalues, then by Theorem 2 the n associated eigenvectors are linearly independent, so Theorem 1 implies that the matrix A is diagonalizable. Thus we have the sufficient condition for diagonalizability given in Theorem 3.
In general, however, an matrix A can be expected to have fewer than n distinct eigenvalues . If , then we may attempt to diagonalize A by carrying out the following procedure.
Step 1. Find a basis for the eigenspace associated with each eigenvalue .
Step 2. Form the union S of the bases . According to Theorem 4 in this section, the set S of eigenvectors of A is linearly independent.
Step 3. If S contains n eigenvectors , then the matrix
diagonalizes A: that is, , where the diagonal elements of D are the eigenvalues (repeated as necessary) corresponding to the n eigenvectors .
If the set S—obtained by “merging” the bases for all the eigenspaces of A—contains fewer than n eigenvectors, then it can be proved that the matrix A is not diagonalizable.
In Example 7 of Section 6.1, we saw that the matrix
has only two distinct eigenvalues, and . We found that the eigenvalue corresponds to a 2-dimensional eigenspace with basis vectors and and that corresponds to a 1-dimensional eigenspace with basis vector By Theorem 4 (or by explicit verification), these three eigenvectors are linearly independent, so Theorem 1 implies that the matrix A is diagonalizable. The diagonalizing matrix
has inverse matrix
so we obtain the diagonalization
of the matrix A.
In Problems 1 through 28, determine whether or not the given matrix A is diagonalizable. If it is, find a diagonalizing matrix P and a diagonal matrix D such that .
Prove: If the matrices A and B are similar and the matrices B and C are similar, then the matrices A and C are similar.
Suppose that the matrices A and B are similar and that n is a positive integer. Prove that the matrices and are similar.
Suppose that the invertible matrices A and B are similar. Prove that their inverses and are also similar.
Show that if the matrices A and B are similar, then they have the same characteristic equation and therefore have the same eigenvalues.
Suppose that the matrices A and B are similar and that each has n real eigenvalues. Show that and that . See Problems 38 and 39 in Section 6.1.
Consider the matrix
and let . Then show that
A is diagonalizable if ;
A is not diagonalizable if ;
If , then A may be diagonalizable or it may not be.
Let A be a matrix with three distinct eigenvalues. Tell how to construct six different invertible matrices and six different diagonal matrices such that for each .
Prove: If the diagonalizable matrices A and B have the same eigenvalues (with the same multiplicities), then A and B are similar.
Given: The diagonalizable matrix A. Show that the eigenvalues of are the squares of the eigenvalues of A but that A and have the same eigenvectors.
Suppose that the matrix A has n linearly independent eigenvectors associated with a single eigenvalue . Show that A is a diagonal matrix.
Let be an eigenvalue of the matrix A, and assume that the characteristic equation of A has only real solutions. The algebraic multiplicity of is the largest positive integer p(i) such that is a factor of the characteristic polynomial . The geometric multiplicity of is the dimension q(i) of the eigenspace associated with . It can be shown that for every eigenvalue . Taking this as already established, prove that the given matrix A is diagonalizable if and only if the geometric multiplicity of each eigenvalue is equal to its algebraic multiplicity.
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