# 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.

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 $$\begin{bmatrix} 0 &-I\\ I&0 \end{bmatrix}$$.