For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues. Reading a linear algebra textbook, I encounter

For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues.

I noticed in Determinant-free proof that a real $n \times n$ matrix has at least one real eigenvalue when $n$ is odd., the proof without determinant are posts.
It seems for me with the trick of determinant, this problem might be easy.  The first thing I came up with is $\det(A_n-\lambda I_n)$ but this does not guarantee that the solution has at least one real eigenvalues. i.e., setting characteristic polynomial as $f_{A}(\lambda) =\lambda^n + a_{n-1} \lambda^{n-2} + \cdots + a_0$ with $a_i \in \mathbb{R}$, we do not know one of $\lambda$ should be real.
What can be the simple proof with determinants?
 A: Setting the characteristic polynomial as $f_{A}(\lambda) =\lambda^n + a_{n-1} \lambda^{n-2} + \cdots + a_0$ with $a_i \in \mathbb{R}$, we know that $f_A$ has at least one real root when $n$ is odd by the intermediate value theorem.
As $\lambda \to -\infty$, $f_A(\lambda) \to - \infty$; as $\lambda \to +\infty$, $f_A(\lambda) \to +\infty$. We know both of these, because that's the behavior of $\lambda^n$ for large $|\lambda|$, and $\lambda_n$ will eventually dominate over the other terms.
Therefore there must be some very negative value of $\lambda$ where $f_A(\lambda)<0$, and some very positive value of $\lambda$ where $f_A(\lambda)>0$. Between these, there must be a value of $\lambda$ where $f_A(\lambda)=0$, and this gives us a real eigenvalue.
A: If $n$ is odd, then
$$f_A( \lambda) \to \infty$$
as $ \lambda \to \infty$
and
$$f_A( \lambda) \to -\infty$$
as $ \lambda \to -\infty.$
The IVT shows now that there is $ \lambda_0 \in \mathbb R$ such that $f_A( \lambda_0=0.$
A: For odd $n$, the characteristic polynomial of an $n\times n$ real matrix has real coefficients and odd degree, but an odd degree polynomial with real coefficients has a real root by the intermediate value theorem.
A: I think you can use your characteristic polynomial. In fact, if you factorize $f_A$, you get a result like this :
$$f_A(\lambda)=\prod \underbrace{(\lambda-\lambda_i)^{a_i}}_{\text{real eigenvalues}}\times\prod \underbrace{(b_i\lambda^2+c_i\lambda+d_i)^{2e_i}}_{\text{corresponds to complex eigenvalues}} $$
If you had only complex eigenvalues, it would mean that $\forall i, a_i=0 $, so that $\sum 2e_i=n $. But n is odd, and $2e_i$ is always even, so $\sum 2e_i $ is also even, which is a contradiction.
$$ $$
So you get that $ \exists i, a_i\neq 0$, so that there is at least one real eigenvalue
A: The characteristic polynomial is of odd degree.  Since complex roots come in conjugate pairs, and by the fundamental theorem of algebra there are an odd number of roots over $\Bbb C$ (counting multiplicity), there must be a real root.
