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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.

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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} $.

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