###### Example 4.3.4

We will try to find a diagonalization of \(A = \left[\begin{array}{rr} -5 \amp 6 \\ -3 \amp 4 \\ \end{array}\right] \text{.}\)

First, we find the eigenvalues of \(A\) by solving the characteristic equation

This shows that the eigenvalues of \(A\) are \(\lambda_1 = -2\) and \(\lambda_2 = 1\text{.}\)

By constructing \(\nul(A-(-2)I)\text{,}\) we find a basis for \(E_{-2}\) consisting of the vector \(\vvec_1 = \twovec{2}{1}\text{.}\) Similarly, a basis for \(E_1\) consists of the vector \(\vvec_2 = \twovec{1}{1}\text{.}\) This shows that we can construct a basis \(\bcal=\{\vvec_1,\vvec_2\}\) of \(\real^2\) consisting of eigenvectors of \(A\text{.}\)

We now form the matrices

and verify that

There are, of course, many ways to diagonalize \(A\text{.}\) For instance, we could change the order of the eigenvalues and eigenvectors and write

If we choose a different basis for the eigenspaces, we will also find a different matrix \(P\) that diagonalizes \(A\text{.}\) The point is that there are many ways in which \(A\) can be written in the form \(A=PDP^{-1}\text{.}\)