Prove that all square matrices with odd number of rows or columns have an eigenvalue

Question

Let $A∈\mathbf R^{n×n}$, with $n$ odd. How can I prove that $\det(A-xI)$ equals zero for some $x∈\mathbf R$?

I have already known the way to solve this by means of polynomial's properties. Could you please provide a new angle to solve this?

Thx.

Answer 1

Any polynomial of odd degree on R will have at least one zero, as it goes to -infinity (hence becomes negative) when x goes to -infinity and to +infinity (hence becomes positive) when x goes to +infinity. By continuity, it must cancel at some point.