Let $A$ be an $n \times n$ invertible complex matrix. Let $Gr(k)=Gr(k,\mathbb{C}^n)$ be the complex $k$-Grassmannian, $1\leq k \leq n$. Since $A$ is invertible, it maps a $k$-dimensional subspace to a $k$-dimensional subspace, so it gives a function (which I'll call $A\_k$) on $Gr(k)$. The fact that matrices have eigenvalues lets us deduce that $A$ has a $k$-dimensional invariant subspace - i.e. this map $A\_k$ has a fixed point.


My question is this: is there a fixed point theorem on $Gr(k)$ (which ideally does not in any way depend on the existence of eigenvalues) that we can invoke to arrive at the same conclusion?



Invariant subspaces for matrices via fixed points on Grassmannians
I think the Lefschetz fixed point theorem still applies. If a self-map M→M of a compact orientable manifold M has no fixed points than the Euler characteristic of M is zero. But if M is a complex Grassmannian then its odd Betti numbers vanish so the Euler characteristic is positive (since b\_0=1).


