I want to prove that all orthogonal matrices are diagonalizable over $C$. I know that a matrix is orthogonal if $Q^TQ = QQ^T = I$ and $Q^T = Q^{-1}$, and that a matrix $A$ is diagonalizable if $A = PDP^{-1}$ where $D$ is a diagonal matrix. How can I start this proof?

  • $\begingroup$ Maybe, in the first row, you meant to write "all orthogonal m. are diagonalizable ..." ? $\endgroup$ – G Cab Apr 11 '16 at 13:34
  • $\begingroup$ I don't understand how that's different from what I've already written...? $\endgroup$ – jackwise Apr 11 '16 at 13:37
  • $\begingroup$ In first line (not title) I read "..all diagonalizable m. are diagonalizable .." $\endgroup$ – G Cab Apr 11 '16 at 13:45
  • $\begingroup$ ^I edited it to fix that. To the OP, do you know how to prove that a normal matrix is diagonalizable? If so, orthogonal matrices are normal, which would finish the proof. $\endgroup$ – Nicholas Stull Apr 11 '16 at 13:48
  • $\begingroup$ Ack, sorry, it's early and I missed that. And @NicholasStull I do know it to some degree but I would still appreciate seeing it written out. $\endgroup$ – jackwise Apr 11 '16 at 13:54

As people have indicated, you could simply apply the spectral theorem. Here I run through a specialized argument to the orthogonal case:

Since $Q$ is orthogonal we have $\langle Qv, Qw \rangle = (Qv)^*Qw = v^* Q^T Q w = \langle v, w \rangle$.

Given any eigenvector $v$ with eigenvalue $\lambda$, if we have some vector $w$ orthogonal to $v$ then we have $\lambda \langle v, Qw \rangle = \langle Qv, Qw \rangle = \langle v, w \rangle = 0$, so $Q$ maps $v^\perp$ into itself. We can induct on the dimension of our space to show $Q$ acts diagonalizably on $v^\perp$ so it acts diagonalizably on $v \oplus v^\perp$

We can infact say more:

Note that if $\lambda$ is an eigenvector of $Q$ then we have $|\lambda|\|v\| = \langle \lambda v, \lambda v \rangle = \langle Qv, Qv \rangle = \|v\|$. We conclude all the eigenvalues have norm $1$.

If $v,w$ are eigenvectors with different eigenvalues then we have $\langle v, w \rangle = \langle Qv, Qw \rangle = \langle \lambda v, \mu w \rangle = \lambda \mu^* \langle v, w \rangle$. Thus if $\lambda \mu^* \neq 1$ then $v$ and $w$ are orthogonal.

Combining these one can show that $Q = PRP^{-1}$ where $P$ is an orthogonal matrix and $R$ is a block diagonal matrix with $1,-1$ and $2 \times 2$ rotation matrices down the diagonal.

  • $\begingroup$ What do you mean by "$Q$ acts diagonalizably on $v^\perp$ and on $v\oplus v^\perp$"? $\endgroup$ – rmdmc89 Feb 20 at 17:32
  • $\begingroup$ Since Q maps $v^\perp$ to itself we get an induced linear map, $Q|_{v^\perp}: v^\perp \to v^\perp$. It's easy to see that this map is orthogonal and on a vector space of one less dimension, so by induction this map is diagonalizable. $\endgroup$ – Nick R Feb 20 at 20:54

Note that if $Q$ is orthogonal then $Q$ is normal, because \begin{equation*} Q Q^T = Q^T Q = I. \end{equation*} So the spectral theorem implies that $Q$ is diagonalizable.


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.