Determinant-free proof that a real $n \times n$ matrix has at least one real eigenvalue when $n$ is odd. Is there a determinant-free proof that $\textbf{A} \in \mathbb{R}^{n \times n}$ must have at least one real eigenvalue when $n$ is odd? 
I have seen a few other definitions of "the set of eigenvalues" which do not invoke the determinant, specifically the complement of the resolvent set, or the set of points for which $\lambda\textbf{I}_{n \times n} - \textbf{A}$ is singular.
 A: 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 $\langle A\vec x, \vec x\rangle$.
A: 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.
