Is a complex symmetric matrix self-adjoint? Is it normal?
Solution
1. A complex matrix \(A\) is symmetric if \(A^t = A\), and self-adjoint (Hermitian) if \(A^* = \overline{A}^t = A\).
definitions.
2. A complex symmetric matrix need not be self-adjoint.
e.g. \(A = \begin{pmatrix} i & 0 \\ 0 & i \end{pmatrix}\) is symmetric (\(A^t = A\)) but \(A^* = \begin{pmatrix} -i & 0 \\ 0 & -i \end{pmatrix} \neq A\).
3. A complex symmetric matrix need not be normal.
e.g. \(A = \begin{pmatrix} 1 & i \\ i & 0 \end{pmatrix}\) is symmetric, but \(A A^* \neq A^* A\) (direct computation shows \(AA^* = \begin{pmatrix} 2 & -i \\ i & 1 \end{pmatrix}\) while \(A^*A = \begin{pmatrix} 2 & i \\ -i & 1 \end{pmatrix}\)).
4. Hence the answer to both questions is: not necessarily.
2 and 3.
5. Q.E.D.
4.