$$I = \begin{pmatrix} 1 \amp 0 \amp ...\amp 0\\ 0 \amp 1 \amp ...\amp 0\\ \vdots \amp \vdots \amp \ddots \amp \vdots\\ 0 \amp 0 \amp ...\amp 1 \end{pmatrix}\text{.}\tag{5.5.1}$$
$$A = \begin{pmatrix} \lambda \amp 0 \amp 0\\ 0 \amp \mu \amp 0\\ 0 \amp 0 \amp \nu \end{pmatrix}\text{.}\tag{5.5.2}$$
The eigenvalues of $A$ are clearly $\{\lambda,\mu,\nu\}\text{,}$ and the corresponding eigenvectors are clearly just the standard basis $\left\{\begin{pmatrix}1\\0\\0 \end{pmatrix} , \begin{pmatrix}0\\1\\0 \end{pmatrix} , \begin{pmatrix}0\\0\\1 \end{pmatrix} \right\}\text{.}$