TITLE: Prove that if V is finite dimensional then V is even dimensional?
QUESTION [38 upvotes]: Let $f:V \to V$ be a linear map such that $(f\circ f)(v) = -v$. Prove that if $V$ is a finite dimensional vector space over $\mathbb R$, $V$ is even dimensional.

From what I can figure out for myself, if $V$ is finite dimensional, then every basis of $V$ is finite, i.e. a linearly independently subset of $V$ has a finite number of vectors. 
And I figure that if $V$ is even dimensional, then every basis of $V$ is even, i.e. a linearly independently subset of $V$ has an even number of vectors. 
But I'm not sure how to connect these two points.

REPLY [8 votes]: Since $f^2 = -I$, we see that the minimal polynomial of $f$ over $\mathbb{R}$ is $x^2 + 1 \in \mathbb{R}[x]$. The minimal polynomial and characteristic polynomial share their root sets, so the characteristic polynomial of $f$ must be $(x^2 + 1)^k$ for some $k\ge 1$. But the degree $2k$ of the characteristic polynomial is the dimension of $V$, hence $V$ is even-dimensional.