Intuitive geometric explanation: existence of eigenvalue in odd dimension real vector space. 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.
 A: 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.
A: I guess deep down it boils down to the fact that 'basic' rotations are two-dimensional, so that general rotations (as combinations of 'disjoint' basic rotations) are always 'even dimensional'.
This means there's always at least one left-over dimension in spaces of odd dimension.

To make it more precise, consider the polar decomposition of $T$.
More generally, let $\mathbb{K}$ be either $\mathbb{R}$ or $\mathbb{C}$. Every linear transformation $M:\mathbb{K}^n\longrightarrow \mathbb{K}^m$ admits a singular value decomposition
$$M=U\Sigma V^*$$
where $U:\mathbb{K}^m\longrightarrow \mathbb{K}^m$ and $V:\mathbb{K}^n\longrightarrow \mathbb{K}^n$ are unitary and $\Sigma:\mathbb{K}^n\longrightarrow \mathbb{K}^m$ is diagonal with non-negative real entries.
In particular if $\mathbb{K}=\mathbb{R}$, all entries involved are real $($and $V^*$ boils down to $V^T)$.
Observing that $VV^T=V^TV=I$, we may then consider the decomposition
$$T=V\left[\underbrace{\left(V^TU\right)}_ {Q}\Sigma\right]V^T$$
The $V$ and $V^T=V^{-1}$ outside brackets merely indicate a change of basis.
Therefore, in essence $T$ can be written as a composition $Q\Sigma$, where $Q$ is orthogonal as is easily checked.
This is a polar decomposition of $T$ $($in analogy to the polar form $z=re^{i\theta}$, except here the order is more like $z=e^{i\theta}r)$.
Now, assume $W\simeq \mathbb{R}^n$ for some odd $n$, and moreover let's assume it has nontrivial kernel otherwise there's nothing really to see.
In this case, $\Sigma$'s diagonal entries are all positive so it is merely some (nonzero) stretching of each of our axes.
Plenty of eigenspaces up till now!
All that's left is to apply the orthogonal transformation $Q$, and here's where the initial observation about rotations being two dimensional comes into play.
