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?
Linear Algebra – Proving Orthogonal Matrices are Diagonalizable
linear algebramatrices
Related Solutions
The proof of this property is not so easy as those of the basic properties of eigenvalues and eigenvectors. It can be shown by induction, or by explicit construction (see eg here)
I like to visualize the property in this way:
We know that an hermitian matrix with $n$ distict eigenvalues has $n$ eigenvectors that are not only LI (as in general matrices) but, more than that, orthogonal. We also know that this matrix is diagonalizable, with unitary $U$ (both properties are easy to prove).
Now, if our hermitian matrix happens to have repeated (degenerate) eigenvalues, we can regard it as a perturbation of some another hermitian matrix with distinct eigenvalues. By a continuity argument, we should see that the matrix perturbation than transforms different (but perhaps close) eigenvalues into coincident ones, cannot make the orthogonal eigenvectors linearly dependent.
Put in other way: an hermitian matrix $A$ with repeated eigenvalues can be expressed as the limit of a sequence of hermitian matrices with distinct eigenvalues. Because all members of the sequence have $n$ orthogonal eigenvectors, by a continuity argument, they cannot end in LD eigenvectors.
This approach leads to a nice intuition, IMO, and it can be formalized. But for a formal proof the other methods are to be preferred.
Your generalization is false. For instance, let $K=\mathbb{Q}$ and $V=\mathbb{Q}(\sqrt{2},\sqrt{3})$, with $A$ acting on $V$ as multiplication by $\sqrt{2}$ and $B$ acting on $V$ as multiplication by $\sqrt{3}$. If a simultaneous block-diagonalization of the sort you ask for exists, then all the blocks must be $2\times 2$, since the minimal polynomials of $A$ and $B$ are both irreducible of degree $2$. It follows that the span of the first two basis vectors must be invariant under both $A$ and $B$. But $V$ is irreducible with respect to $A$ and $B$--the only nontrivial subspace of $V$ which is invariant under both $A$ and $B$ is $V$ itself (since such a subspace is then invariant under multiplication by any element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$). So no such simultaneous block-diagonalization can exist.
In general, what will happen is that you can decompose your vector space $V$ as a direct sum of subspaces $V_i$ which have the form of the example above: each $V_i$ is a finite separable field extension of $K$, with $A$ and $B$ corresponding to elements of the field which generate the field over $K$ and act by multiplication. (Proof sketch: If $f$ and $g$ are the minimal polynomials of $A$ and $B$, then $V$ is a $K[x,y]/(f(x),g(y))$-module via $A$ and $B$. Since $A$ and $B$ are semisimple, $f$ and $g$ are separable and it follows that the ring $K[x,y]/(f(x),g(y))$ is semisimple, so any module over it is a direct sum of simple modules which correspond to field quotients of $K[x,y]/(f(x),g(y))$.)
In the case $K=\mathbb{R}$, these field extensions can only be $\mathbb{R}$ or $\mathbb{C}$, and in particular each of $A$ or $B$ is either just an element of $\mathbb{R}$ or else generates the entire field. It is then easy to see you can choose a basis for $V_i$ such that both $A$ and $B$ have the form you ask for. But in general, you can do no such thing, as the counterexample above shows.
Best Answer
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.