##### Example4.3.6

Suppose we know that \(A=PDP^{-1}\) where

\begin{equation*} D = \left[\begin{array}{rr} 2 \amp 0 \\ 0 \amp -2 \\ \end{array}\right],\qquad P = \left[\begin{array}{cc} \vvec_2 \amp \vvec_1 \end{array}\right] = \left[\begin{array}{rr} 1 \amp 1 \\ 1 \amp 2 \\ \end{array}\right] \text{.} \end{equation*}In this case, we know that the columns of \(P\) form eigenvectors of \(A\text{.}\) For instance, \(\vvec_1 = \twovec{1}{1}\) is an eigenvector of \(A\) with eigenvalue \(\lambda_1 = 2\text{.}\) Also, \(\vvec_2 = \twovec{1}{2}\) is an eigenvector with eigenvalue \(\lambda_2=-2\text{.}\)

We can verify this by computing

\begin{equation*} A = PDP^{-1} = \left[\begin{array}{rr} 6 \amp -4 \\ 8 \amp -6 \\ \end{array}\right] \text{.} \end{equation*}Then, we can compute that \(A\vvec_1 = \twovec{1}{1}=2\vvec_1\) and \(A\vvec_2 = \twovec{1}{2} = -2\vvec_2\text{.}\)