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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?

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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.

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