TITLE: Intuitive geometric explanation: existence of eigenvalue in odd dimension real vector space.
QUESTION [13 upvotes]: I'm looking for an intuitive geometric explanation for the fact that given an odd dimensional real vector space $W$ and an endomorphism $T:W \rightarrow W$, there exists a real eigenvalue of $T$. I'm not looking for an algebraic explanation which involves the degree of characteristic polynomial.

REPLY [5 votes]: Warning: this may not satisfy many readers' definition of "intuitive".
Let's suppose $T$ is a endomorphism of $V=\Bbb R^n$ which vanishes nowhere.
Then $T$ determines a continuous map $f$ from the sphere $S^{n-1}$ to
itself by $f(u)=T(u)/|u|$. Moreover, the map is bijective so has degree
$\pm1$.
If $n$ is odd, $n-1$ is even and $S^{n-1}$ has Euler characteristic
$2$. Replacing $T$ by $-T$ is necessary, we may assume $g$ has degree
$1$ and so $f$ acts trivially on the singular homology of $S^{n-1}$.
As the Euler characteristic of $S^{n-1}$ is nonzero, by the Lefschetz Fixed Point Theorem, $f$
has a fixed point, and this means that $T$ has an eigenvector.
Alternatively, if $T$ has no eigenvector then neither $f$ nor $-f$ has
a fixed point. If we define $g(u)$ as the component of $f(u)-u$
orthogonal to $u$, then $g$ is a nowhere vanishing vector field on $S^{n-1}$. By the generalised Hairy Ball Theorem, this is only possible
if $S^{n-1}$ has zero Euler characteristic, that is if $n$ is even.