Eigenvalues and eigenvectors of an integral operator We have the following integral operator
$$
Ku(t)=\int_0^1 G(t,s)\, u(s)\, ds,\,\, u\in L^2[0,1],
$$
where $$G(t,s)=\begin{cases} s(1-t)~ 0\leq s\leq t\leq 1\\ t(1-s)~ 0\leq t\leq s\leq 1\end{cases}$$if the eingenvalues of $K$ are $1/k^2\pi^2$ what can we say about the eigenvalues of $K^2=K\circ K$?
Thank you. 
 A: Let us denote by $\sigma_p(T)$ the set of all eigenvalues of an operator $T$. Then, for any operator $K$ acting on a complex space, we have 
$$\sigma_p(K^2)=\{ \lambda^2;\; \lambda\in \sigma_p(K)\}.$$
As observed by Yiorgos, one inclusion is obvious: if $\lambda\in\sigma_p(K)$ and if $u$ is an associated eigenvector, then $K^2(u)=K(Ku)=K(\lambda u)=\lambda\, Ku=\lambda^2 u$ and hence $\lambda^2\in\sigma_p(K^2)$. This part works on a real or complex space.
Conversely, let $\mu\in\sigma_p(K^2)$, and let $u$ be an associated eigenvector. Choose $\alpha\in\mathbb C$ such that $\alpha^2=\mu$. Then we have 
$$ 0=(K^2-\mu Id)u=(K-\alpha Id)(K+\alpha Id)u\, .$$
If $v:=(K+\alpha Id)u\,$  is not $0$, it follows that $\alpha$ is an eigenvalue of $K$ with eigenvector $v$. If $v=0$, then $-\alpha$ is an eigenvalue of $K$ with eigenvector $u$. In both cases, we have $\mu\in\{ \lambda^2;\; \lambda\in\sigma_p(K)\}$.
For operators acting on real spaces, this may not be true. For example, consider the operator $K$ acting on $\mathbb R^2$ whose matrix is 
$$\left(\begin{matrix}0&-1\\1&0
\end{matrix}\right) $$
Then $K^2=-Id$, so $-1$ is an eigenvalue of $K^2$; but $K$ does not have any eigenvalue.
A: If $Ku=\lambda u$, then $K^2 u=\lambda^2 u$. Hence the eigenvalues of $K^2$ are
$$
\frac{1}{\pi^4n^4}, \quad n\in\mathbb N,
$$
