Show that $A$ is similar to this matrix

Question

Let $A \in M_n(\mathbb{R})$ such that $A^2 = -I_n$. Show that there exists a non singular $S \in M_n(\mathbb{R})$ such that

$$S^{-1}AS = \begin{bmatrix} 0 && -I_{\frac{n}{2}} \\ I_{\frac{n}{2}} && 0 \\ \end{bmatrix}$$

So far what I have done is noted that $A^2 +I_n = 0$ and so $p(t)=t^2+1$ is its minimal polynomial. Therefore the eigenvalues of $A$ are $\{i,-i\}$. Thus the characteristic polynomial is

$$(x-i)^r(x+i)^s$$

such that $r+s=n$. Now if $r>s$ or $s>r$ the characteristic polynomial would not be real which is a contradiction. Thus $r=s$ so $2r=n$ which is to say that $n$ is even. Now by jordans real form we can find a nonsingular $X \in M_n(\mathbb{R})$ such that

$$X^{-1}AX = J_n(0)^2 + (J_2(0) - J_2(0)^T) \oplus \dots \oplus(J_2(0) - J_2(0)^T)$$

From here I have no idea how to get to the desired form, thanks in advance for any suggestions!

Answer 1

First note that $n$ must be even as $\det A^2 > 0$ and $\det (-I) = (-1)^n$.

Since $A^2+I =0$ you know that the eigenvalues are $\pm i$ and there is a (complex) basis such that $A$ is diagonalisable. Since $A$ is real, we know that the eigenvalues occur in conjugate pairs so there are ${ n \over 2}$ vectors $u_k+iv_k$ such that $A (u_k+iv_k) = i (u_k+iv_k)$. From this we see that $A u_k = - v_k, A v_k = u_k$. It is straightforward to check that the vectors $u_1,...,u_{n \over 2}, v_1,..., v_{n \over 2}$ are linearly independent.

In this basis we see that $A$ has the desired form.

Answer 2

Since the minimal polynomial $p(t) = t^2+1 = (t+i)(t-i)$ splits into linear factors, we conclude that $A$ is diagonalizable over $\mathbb{C}$.

You already determined that the characteristic polynomial of $A$ is given by $(x+i)^{\frac{n}2}(x-i)^{\frac{n}2}$ so $A$ diagonalizes to $$D = \begin{bmatrix} iI_{\frac{n}{2}} & 0 \\ 0 & -iI_{\frac{n}{2}} \\ \end{bmatrix}$$

Notice that the matrix $\begin{bmatrix} 0 & -I_{\frac{n}{2}} \\ I_{\frac{n}{2}} & 0 \\ \end{bmatrix}$ has exactly the same minimal and characteristic polynomials as $A$, so by the same reasoning it also diagonalizes to $D$ over $\mathbb{C}$.

Hence, $A$ and $\begin{bmatrix} 0 & -I_{\frac{n}{2}} \\ I_{\frac{n}{2}} & 0 \\ \end{bmatrix}$ are similar over $\mathbb{C}$. Since both are real, they are also similar over $\mathbb{R}$.

Answer 3

You can give the $\Bbb R$-vector space $\Bbb R^n$ the structure of a complex vector space by having the imaginary unit $\def\ii{\mathbf i}\ii$ act as multiplication by$~A$ (the necessary checks for this claim are all easy). This shows in particular that $n$ is even; the dimension of the space as $\Bbb C$-vector space is $n/2$. Now let $b_1,\ldots,b_{n/2}$ be a basis of that $\Bbb C$-vector space, and $b_{i+n/2}=\ii\,b_i$ (which means that $b_{i+n/2}=A\cdot b_i$, but I just thought it was cute to have two i's in the same formula) for $i=1,,\ldots,n/2$. Then $b_1,\ldots,b_n$ is an $\Bbb R$-basis of$~\Bbb R^n$, and change of basis to this basis transforms $A$ into the matrix displayed in the question.