 Solution

Let $Az=\lambda z$ with $z\ne 0$, allowing complex eigenvectors. A real symmetric matrix is Hermitian, so $z^{\ast }Az$ is real and $\lambda =z^{\ast }Az/(z^{\ast }z)$ is real. For eigenvectors $u,v$ of distinct real eigenvalues $\lambda ,\mu$, self-adjointness gives $\lambda u^{\ast }v=u^{\ast }Av=\mu u^{\ast }v$, hence $u^{\ast }v=0$. Thus the distinct eigenspaces are orthogonal.

Here $A=3I-J$, where $J$ is the all-ones matrix. Since $J(1,1,1{)}^T=3(1,1,1{)}^T$ and $J$ vanishes on the plane of coordinate sum zero,Every nonzero vector in the indicated eigenspace is an eigenvector. The repeated eigenvalue three has a two-dimensional eigenspace.

A complex symmetric nilpotent matrix provides the requested counterexample:Its characteristic polynomial is ${\lambda }^2$, so its only eigenvalue is zero. It is not diagonalizable: a diagonalizable matrix with only zero eigenvalues would itself be zero. Complex symmetry $N^T=N$ is weaker than the Hermitian condition $N^{\ast }=N$.