Solution
= Solution
\b[False.] The <matrix> $A=\begin{pmatrix}1&1\\0&2\end{pmatrix}$ has real <eigenvalues> $1$ and $2$, because its <characteristic polynomial> is $(t-1)(t-2)$, but $A^*=\begin{pmatrix}1&0\\1&2\end{pmatrix}\ne A$. Real spectrum alone does not imply that it is a <Hermitian matrix>.