= Solution
The <characteristic polynomial> of $A$ is
$$
\det(A-\lambda I)=(-5-\lambda)(5-\lambda)+16=\lambda^2-9.
$$
Its distinct <eigenvalues> are $3$ and $-3$. For $3$, the <eigenvector> relation gives $y=-2x$, so choose $(1,-2)^T$. For $-3$, it gives $x=-2y$, so choose $(2,-1)^T$. Put these independent <eigenvectors> into columns:
$$
\boxed{B=\begin{pmatrix}1&2\\-2&-1\end{pmatrix},\qquad C=\begin{pmatrix}3&0\\0&-3\end{pmatrix}.}
$$
Then $AB=BC$ column by column. Also $\det B=3\ne0$, so this choice supplies the <diagonalization of a matrix> $B^{-1}AB=C$. Scaling columns or interchanging their order gives other valid choices, with the matching order in $C$.
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