Page 555 - Elementary_Linear_Algebra_with_Applications_Anton__9_edition

if we had written
in Example 1, we would have obtained

EXAMPLE 2 A Matrix That Is Not Diagonalizable
Find a matrix P that diagonalizes

Solution

The characteristic polynomial of A is

so the characteristic equation is

Thus the eigenvalues of A are and . We leave it for the reader to show that bases for the eigenspaces are

Since A is a matrix and there are only two basis vectors in total, A is not diagonalizable.

Alternative Solution

If one is interested only in determining whether a matrix is diagonalizable and is not concerned with actually finding a

diagonalizing matrix P, then it is not necessary to compute bases for the eigenspaces; it suffices to find the dimensions of the

eigenspaces. For this example, the eigenspace corresponding to is the solution space of the system

The coefficient matrix has rank 2(verify). Thus the nullity of this matrix is 1 by Theorem 5.6.3, and hence the solution space is
one-dimensional.

The eigenspace corresponding to is the solution space of the system