For $n >1$,let $\displaystyle f(n)$ be the number of $n \times n$ real matrices $A$ such that $A^2+I=0.$

I came across the following problem that says:

For $n >1$,let $\displaystyle f(n)$ be the number of $n \times n$ real matrices $A$ such that $A^2+I=0.$ Then which of the following options is correct?
$1.\displaystyle f(n) \equiv 0$
$2.\displaystyle f(n) \equiv \infty$
$3.\displaystyle f(n)=0$ iff $n$ is even
$4.\displaystyle f(n)=0$ iff $n$ is odd.

Can someone throw light on it? Thanks in advance for your time.

• Notice that $\begin{pmatrix} 0 &-1 \\ 1 & 0 \end{pmatrix}$ is one such matrix of size $2$. Further notice that, since $A²=-I$, $A$ must be invertible, so lies in $GL_n(R)$. Now a theorem says that a group in which every element is of ordr at most $2$ is of order $2^k$ for some $k$. – awllower Feb 11 '13 at 9:56
• Hint: look at the determinant of such a matrix. – Davide Giraudo Feb 11 '13 at 9:58

Note that $A^2 = -{\rm I}_n$, $~{\sf det}(-{\rm I}_n)=(-1)^n~$ and ${\sf det}(A^2) = ({\sf det}A)^2$. Since a real matrix has real determinant, then $f(n)=0$ if $n$ is odd. Conversely if $n$ is even, then the matrix $$A=\begin{bmatrix} &&&&& -1 \\ &&&&\cdots \\ &&&-1 \\ && 1 \\ &\cdots \\ 1 \end{bmatrix}$$ satisfies $A^2+{\rm I}=0$. Thus $f(n)\geq 1$ if $n$ is even so that among the options listed, number 4 (f(n)=0 iff n is odd) is the only one true.

Consider the matrix $$\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$ and all of its conjugates.

Then consider what you know about the real roots of a real polynomial of odd degree.

(a) is wrong because for $n=2$ we have $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

(b) is wrong because for $n=3$, the polynomial $\det(A-xI)$ is of degree $3$ and hence $A$ has a real eigenvalue $\lambda$. But $A^2+I=0$ would imply $\lambda^2+1=0$, contradiction.

(c) is wrong, again by the esample $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$.

(d) is correct, again by observing that $\det(A-xI)$ is of odd degree and hence has a real root.

You’ve not looked hard enough: if $$A=\pmatrix{0&b\\-\frac1b&0}\;,$$ then

$$A^2=\pmatrix{-1&0\\0&-1}\;.$$

Added: Remember that the characteristic polynomial of $A$ is of degree $n$, that $A$ satisfies its own characteristic polynomial, and that every real polynomial of odd degree has at least one real root. Suppose that $n$ is odd. Then $A$ has a real eigenvalue $\lambda$. Let $v$ be a non-zero eigenvector for $\lambda$. Then

$$\left(A^2+I\right)v=\lambda^2v+v=0$$

if and only if $\lambda^2=-1$, so ... ?

• Hence $f(n)=\infty$? Also, you probably should add that $b \neq 0$, though it matters not really. – awllower Feb 11 '13 at 10:00
• @awllower: What I wrote eliminates (1) and (3), but it still leaves (4) as a possibility. You still have to decide whether $f(n)$ is $0$ or infinite for odd $n$. – Brian M. Scott Feb 11 '13 at 10:02
• Thanks for pointing out this. – awllower Feb 11 '13 at 10:05
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