# 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.

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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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If you want $n$ to be minimal, you just need to require that at least one of the $f_i$ has a primitive $2n^{th}$ root of unity as one of its roots. –  Qiaochu Yuan Feb 10 '12 at 5:14
I was led to believe that we could obtain more, that for example if $n=2k$, we could say that $V=V_1\oplus\cdots\oplus V_{2k}$, where each $V_i$ is invariant under $\phi$. I may have misunderstood, or maybe representation theory isn't what I'm looking for. –  Eric Gregor Feb 10 '12 at 5:15
@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. –  Qiaochu Yuan Feb 10 '12 at 5:23

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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The question might be more interesting if one knew of an $n$ and a $\phi$ such that $\phi^n=-I$ but if $0\lt m\lt n$ then $\phi^n\ne-I$. –  Gerry Myerson Feb 10 '12 at 4:47
Even then, the $\mathbb C^k$ example still works for even $n$, and for odd $n$ we could still have any dimension $k \ge 2$ (namely $V=\mathbb C\oplus \mathbb R^{k-2}$ and $\phi=e^{\pi i/n}\oplus -I_{k-2}$). –  Henning Makholm Feb 10 '12 at 4:55
I should have written that $n$ was the minimal such integer. –  Eric Gregor Feb 10 '12 at 5:03
Taking the case where $n$ is even then, can't we say anything else about its structure? It seems the condition on $V$ is strong. –  Eric Gregor Feb 10 '12 at 5:04
Indeed. Henning, I ignored all mention of $\bf R$ and thought we were working over $\bf Q$. My apologies to all. –  Gerry Myerson Feb 10 '12 at 5:06