Let $A$ be an $n\times n$ matrix with real entries such that $A^2+I=0$. Prove that $n$ is even, and if $n=2k$, then $A$ is similar over the field of real numbers to a matrix of the block form $\begin{bmatrix}0&-I\\I&0\end{bmatrix}$ where $I$ is the $k\times k$ identity matrix.
My attempt: Let $m$ and $f$ be minimal and characteristic polynomial of $A$, respectively. $A^2+I=0$ implies $x^2+1$ annihilates $A$. So $m|(x^2+1)$. Since $x^2+1$ is irreducible over $\Bbb{R}$, we have $m=x^2+1$. By theorem 4 section 7.2, $f=(x^2+1)^k$ for some $k\geq 1$. Then $\deg (f)=k\cdot \deg (x^2+1)=2k=n$. Thus $n$ is even.
Suppose $n=2k$. Then $m=x^2+1$ and $f=(x^2+1)^k$. By theorem 3 section 7.2, $\exists \alpha_1,…,\alpha_r\in V\setminus \{0\}$ with respective $T$-annihilators $p_1,…,p_r$ such that $V=Z(\alpha_1;T)\oplus \cdots Z(\alpha_r;T)$, and $p_{i+1}|p_i$, and $p_1=m$, and $p_1\cdots p_r=f$. Since $p_i|p_1=x^2+1$, we have $p_i=x^2+1$ for all $i\in J_r$. So $p_1\cdots p_r=(x^2+1)^r=f=(x^2+1)^k$ and $r=k$. Then $\exists B$ basis of $V$ such that $$[T]_B=\begin{bmatrix}A_1 & & \\ &\ddots & \\ & & A_k\end{bmatrix}$$ where $A_i= \begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\in M_2(\Bbb{R})$ is companion matrix of $p_i$. How to progress from here? I think, we can switch rows and columns of $[T]_B$ in a specific way to reach $\begin{bmatrix}0&-I\\ I&0\end{bmatrix}$. Switching columns of $[T]_B$ is like switching elements of $B$. So we obtain ordered basis $B’$. Therefore similarity is preserved. But row permutation don’t preserve similarity (here is counter example).
We can apply a permutation similarity to get from your $[T]_B$ to the desired form. In particular: let $e_1,e_2,\dots,e_{2k}$ denote the columns of the size $2k$ identity matrix. We can take $$ P = \pmatrix{e_1 & e_3 & \cdots & e_{2k-1} & e_2 & e_4 & \cdots & e_{2k}} $$ and find that if $M$ denotes the desired block form, $M = P^T[T]_B P$.
This permutation matrix $P$ is the commutation matrix $K^{(k,2)}$. To see that this works, it suffices to note that $[T]_B$ is the Kronecker product $I \otimes A$, and the desired form is $M = A \otimes I$, where $A = \left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix} \right]. $ As is noted on the "properties" section of the commutation matrix wiki page, we have $$ M = A \otimes I = K^{(2,k)} (I \otimes A)K^{(k,2)} = P^T [T]_B P. $$
Here's an outline of an alternative proof that we get the desired form. Consider the product $e_i^T(P^T[T]_BP)e_j = (Pe_i)^T[T]_B(Pe_j)$, which is the $i,j$ entry of the result $M = P^T[T]_B P$. You need to show 3 facts:
Note that for the original matrix $[T]_B$, we have $$ e_p [T]_B e_q = \begin{cases} 1 & i \text{ is even and } j = i-1\\ -1 & i \text{ is odd and } j = i+1\\ 0 & \text{ otherwise} \end{cases} $$