I have one simple question. How I suppose to show that $\lambda =0$ is an eigenvalue of some problem. Does it mean that I must have non-trivial solution for $\lambda=0 $?
UPD:I mean by problem some Sturm–Liouville equation...
Suppose that $A$ is a linear operator of some vector space. Then $\lambda$ is an eigenvalue if and only if there exists non-zero vector $x$ such that $Ax= \lambda x$. In the case of $\lambda=0$, we must have $Ax=0, x \neq 0$.
Find a vector $v$ such that :
1) $v$ is not zero ;
2) $Av = 0$, where $A$ is the matrix in your problem.