= Solution
Let $Az=\lambda z$ with $z\ne0$, allowing complex <eigenvectors>. A real <symmetric matrix> is <Hermitian>, so $z^*Az$ is real and $\lambda=z^*Az/(z^*z)$ is real. For eigenvectors $u,v$ of distinct real <eigenvalues> $\lambda,\mu$, self-adjointness gives $\lambda u^*v=u^*Av=\mu u^*v$, hence $u^*v=0$. 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,
$$
\boxed{\lambda=0:\ E_0=\operatorname{span}\{(1,1,1)^T\};\qquad\lambda=3:\ E_3=\operatorname{span}\{(1,-1,0)^T,(1,1,-2)^T\}.}
$$
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:
$$
\boxed{N=\begin{pmatrix}1&i\\i&-1\end{pmatrix},\qquad N\ne0,\quad N^T=N,\quad N^2=0.}
$$
Its characteristic polynomial is $\lambda^2$, so its only eigenvalue is zero. \b[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 $N^*=N$.
Back to article page
