\(A = P D P^{-1}\): \[\begin{bmatrix} 5 & -50 \\ 0 & -5 \end{bmatrix} = \begin{bmatrix} -5 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} -0.0 & -1.0 \\ 1.0 & -5.0 \end{bmatrix}\]

Your answer may differ by scaling eigenvectors (columns of \(P\)) or otherwise choosing a different basis for the eigenspace.

Problem:

\[
\begin{bmatrix} 
5 & -50   \\ 
0 & -5
\end{bmatrix}
\]

Diagonalization:

\begin{solution}
\[
\begin{bmatrix} 
5 & -50   \\ 
0 & -5
\end{bmatrix} = 
\begin{bmatrix} 
-5 & 1   \\ 
-1 & 0
\end{bmatrix} 
\begin{bmatrix} 
-5 & 0   \\ 
0 & 5
\end{bmatrix} 
\begin{bmatrix} 
-0.0 & -1.0   \\ 
1.0 & -5.0
\end{bmatrix}
\]
\end{solution}