Solution
= Solution
For a real <eigenvalue> $\lambda$, choose a nonzero real <eigenvector> $v$ with $Av=\lambda v$. By (e),
$$
0=v^\top Av=\lambda v^\top v.
$$
Since $v^\top v=\|v\|^2>0$, \b[$\lambda=0$]. If <eigenvectors> were initially allowed to be complex, the real <matrix> $A-\lambda I$ has a nonzero real <null space> whenever it has a nonzero complex <null space>, so this choice of $v$ remains justified.