Understanding the structure of a finite dimensional vector space based on the properties of linear maps to itself Let $V$ be a finite dimensional vector space over $\mathbb{R}$. What can we say about the dimension of $V$ if we know that there exists some linear map $\phi: V\to V$ such that $\phi^n=-I$, where $I$ is the identity and $n>1$. Shouldn't we be able to infer structural information about our vector space based on such a map? Would you need to use representation theory to understand such a thing?
Edit: I'm supposing $n$ is the minimal such integer satisfying the above.
 A: If $n$ is odd, we can't conclude anything because then $\phi$ might be $-I$ itself. Or, more ambitiously, we can make $n$ be the least power such that $\phi^n=-I$ for any dimension $\ge 2$ by setting
$$\phi=\begin{pmatrix}\cos(\pi/n)&\sin(\pi/n)\\-\sin(\pi/n)&\cos(\pi/n)\\ &&-1\\&&&\ddots\\&&&&-1\end{pmatrix}$$
On the other hand, if $n$ is even, then $(-1)^{\dim V}=\det(-I)=(\det \phi)^n$ which is positive, and therefore the dimension of $V$ is even. But in this case $V$ can still have any even dimension, by letting $\phi$ be a block diagonal matrix with $\begin{pmatrix}\cos(\pi/n)&\sin(\pi/n)\\-\sin(\pi/n)&\cos(\pi/n)\end{pmatrix}$ blocks on the diagonal.
A: For the sake of completeness, here's what representation theory has to say (although since the relevant algebra is commutative it really becomes commutative algebra). You want to study finite-dimensional representations of (finitely-generated modules over) the algebra $R = \mathbb{R}[x]/(x^n + 1)$. Now, $\mathbb{R}[x]$ is a principal ideal domain, so by the structure theorem any such representation decomposes into a finite direct sum
$$\bigoplus (R/f_i(x))^{e_i}$$
where $f_i(x)$ is an irreducible factor of $x^n + 1$ over $\mathbb{R}$. (One can also deduce this using Jordan normal form, but this is a special case of the structure theorem anyway.) Now, the identity
$$x^n + 1 = \frac{x^{2n} - 1}{x^n - 1}$$
shows that the roots of $x^n + 1$ are the $2n^{th}$ roots of unity which are not $n^{th}$ roots of unity; these are precisely the roots of unity of the form $e^{ \frac{\pi i k}{n} }$ with $k$ odd. If $n$ is even, these come in complex conjugate pairs, and so all the $f_i$ are quadratic; if $n$ is odd, $x^n + 1$ has a unique linear factor $x + 1$ and the remaining $f_i$ are all quadratic. 
Hence if $n$ is odd the finite-dimensional representations can have any finite dimension, and if $n$ is even the finite-dimensional representations can have any even dimension. 
