If $V$ finite dimensional show that $f\circ g$ and $g\circ f$ have the same eigenvalues 
Let $f,g:V\to V$ be $2$ endomorphisms, $V$ finite dimensional show that $f\circ g$ and $g\circ f$ have the same eigenvalues

If $\lambda$ is eigenvalue of $f\circ g$ to eigenvector $v$ then $f(g(v))=\lambda v$, then $\lambda$ is the eigenvalue of $g\circ f$ to the eigenvector $g(v)$, but where is the problem, if the dimension is not finite ?
 A: Let $\lambda\ne0$ be an eigenvalue of $f\circ g$; then there exists $v\ne0$ such that
$$
f(g(v))=\lambda v
$$
Then
$$
g(f(g(v))=\lambda g(v)
$$
and $g(v)\ne0$, otherwise $f(g(v))=0$ contrary to $\lambda v\ne0$. Thus $g(v)$ is an eigenvector for $g\circ f$ relative to $\lambda$.
Similarly, nonzero eigenvalues of $g\circ f$ are eigenvalues of $f\circ g$.
This shows that $f\circ g$ and $g\circ f$ share their nonzero eigenvalues and this does not depend on the dimension of $V$.
The problem is thus with a possible zero eigenvalue.
Assume $V$ is finite dimensional. If $0$ is an eigenvalue of $f\circ g$, then the rank of $f\circ g$ is less than $\dim V$. Therefore $g\circ f$ cannot be invertible, because otherwise $g$ would be surjective (hence invertible) and $f$ would be injective (hence invertible). Thus $g\circ f$ is not invertible and so it has $0$ as eigenvalue.
Note that the two clauses “hence invertible” depend on the fact that $V$ is finite dimensional. It's easy to find examples of a pair of linear maps $f$ and $g$ such that $g\circ f$ is not invertible and $f\circ g$ is invertible, assuming $V$ is not finite dimensional.
A: Consider $V=C^{\infty}(\Bbb R)$ and for $\phi\in V$ let $f(\phi)(x)=\phi'(x)$, $g(\phi)(x)=\int_0^x\phi(t)\,\mathrm dt$. Then for $\phi(x)=1$ we have $(g\circ f)(\phi)=0$, so $\phi$ is an eigenvector to eigenvalue $\lambda=0$. However, $f\circ g$ is the identity map (Fundamental Theorem), hence there is no (non-zero!) eigenvector with eigenvalue $0$.
A: Your argument has a gap: $g(v)$ could be $0$ in which case it takes more justification to show that $\lambda$ is an eigenvalue of $g \circ f$.  This can be repaired if the vector space is finite dimensional (see https://en.wikipedia.org/wiki/Characteristic_polynomial#Characteristic_polynomial_of_a_product_of_two_matrices), but it fails in any infinite-dimensional vector space.
A standard example is $V$ being the vector space of sequences of real numbers.  Take the right- and left-shift maps:
$$\begin{align}
f(x_1,x_2,x_3,\ldots) &:= (0,x_1,x_2,\ldots); \\
g(x_1,x_2,x_3,\ldots) &:= (x_2,x_3,x_4,\ldots).
\end{align}$$
Then $g(f(x)) = x$, so $g \circ f$ has only one eigenvalue $1$, but $f(g(x))$ zeroes out the first element of $x$, so it has eigenvalues $0$ and $1$.
Hagen von Eitzen's example is essentially the same pair of maps if you consider the power series of just the real-analytic functions in $C^\infty(\mathbb R)$.  You could also restrict to just polynomials (corresponding to sequences with finitely many non-zero terms).
