Spectrum of $Tu=\int^1_0 (x+y)u(y)dy$ Given the operator 
$$Tu(x)=\int^1_0 (x+y)u(y)dy$$ 
on $L^2(0,1)$, find the spectrum of $T$.  For all eigenvalues, find their multiplicities and the eigenfunctions.   
The kernel is Hilbert Schmidt and symmetric, so we know $T$ is compact and self adjoint.
The self adjointness tells us that $\sigma_r(T)=\emptyset$. 
I have found so far that for $\lambda_\pm=1/2\pm 1/\sqrt 3$  we have eigenfunctions $u_\pm(x)=c(x\pm 1/\sqrt 3)$ respectively- both with multiplicity one.  (Do you agree?)
$\lambda=0$ is giving me some trouble.  By compactness, it is either in the point spectrum or continuous spectrum, but which?  I answered this question with what I think so far.  Please comment.
 A: Consider $\lambda=0$.
$$\int^1_0 (x+y)u(y)dy = 0 $$ 
or 
$$x\int^1_0 u(y)dy+\int^1_0 yu(y)dy = 0 $$ 
So $u(x)$ is an eigenfunction if both $<y,u>=0$ and $<1,u>=0$ (by linear independence).   It seems me that this implies $0$ is an eigenvalue of infinite multiplicity because any polynomial of degree greater than 2 can be made orthogonal to both $1$ and $y$. 
A: The subspace $M$ spanned by $\{ 1, x\}$ is invariant under this operator. And, if $u \perp M$, then $Tu=0$. So this completely reduces to a $2\times 2$ matrix problem. You could use an orthonormal basis for $M$, but there's no real need that I can tell. Using the ordered basis $b=\{ 1,x\}$,
$$
         Tu = x(u,1)+1(u,x),\;\;\; (f,g) = \int_{0}^{1}f(t)g(t)\,dt.\\
               T1 = x(1,1)+1(1,x) = \frac{1}{2}1+x,\\
               Tx = x(x,1)+1(x,x) = \frac{1}{3}1+\frac{1}{2}x.
$$
So $T$ has the following matrix representation with respect to the basis $\{ 1,x\}$:
$$
              [T]_{b,b} = \left[\begin{array}{cc}\frac{1}{2} & \frac{1}{3} \\ 1 & \frac{1}{2}\end{array}\right]
$$
That definitely has characeristic polynomial $(\lambda-1/2)^{2}-1/3$ with roots as you stated. So I agree that those are eigenvalues. The remaining eigenvalue is $0$ because $T=0$ on $M^{\perp}$.
To find the eigenvectors of $T$ in $M$, first consider
$$
           [T]-\left(1/2-\frac{1}{\sqrt{3}}\right)I=
           \left[\begin{array}{cc}
                 \frac{1}{\sqrt{3}} & \frac{1}{3} \\
                    1 & \frac{1}{\sqrt{3}}
                 \end{array}\right],
$$
whose null space is spanned by
$$
        \left[\begin{array}{c} \frac{1}{\sqrt{3}} \\ -1\end{array}\right]
$$
In terms of polynomials, $\frac{1}{\sqrt{3}}-x$ is an eigenvector of $T$ with eigenvalue $1/2-1/\sqrt{3}$. Similarly $\frac{1}{\sqrt{3}}+x$ is an eigenvector of $T$ with eigenvvalue $1/2+1/\sqrt{3}$. It can be checked that these eigenvectors are orthogonal because
$$
   (1/\sqrt{3}-x,1/\sqrt{3}+x) = \int_{0}^{1}\left(\frac{1}{3}-x^{2}\right)\,dx=0.
$$
