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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.

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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.

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    @Eric: well, each $R/f_i(x)$ is invariant under $\phi$ (that's part of what it means to be a direct summand as an $R$-module). But I don't know why you expect there to be $2k$ summands. There may be any positive integer number of irreducible summands. Perhaps by $V_i$ you mean the $i^{th}$ isotypic component? That is exactly the summand $(R/f_i(x))^{e_i}$ I used above. I'm not sure what more you are expecting exactly.2012-02-10
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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.

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    @Eric: So you want a structure relative to the fixed linear transformation $\phi$. That would have been clearer if you had written something like "Given $V$ and $\phi$ such that ..." instead of "Given $V$ such that a there exists a $\phi$ ...".2012-02-10