Why are this all the eigenvalues/eigenvectors of this transformation? Consider the linear transformation: $$ f(x)\longmapsto \int_{-x}^x f(t)dt,$$ in the linear space (over $\mathbb{R}$) that is span by $ \langle 1,\cos(x),sin(x),\dots,\cos(mx),\sin(mx)\rangle$ for $m\in \mathbb{N}$. My professor in his solution claims that the only eigenvalue of this transformation is $0$ and is the same eigenvalue for all eigenvectors and these have the form $\sin(kx)$ for some $k\in \mathbb{N}$.
Is not hard to calculate the integrals from all elements of the set given and see that is true that $0$ is an eigenvalue and indeed some eigenvectors have this form, but I don't understand why these are the ONLY eignevalues/vectors. Can you give me some help here?
 A: Let $T$ denote the transformation. Note that the eigenspace associated with the eigenvalue $0$ consists of the set of all odd functions within the domain. Now, show that $Tf$ is itself an odd function for all $f$. It follows that for all functions $f$ from the domain, we have $T(T(f)) = 0$.
Now, suppose that $\lambda$ is an eigenvalue of $T$ and that $f \neq 0$ is an associated eigenvector. We have $T(T(f)) = 0 \implies \lambda^2 f = 0$. This implies that $\lambda^2 = 0$, which means that $\lambda = 0$. So, the only possible eigenvalue of $T$ is $0$.
A: First, suppose $f \in \mathrm{span}\left(1,\cos,\sin,\ldots,x\mapsto\cos(nx),x\mapsto \sin(nx)\ldots \right)$ and write $u(f) : x \mapsto \int_{-x}^x f$. Then $u(f)$ is an odd function because:
\begin{align}
\forall x\in \mathbb{R},~u(f)(-x) &= \int_{x}^{-x} f \\
&= - \int_{-x}^x f\\
&= -u(f)(x)
\end{align}
Then, if $\lambda\neq 0$ is an engeinvalue with eigenvector $f$ of $u$, $u(f) = \lambda f$, and $f$ is forced to be an odd function. Thus, $f$ is an odd function in $\mathrm{span}\left(1,\cos,\sin,\ldots,x\mapsto\cos(nx),x\mapsto \sin(nx)\ldots \right)$. There exists $n_1,\ldots, n_k$ integers and $\lambda_1,\ldots,\lambda_k$ reals so that
\begin{align}
\forall x \in \mathbb{R},~f(x)= \sum_{j=1}^k \lambda_j\sin\left(n_j x \right)
\end{align}
But every odd function in the above vector space is an eigenvector with eigenvalue $0$! This is an easy computation. So the only eigenvalue for $u$ is $0$.
