Determinant-free proof that a real $n \times n$ matrix has at least one real eigenvalue when $n$ is odd.

This is a consequence of the hairy ball theorem.

Given a real $n \times n$ matrix $A$, define a function on the unit sphere in $\mathbb R^n$ by mapping a unit vector $\vec x$ to the component of $A\vec x$ perpendicular to $\vec x$: that is, $$\vec x \mapsto A \vec x - \langle A\vec x, \vec x\rangle \vec x.$$ Being perpendicular to $\vec x$, the result is always tangent to the sphere.

When $n$ is odd, the unit sphere in $\mathbb R^n$ is even-dimensional, and so (by the hairy ball theorem) there is no nonvanishing continuous function with the above property. This means there must be some unit vector $\vec x \in \mathbb R^n$ such that $A \vec x - \langle A\vec x, \vec x\rangle \vec x = \vec 0$: that is, $\vec x$ is an eigenvector of $A$ with eigenvalue $\langle A\vec x, \vec x\rangle$.

If $A$ is a matrix and $$ Av=\lambda v $$ then we also have $$ A\overline{v}=\overline{A}\overline{v}=\overline{Av}=\overline{\lambda v}=\overline{\lambda}\overline{v} $$ where we use the equality $A=\overline{A}$ because $A$ has real entries. Thus the eigenvalues of $A$ come in complex conjugate pairs, so in particular if $A$ has $2k+1$ eigenvalues, at least $1$ must be real.