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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,

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    $\begingroup$ Have you tried checking this using the definitions? $\endgroup$ – Jonas Meyer Jul 2 '15 at 3:52
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    $\begingroup$ 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. $\endgroup$ – Omnomnomnom Jul 2 '15 at 4:56
  • $\begingroup$ 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. :-) $\endgroup$ – User001 Jul 2 '15 at 5:19
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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.

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  • $\begingroup$ Thanks so much @Omnomnomnom. I actually have two quick questions on your proof: recently, I saw (through an exercise) that real matrices can have complex eigenvalues. This shouldn't change your first proof at all, except for the fact that we should say a diagonal matrix D instead of real diagonal matrix D, unless there is some result that guarantees the eigenvalues are real - A being symmetric does just that, but that is the result we were going after, so we can't assume that, I think? $\endgroup$ – User001 Jul 5 '15 at 21:14
  • $\begingroup$ second question is: is D*D = |D|^2 ... essentially a definition of the operator norm? (And then it'd also be true for $D^tD$.) $\endgroup$ – User001 Jul 5 '15 at 21:15
  • $\begingroup$ adding to my first comment, even if the ground field of scalars is real, making the vector space a real (inner product) vector space, I think the matrices can still have complex eigenvalues - irrespective of the ground field. I'm not 100% sure, but it seems that way, though...again through a recent problem that I worked on... $\endgroup$ – User001 Jul 5 '15 at 21:18
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    $\begingroup$ The spectral theorem guarantees that real symmetric matrices have real eigenvalues. So, in the converse (of the real case), we assume $D$ is real. $\endgroup$ – Omnomnomnom Jul 5 '15 at 22:41
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    $\begingroup$ $|D|^2$ was lazy notation for taking the magnitude squared of the entries of $D$. $\endgroup$ – Omnomnomnom Jul 5 '15 at 22:55

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