Solution
= Solution
The <eigenvector> equations give $AB=B\operatorname{diag}(4,-2,-2)$. Multiplying on the left by $B^{-1}$ therefore gives
$$
\boxed{B^{-1}AB=\begin{pmatrix}4&0&0\\0&-2&0\\0&0&-2\end{pmatrix}}.
$$
This is an <orthogonal diagonalization>: the <linear map> acts independently by multiplication by its <eigenvalues> in the new <orthonormal basis>. The repeated <eigenvalue> explains the freedom to rotate the last two basis directions.