Show that non-diagonal symmetric matrices must have at least two distinct eigenvalues. "Food for thought" problem in university math class.  I know that symmetric matrices are diagonalizable and thus they are similar to a diagonal matrix which would have the same eigenvalues, but how would I show that these eigenvalues are necessarily distinct?

By contradiction, assume $A$ is non-diagonal and symmetric, and has only one eigenvalue $\lambda$. Then, since $A$ can be diagonalized (it is symmetric): there exist $P, \Delta$ such that
$$
A = P^{-1}\Delta P
$$
But the assumption that $A$ only has one eigenvalue means that $\Delta=\lambda I$, and therefore 
$$
A = P^{-1}\lambda I P = \lambda P^{-1} I P = \lambda P^{-1} P = \lambda I
$$
and $A$ is diagonal, contradicting the hypothesis.
