Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even.

Question

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. Since $\det(A^2) = \det (-I)$ we must have $\det (-I)$ non negative and hence $n$ must be even.

Need Help in the second part.

Thank You.

Answer 1

You can solve both parts of the question at the same time as follows: Given $A$, define a map $\mathbb C\times\mathbb R^n\to\mathbb R^n$ by $(u+iv,x)\mapsto (u+iv)\cdot x:=ux+vAx$. Evidently, this is linear over $\mathbb R$ in both components and satisfies $(1,x)\mapsto x$. Using that $A^2=-I$, one easily verifies that it is multiplicative, i.e. $z\cdot (w\cdot x)=(zw)\cdot x$. Hence is satisfies all properties required from scalar multiplication, and hence makes $\mathbb R^n$ into a complex vector space, so $n=2m$ for some $m\in\mathbb N$. Moreover, choosing a complex basis $\{x_1,\dots,x_m\}$ for this vector space, $\{x_1,\dots,x_m,i\cdot x_1,\dots,i\cdot x_m\}$ is a real basis for $\mathbb R^n$. By definition the matrix representation of $A$ with respect to this real basis is the block matrix you have indicated.

Answer 2

Let $T$ be an operator on a $n$-dimensional real vector space $V$ satisfying $T^2+I=0$. We will prove that all such operators have a distinguished cyclic decomposition.

If $T^2+I=0$, then the polynomial $x^2+1$ annihilates $T$. Since this polynomial is irreducible (over the reals), it must be the minimal polynomial. The characteristic polynomial of $T$ must have the same irreducible factors as the minimal polynomial, so it must have the form $(x^2+1)^k$ for some positive integer $k$. Therefore its degree (the dimension of $V$) must be $n=2k$, that is, even. Now we consider the cyclic decomposition of $T$. In other words, we apply the cyclic decomposition theorem to $T$ to get a decomposition of $V$ as a (finite) direct sum of $T$-cyclic subspaces $Z(v,A)$ of $V$ with certain non-zero vectors $v\in V$. The $T$-annihilator of $v$ must divide the minimal polynomial, therefore it must equal $x^2+1$. Hence $(v,Tv)$ is a basis of $Z(v,T)$ such that the matrix of the compression of $T$ to $Z(v,T)$ is the companion matrix of the polynomial $x^2+1$, that is, the matrix $$\left(\begin{array}{cc} 0 & -1\\ 1 & 0\end{array}\right).$$

Therefore the operator $T$ has the representation as a block diagonal matrix with $k$ blocks equal to the above matrix.

This argument shows that all operators $T$ on an $n$-dimensional real vector space satisfying $T^2+I=0$ have the same matrix in a direct sum of $k=2/n$ cyclic basis.

Therefore all $n\times n$ matrices $A$ satisfying $A^2+I=0$ are similar, in particular they are all similar to a $n\times n$-matrix of the form $$B=\left(\begin{array}{cc} 0 & -I\\ I & 0\end{array}\right),$$ as $B$ also satisfies $B^2+I=0$, where $I$ is the $k\times k$ identity matrix and $n=2k$.

Answer 3

$A^2=-I$. If $n$ is odd, taking determinants gives us $$ (\det A)^2=\det (A^2)=\det(-I)=(-1)^n=-1. $$ As $\det A$ is a real number, this is a contradiction.

Answer 4

We can do both parts in one go, using some module theory (essentially @Alcides Buss's answer with a bit more detail towards the end).

Take $\mathbb F = \mathbb R$ and $V = \mathbb R^n$. Let $\alpha:V \to V$ be the endomorphism corresponding to the matrix $A $ and $V_\alpha$ the corresponding $\mathbb R[X]$-module. The rational canonical form theorem says that, as $\mathbb R$-modules, $$ V_\alpha \cong \frac{\mathbb R[X]}{(f_1)} \oplus \frac{\mathbb R[X]}{(f_2)} \oplus \cdots \oplus \frac{\mathbb R[X]}{(f_s)}, $$ where $f_1 | f_2 | \cdots | f_s$. The minimal polynomial is $f_s$, and we know that since $\alpha^2 + \imath = 0$, the degree of the minimal polynomial cannot exceed 2. We can quite easily argue that the minimal polynomial cannot be degree 1, so $X^2 + 1$ is a minimal polynomial, and since minimal polynomials are unique up to a unit, $X^2 + 1$ is in fact the (monic) minimal polynomial.

So $f_s = X^2 + 1$. Since $X^2 + 1$ is irreducible in $\mathbb R[X]$ and we need $f_1 | f_2 | \cdots | f_s$, we can WLOG conclude that $f_1 = f_2 = \cdots = f_s = X^2 + 1$ (technically, we could add as many $1$'s as we like to the front, but since $\mathbb R[X]/(1) \cong 0$, it doesn't change the direct sum).

The companion matrix of $X^2 + 1$ is $$ c(X^2 + 1) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. $$ Hence, there exists a basis of $V$ in which the matrix representing $\alpha$ is $$ \begin{pmatrix} c(f_1) & & & \\ & c(f_2) & & \\ & & \ddots & \\ & & & c(f_s) \end{pmatrix} = \begin{pmatrix} 0 & -1 & & & & \\ 1 & 0 & & & & \\ & & 0 & -1 & & \\ & & 1 & 0 & & \\ & & & & \ddots & \\ & & & & & 0 & -1 \\ & & & & & 1 & 0 \end{pmatrix}. \tag{$*$} $$ It's clear then that $n$ is even.

At this point we can proceed in two ways:

1. We can see from the matrix $(*)$ that in this basis, say $\{v_1, \dots, v_k, v_{k+1}, \dots, v_{2k}\}$, the action of $\alpha$ is to map $v_{2i-1} \mapsto v_{2i}$ and $v_{2i}\mapsto -v_{2i-1}$ for each $i$ between $1$ and $k$. Now consider a new basis $\{v_1', \dots, v_k', v_{k+1}', \dots, v_{2k}'\}$ in which $v_1', \dots, v_k'$ are $v_1, v_3, \dots, v_{2k-1}$ (the odd-indexed basis vectors) respectively and $v_{k+1}', \dots, v_{2k}'$ are $v_2, v_4, \dots, v_{2k}$ (the even-indexed basis vectors) respectively. In the primed basis, you get the desired form $$ M = \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix}, $$ and since this is related to $A$ by a change of basis, they are similar.

2. Alternatively, $M$ satisfies the equation $M^2 + I = 0$. Hence $A$ and $M$ are both similar to $(*)$, so they are similar to each other.

Answer 5

I solve it another method : $det(A^2)=det(-I)=(-1)^n ....(1)$.

Now $\lambda$ is an eigenvalue of $A$ then ${\lambda}^2+I^2=0\Rightarrow \lambda =i,-i.$ So $detA=i(-i)=1\Rightarrow det(A^2)=(detA)^2=1 ....(2)$

So from $(1)$ and $(2)$ we get $n=2m$ for dome $m\in \mathbb{N} $