Matrix, complex vector spaces Suppose $M$ is a real 2 by 2 matrix with the property that $M^2+I=0$, and the eigenvalues of $M$ is $i$ and $-i$.
Prove that there exists a complex matrix P with the property that $$P^{-1}(aI+bM)P=\begin{bmatrix}
a & -b\\ b
 &a 
\end{bmatrix}$$ for all $a,b\in\mathbb{R}$
I have some progress of this question, so I am not sure if this is correct $$P^{-1}(aI+bM)P=aP^{-1}IP+bP^{-1}MP=aI+bP^{-1}MP$$ ???? how do you show this?well, if that is correct, I some how want to show that $$P^{-1}MP$$ is the matrix \begin{bmatrix}
0 &-1 \\1 
 &0 
\end{bmatrix}
then something about the two matrix are similar, I am thinking really hard, all of this now looks like a mess in my head, can someone please help me? Thank you.
 A: Note that $$Q^{-1} \underbrace{\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}}_{:=J} Q = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} \quad \textrm{where } Q := \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}$$ and then, since there exists a complex matrix $R$ with $$R^{-1}MR = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}$$ we have $$QR^{-1}MRQ^{-1} = Q \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} Q^{-1} = J.$$ Thus, for $P := RQ^{-1}$, $$P^{-1}(aI+bM)P = aI+bP^{-1}MP = aI + bJ = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}.$$
A: As you have correctly established, what we're looking for is $P$ such that $PMP^{-1} = J$, where
$$
J = \pmatrix{0&-1\\1&0}.
$$
One standard approach to this is to use diagonalization over the complex numbers: because both $M$ and $J$ are diagonalizable with eigenvalues $\pm i$, they are similar to the same complex matrix and therefore similar to each other. This tells us that there is an invertible matrix $P$ with possibly complex entries for which $PMP^{-1} = J$.
Here is an approach which guarantees that $P$ is real. The operation $M\mapsto PMP^{-1}$ is a change of basis, so what we want to show is that there is a basis $\mathcal B = \{v_1,v_2\}$ such that the matrix of $M$ relative to $\mathcal B$ is $J$.
With that in mind, take $v_1$ to be any non-zero vector. I claim that $Mv_1$ and $v_1$ are linearly independent. Otherwise, $Mv_1$ would be a real vector that is a multiple of $v_1$, which would mean that $v_1$ is an eigenvector of $M$ corresponding to a real eigenvalue, which is impossible. Define $v_2 = Mv_1$. Note that
$$
Mv_2 = M(Mv_1) = M^2v_1 = (-I)v_1 = -v_1.
$$
Thus, we have $Mv_1 = 0\cdot v_1 + 1 \cdot v_2$, and $Mv_2 = -1 \cdot v_1 + 0 \cdot v_2$. It follows that the matrix $[M]_{\mathcal B}$ of $M$ with respect to $\mathcal B$ is $J$. Equivalently, if we take $P$ to be the matrix whose columns are $v_1,v_2,$ then $PMP^{-1} = J$, which was what we wanted.
