Eigenvalues of Operator on $L^{2}[0, 1]$ Let $X=L^{2}[0, 1]$ and $$Ax(t)=\int_{0}^{1}{ts(1-ts)x(s)}ds.$$ I have shown that this operator is compact, hence all non-zero elements of the spectrum $\sigma(A)$ are eigenvalues of $A$. But then I am stuck to find the eigenvalues in order to find the spectrum, i.e. all $\lambda \ne 0$ such that $$\int_{0}^{1}{ts(1-ts)x(s)}ds=\lambda x(t).$$ Can someone help?
 A: Old, Naive, Too Complicated Solution.
Suppose that $x$ is an eigenfunction of $A$ w/ eigenvalue $\lambda$.  That is,
$$\lambda x(t)=Ax(t)=\int_0^1 ts(1-ts)x(s)\ ds.$$
See a remark in the comment section below that proves differentiability of $x$.  Note that
$$\lambda x'(t)=\int_0^1 s(1-ts)x(s)\ ds-t\int_0^1s^2x(s)\ ds.$$
That is,
$$\lambda t x'(t)=\lambda x(t)-t^2\gamma(x)\tag{1}$$
where $\gamma:L^2[0,1]\to \Bbb C$ is given by $\gamma(x)=\int_0^1 s^2 x(s)\ ds$.  
Observe that the eigenspace with the eigenvalue $0$ is given by
$$\ker A=\left\{x\in L^2[0,1]:\int_0^1sx(s)\ ds=0\wedge \int_0^1s^2x(s)\ ds=0\right\}.$$
From now on suppose that $\lambda\ne 0$.
From (1), we get
$$\frac{d}{dt}\frac{x(t)}{t}=-\frac{1}{\lambda}\gamma(x).$$
Therefore,
$$\frac{x(t)}{t}=C-\frac{t}{\lambda}\gamma(x)$$
or 
$$x(t)=Ct-\frac{t^2}{\lambda}\gamma(x).$$
Since $\gamma(x)=\int_0^1s^2x(s)\ ds$, we need
$$\gamma(x)=\int_0^1s^2\left(Cs-\frac{s^2}{\lambda}\gamma(x)\right)\ ds.$$
So
$$\gamma(x)=\frac{C}{4}-\frac{1}{5\lambda}\gamma(x)\implies C=\left(4+\frac4{5\lambda}\right)\gamma(x).$$
Hence,
$$x(t)=\gamma(x)\Biggl(\left(4+\frac4{5\lambda}\right){t}-\frac{t^2}{\lambda}\Biggr).\tag{2}$$
Therefore, the eigenspace of $A$ w/ eigenvalue $\lambda \ne 0$ is a subspace of the span of
$$x_\lambda(t)=\left(4+\frac4{5\lambda}\right){t}-\frac{t^2}{\lambda}.$$
Plugging this into $\lambda x_\lambda (t)=Ax_\lambda (t)$, it turns out that $\lambda$ must satisfy
$$\lambda\left(4+\frac{4}{5\lambda}\right)=\frac{\frac{1}{\lambda}+80}{60}.$$   That is, $\lambda=\frac{4\pm\sqrt{31}}{61}$.

Simpler Solution.
Suppose that $x$ is an eigenfunction of $A$ w/ eigenvalue $\lambda\neq 0$.  Then,
$$x(t)=\frac{Ax(t)}{\lambda}=\left(\frac{\int_0^1 sx(s)\ ds}{\lambda}\right)t+\left(-\frac{\int_0^1s^2x(s)\ ds}{\lambda}\right)t^2.$$
That is, $x(t)=at+bt^2$ for some constants $a,b$.  Plugging this into $\lambda x=Ax$, we get
$$\lambda(at+bt^2)=\left(\frac{a}{3}+\frac{b}{4}\right) t +\left(-\frac{a}{4}-\frac{b}{5}\right)t^2.$$
Hence,
$$\frac{a}{3}+\frac{b}{4}=\lambda a\wedge -\frac{a}{4}-\frac{b}{5}=\lambda b.$$
So $\lambda$ is an eigenvalue of the matrix
$$\begin{pmatrix}\frac13 & \frac14\\ -\frac14 &-\frac15\end{pmatrix},$$ which means $\lambda=\frac{4\pm\sqrt{31}}{60}$.
A: Here is how I would approach this question: Keep differentiating both sides with respect to $t$ until you reach zero on one side. This will show that an eigenfunction has to be a polynomial of a certain degree. Work out the equations for the coefficients of that polynomial from the equation we got so far…
By the way: This operator seems to have a pretty large null space.
