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?


  • 3
    $\begingroup$ Have you tried checking this using the definitions? $\endgroup$ – Jonas Meyer Jul 2 '15 at 3:52
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $|D|^2$ was lazy notation for taking the magnitude squared of the entries of $D$. $\endgroup$ – Omnomnomnom Jul 5 '15 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.