# 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,

• Have you tried checking this using the definitions? – Jonas Meyer Jul 2 '15 at 3:52
• The answer is yes. As JM suggests above, you should try to prove that this is the case using definitions; it's easier than you might expect. – Omnomnomnom Jul 2 '15 at 4:56
• Ok will do - thanks! I wanted something to think about on my walk home - the library was closing so I squeezed in this question before leaving. :-) – User001 Jul 2 '15 at 5:19

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.
• second question is: is D*D = |D|^2 ... essentially a definition of the operator norm? (And then it'd also be true for $D^tD$.) – User001 Jul 5 '15 at 21:15
• The spectral theorem guarantees that real symmetric matrices have real eigenvalues. So, in the converse (of the real case), we assume $D$ is real. – Omnomnomnom Jul 5 '15 at 22:41
• $|D|^2$ was lazy notation for taking the magnitude squared of the entries of $D$. – Omnomnomnom Jul 5 '15 at 22:55