Is every orthogonal matrix orthogonally diagonalizable? Is every orthogonal matrix orthogonally diagonalizable? If so, how do you prove it?
And is it true that the entries of every orthogonal matrix is real?
Since every unitary matrix is unitarily diagonalizable, so is true that every orthogonal matrix orthogonally diagonalizable?
Thank you.
 A: The short answer is no.
Any orthogonally diagonalizable matrix must be symmetric. Indeed, if $A = UDU^T$ where $U$ is orthogonal and $D$ diagonal, then it is easy to see $A^T = A$. On the other hand, there are plenty of orthogonal matrices which aren't symmetric. For example,
$$A = \begin{bmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$$
is such a matrix.
As for the question "must the entries of orthogonal matrices be real?"--yes and no. When people say "orthogonal matrix" they mean a real orthogonal matrix.
On the other hand, one could define a set
$$O(n,\Bbb C) = \{A\in M(n,\Bbb C): A^TA=AA^T = I\}$$
but there isn't a good reason to look at such matrices. They don't preserve the complex inner product, so they're not a natural generalization of real orthogonal matrices (the unitary matrices are though, since they do preserve the complex inner product).
A: Remark 1: Yes, orthogonal matrix is a real square matrix $A$ such that $AA^T=I$.

Claim: Orthogonal matrix is orthogonally diagonalizable if and only if all its eigenvalues are real.
Proof: Every orthogonal matrix $A$ is normal (i.e. commutes with its conjugate transpose) and the set of normal matrices is precisely the set of matrices that are unitarily diagonalizable. Therefore, there exists a unitary matrix $U$ and a diagonal matrix $D$ such that
$$
A = U D U^H\tag1.
$$
Notice that the columns of $U$ are the eigenvectors of A. If $v$ is an eigenvector of $A$ associated to eigenvalue $\lambda$ then
$$
Av = \lambda v\\
A\overline{v} = \lambda \overline{v}
$$
where we used the fact that $\lambda$ and all entries in $A$ are real. We see that $\overline{v}$ is an eigenvector of $A$ associated to eigenvalue $\lambda$. Therefore, the real vector $v + \overline{v}$ is an eigenvector of $A$ associated to eigenvalue $\lambda$. Consequently, we can replace $U$ in $(1)$ with a unitary matrix with real entries. Such a matrix is orthogonal.
For the converse, assume $A$ with an eigenvalue $\lambda \notin \mathbb{R}$ is orthogonally diagonalizable and write $(1)$ as
$$
AU = UD\tag2
$$
with $U$ orthogonal (and therefore real). Note that the left hand side of $(2)$ contains only real entries. Also, note that the entries in the row on the right hand side which corresponds to $\lambda$ are not all real. The contradiction reveals that the assumption that $A$ is orthogonally diagonalizable is false.

Remark 2: Orthogonal matrices with real eigenvalues have a simple geometric characterization. Since these matrices preserve the 2-norm, their eigenvalues have absolute value $1$. The only real numbers that satisfy this condition are $1$ and $-1$. Therefore, the only orthogonally diagonalizable orthogonal matrices are reflections (including identity as a trivial case).
