Let $E$ an $\mathbb{R}$-vectorial space and $f : E \rightarrow E $ an endomorphism. Prove that:
$(a)$ $dim (E)$ odd $\Rightarrow$ $f$ has at least one eigenvalue.
$(b)$ $dim (E)$ pair and $det(f) < 0 $ $\Rightarrow$ $f$ has at least two different eigenvalues.
$(c)$ It is also possible that $dim (E)$ pair and $f$ has no eigenvalue.
Any ideas to begin with?
Here's a sketch/hints for you to think about:
(a) every odd degree polynomial has a real root.
(b) if $p(t)$ is the characteristic polynomial of $f$ and it has even degree, then $\lim_{t \to \pm \infty}p(t)=+\infty$, but $p(0)<0$, so the intermediate value theorem gives at least two distinct roots (one positive and one negative).
(c) reverse engineer some $f$, in the case $\dim E =2$, with $p(t)=t^2+1$.
