Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even. Not duplicated Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ 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. 
I have done the first part. this question has already been answered here Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even..But This is Exercise 16 of Section 7.2, "Cyclic Decomposition and the Rational Form," in Linear Algebra, second edition, by Hoffman and Kunze,  by Hoffman and Kunze, so I want to know if someone can solve the second part of the exercise using the cyclic decomposition theorem or something similar of this chapter.
 A: Let $V$ be the vector space $\Bbb R^n$, organized as a $\Bbb R[x]$-module by letting $x$ act as $x.v:=Av$. Using the theorem, we have $V=V_1\oplus\dots V_k$ a decomposition of $V$ as a direct sum of cyclic modules.
Let $v_1$ be such that
$$
V_1=\Bbb R[x]\cdot v_1
\ .
$$
Then $v_1$ and $\pm Av_1$ are different and not zero, so $v_1,-Av_1$ is a basis of $V_1$, it has dimension two, and the first part is clear, $n=2k$. Now let us write the matrix of $A$ w.r.t this basis. We have formally (at the first step):
$$
A
\begin{bmatrix}
v_1\\ -Av_1
\end{bmatrix}
:=
\begin{bmatrix}
Av_1\\ -AAv_1
\end{bmatrix}
=
\begin{bmatrix}
Av_1\\ v_1
\end{bmatrix}
=
\begin{bmatrix}
0 &-1\\ 1&0
\end{bmatrix}
\begin{bmatrix}
v_1\\-Av_1
\end{bmatrix}
\ ,
$$
which means that the restriction of $A$ to $V_1$ has the above matrix form.
(Depending on conventions, the transpose is the above matrix, then please take all the time the other value among $\pm Av_1$ as the second base vector.)
Now we do the same with all components, get vectors $v_1,\dots,v_k$, consider the basis $v_1,\dots,v_k,-Av_1,\dots,-Av_k$, and this is doing the job, since the many matrices $\begin{bmatrix}
0 &-1\\ 1&0
\end{bmatrix}
$ come together to deliver the wanted
$\begin{bmatrix}
0 &-I\\ I&0
\end{bmatrix}
$.
