Solution
= Solution
In the <basis> $(v,Av)$, the <linear map> given by $A$ sends $v$ to the second <basis> vector and sends $Av$ to $A^2v=-v$. Equivalently,
$$
AE=(Av,-v)=E\begin{pmatrix}0&-1\\1&0\end{pmatrix}.
$$
Multiplying on the left by the <matrix inverse> $E^{-1}$ gives
$$
\boxed{E^{-1}AE=\begin{pmatrix}0&-1\\1&0\end{pmatrix}}.
$$
This <change of basis> produces the <companion matrix> found in part (a)(ii).