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

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.

• I think that $A^2 + I = 0$, eigenvalues of $A$ are $\pm i$. Since $A$ is an $n \times n$ real matrix, $n$ is even.
– bing
Nov 10, 2016 at 8:07

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.

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 $$n=2k$$.

• This is an Ex of Hoffman Kunze. Your proof is supposed to be "the" correct solution. May 18 at 1:20

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 $m\in \mathbb{N}$

• plz specify reason for downvote..(for any improvement) Apr 21, 2018 at 11:20
• I think $det(A)=i(-i)$ is not true, first of all, it's true that $\lambda = \pm i$, but is not necessarily true that both values must be eigenvalues, it could also be just $i$ or $-i$ with multiplicity. Furthermore, the determinant is the product of the eigenvalues, counted with multiplicity. So even if both $i$ and $-i$ were eigenvalues the determinant would be $det(A)=i^k (-i)^j$ with $k+j=n$. Jun 15, 2021 at 6:13