Is the converse of the Spectral Theorem true? In the book by Friedberg, Insel and Spence, symmetric matrices are orthogonally diagonalizable, and over the complex number field, normal matrices are orthogonally diagonalizable -- this is all from the Spectral Theorem. Is the converse true? Are orthogonally diagonalizable matrices in real space necessarily symmetric? Are orthogonally diagonalizable matrices in complex space necessarily normal? Thanks,

The answer to this question is yes.

If $A$ is real and orthogonally diagonalizable, then $A = UDU^T$ for some orthogonal matrix $U$ and real diagonal matrix $D$. We find that $$ A^T = (UDU^T)^T = UDU^T = A $$ so that $A$ is symmetric.

Similarly, if $A$ is complex and unitarily diagonalizable, then $A = UDU^*$ for some unitary matrix $U$ and (complex) diagonal matrix $D$. We find that $$ \begin{align} A^*A &= (UDU^*)^*(UDU^*) = UD^*(U^*U)DU^* = UD^*DU^* =U|D|^2 U^* \\\ & = UDD^*U^* = UD(U^*U)D^*U^* = (UDU^*)(UDU^*)^* = AA^* \end{align} $$ so that $A$ is normal.