\(A = P D P^{-1}\): \[\begin{bmatrix} -1 & 0 \\ -4 & -5 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -5 \end{bmatrix} \begin{bmatrix} -1.0 & -0.0 \\ -1.0 & -1.0 \end{bmatrix}\]

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

Problem:

\[
\begin{bmatrix} 
-1 & 0   \\ 
-4 & -5
\end{bmatrix}
\]

Diagonalization:

\begin{solution}
\[
\begin{bmatrix} 
-1 & 0   \\ 
-4 & -5
\end{bmatrix} = 
\begin{bmatrix} 
-1 & 0   \\ 
1 & -1
\end{bmatrix} 
\begin{bmatrix} 
-1 & 0   \\ 
0 & -5
\end{bmatrix} 
\begin{bmatrix} 
-1.0 & -0.0   \\ 
-1.0 & -1.0
\end{bmatrix}
\]
\end{solution}