###### 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 $$A\vvec_2 = \twovec{1}{2} = -2\vvec_2\text{.}$$

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