Square root of $-I_2$ I would like to get all matrices $N \in M_2(\mathbb R)$ such that $N^2 = -I_2$.
To start with, I know that 
$N_0=\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}$ works, and we can prove that every matrices $N$ that are similar to $N_0$ work.
$i.e.$ Let $N \in M_2(\mathbb R)$ if $\exists P \in \mathrm{GL}_2(\mathbb R)$ such that $N = PN_0P^{-1}$, then $N^2 = -I_2$.
My question is, is the converse true?
Are all matrices $N \in M_2(\mathbb R)$ such that $N^2 = -I_2$ similar to $N_0$?
 A: For $A, B \in M_n(\mathbb{C})$, write $A \sim B$ if $A$ and $B$ are similar in $\mathbb{C}$.
Assume that $N \in M_2(\mathbb{R})$ satisfies $N^2 + I_2 = 0$. Then the minimal polynomial of $N$ is $X^2+1$ and this factors into distinct linear factors in $\mathbb{C}$. So $N$ is diagonalizable in $\mathbb{C}$ with eigenvalues $i$ and $-i$, i.e., $N \sim \operatorname{diag}(i, -i) $ in $\mathbb{C}$. By the same reason, $N_0 \sim \operatorname{diag}(i, -i)$ and hence $N \sim N_0$. Then the desired claim follows from the following proposition.

Proposition. Let $A, B \in M_n(\mathbb{R})$. Suppose that $A \sim B$. Then $A$ and $B$ are similar in $\mathbb{R}$.

Proof. Choose $P \in \mathrm{GL}_n(\mathbb{C})$ such that $A = PBP^{-1}$. Write $P = Q+iR$ for $Q, R \in M_n(\mathbb{R})$ and define $P_z = Q + zR$ for $z \in \mathbb{C}$.


*

*Since $AQ = QB$ and $AR = RB$, we have $AP_z = P_z B$ for all $z \in \mathbb{C}$,

*$\det(P_i) = \det(P) \neq 0$, hence $\det(P_z)$ is a non-zero polynomial in $\mathbb{R}[z]$.
So we can find $x \in \mathbb{R}$ such that $P_x$ is invertible. Then $P_x \in \operatorname{GL}_2(\mathbb{R})$ and $A = P_x BP_x^{-1}$, hence $A$ and $B$ are similar in $\mathbb{R}$ as required. $\Box$
A: Something more general holds. If $f(t) = t^n + a_{n-1}t^{n-1} + \dots + a_1t + a_0$ is a monic polynomial then the matrix
$$ C_f = \begin{pmatrix}
0 & 0 & 0 & \dots & 0 & -a_0 \\
1 & 0 & 0 & \dots & 0 & -a_1 \\
0 & 1 & 0 & \dots & 0 & -a_2 \\
0 & 0 & 1 & \dots & 0 & -a_3 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 1 & -a_{n-1}
\end{pmatrix}
$$
is called the companion matrix of $f$. A simple calculation shows that $C_f$ satisfies $f$, meaning that
$$ f(C_f) = C_f^n + a_{n-1}C_f^{n-1} + \dots + a_1 C_f + a_0I = 0 $$
Moreover, if $C_f$ satisfies any other polynomial equation $g(C_f) = 0$ then $f$ divides $g$. We say that $f$ is the minimal polynomial of $C_f$.
There's a result in algebra called rational normal form (or "canonical form") that says that every matrix is similar to a block diagonal matrix of companion matrices. Specifically,

If $A$ is a matrix that satisfies a polynomial $f$ (for example its characteristic polynomial) then $A$ is similar to a block diagonal matrix whose blocks are companion matrices. I.e. $$A \sim \operatorname{diag}(C_{g_1},\dots,C_{g_r}).$$
  Moreover:
  
  
*
  
*$g_i$ divides $f$ for $i = 1, \dots, r$
  
*We can make it so that $g_i$ divides $g_{i + 1}$ for $i = 1,\dots,r-1$
  
*If 2. holds, you can check that $g_r$ is the minimal polynomial of $A$ in the sense described above.
  

One idea here is that if $h$ is any polynomial then
$$h(\operatorname{diag}(C_{g_1},\dots,C_{g_r})) = \operatorname{diag}(h(C_{g_1}),\dots,h(C_{g_r})). $$
This is true generally for block-diagonal matrices.
The point is that for the polynomial $f(t) = t^2 + 1$ the only real factors of $f$ are $1$ and $t^2 + 1$. On the other hand, $C_1$ is the empty $0\times 0$ matrix which just leaves us with $C_{t^2 + 1}$ is exactly what you called $N_0$.
