Odd-sized real matrices have a real eigenvalue

Idea. Eigenvalues are roots of the characteristic polynomial $\chi_A(t)=\det(tI-A)$. A real polynomial of odd degree changes sign, so the intermediate value theorem gives a real root.

Step 1. Expanding the determinant, $\chi_A(t)=t^n+c_{n-1}t^{n-1}+\dots+c_0$ with real $c_j$: it is monic of degree $n$.

Step 2: signs at infinity. Write $\chi_A(t)=t^n\bigl(1+c_{n-1}/t+\dots+c_0/t^n\bigr)$. The bracket tends to $1$ as $|t|\to\infty$, and $t^n\to+\infty$ as $t\to+\infty$, $t^n\to-\infty$ as $t\to-\infty$ because $n$ is odd. Hence there are $a<b$ with $\chi_A(a)<0<\chi_A(b)$.

Step 3: IVT. $\chi_A$ is continuous, so it vanishes at some $t_0\in(a,b)$. Then $\det(t_0I-A)=0$, i.e. $t_0$ is a real eigenvalue.

Remark. For $n$ even the statement is false: the rotation matrix $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has eigenvalues $\pm i$ only. Since non-real roots of a real polynomial come in conjugate pairs, an equivalent argument is that an odd number of roots cannot all be paired.

Lesson. Odd degree plus continuity yields a real root.