$\det(xA - B) = 0$ and diagonalization Let $A, B$ be two $3 \times 3$ (complex) symmetric matrices and suppose the equation $\det(xA - B) = 0$ has three distinct solutions. Prove that $A$ is invertible.
Any help appreciated!

Since @user1551 didn't submit an answer, I'm giving one that is inspired on his comment. Nice one!
For starters the matrices do not have to be symmetric. They do not have to be of size $3\times 3$ either, but can be of arbitrary size $n \times n$. We do require that there are exactly $n$ distinct solutions, because otherwise $A=B=0$ is a counter example.
Let $A=U\Sigma V^*$ be a singular value decomposition (SVD) of matrix $A$, which is guaranteed to exist for any complex matrix $A$. Then $U$ and $V$ are unitary matrices and $\Sigma$ is a diagonal matrix of the singular values. 
It follows that:
$$\det(xA-B)=\det(xU\Sigma V^*-B)=\det(U(x\Sigma-U^*BV)V^*)=\det(x\Sigma-U^*BV)$$
Suppose $A$ is singular. Then the diagonal matrix $\Sigma$ has a $0$ on its diagonal. Consequently $(x\Sigma-U^*BV)$ has a row that does not contain $x$. When we apply the Laplace expansion to calculate it, we find that the resulting determinant cannot be of the maximum degree. Therefore there cannot be exactly $n$ distinct solutions.
Thus $A$ cannot be singular and must therefore be invertible.
