Consider a complex unitary matrix $U \in \mathbb{C}^{N\times N}$ and pick two diagonal and two off-diagonal elements from its $m$th and $n$th rows to construct a $2 \times 2$ submatrix: \begin{equation} M= \begin{bmatrix} U_{mm} & U_{mn}\\ U_{nm} & U_{nn} \end{bmatrix} = \begin{bmatrix} U_{mm} & U_{mn}\\ (U_{mn})^* & U_{nn} \end{bmatrix} \end{equation}
(Here I consider the Hermitian case $U_{nm} = (U_{mn})^*$.) Is $M$ always diagonalizable?
Update:
Conjecture: $M$ should be diagonalizable. Because $U$ is unitary, the eigenvectors of $U$ should span a linear space of dimension $N$. If there exists any $M$ not diagonalizable, the eigenvectors of that $M$ does not span a dimension $2$ space.
Brute force proof: find the eigenvectors of this sample $2\times 2$ submatrix. Its eigenvectors are also shown in the figure. ($a_{1,2}, b_{1,2} \in \mathbb{R}$, and $c \in \mathbb{C}$)
For the two eigenvectors to be identical, the following two equations need to be satisfied: \begin{equation} (a_1-b_1)^2 - (a_2-b_2)^2 + 4|c|^2 = 0 \text{ and } (a_1-b_1)(a_2-b_2) = 0. \end{equation} Clearly, both equations can be simultaneously satisfied. When they are both satisfied, $M$ does not have two independent eigenvectors and therefore is not diagonalizable.
This result looks counter-intuitive.
Moreover, I started to think about this question when studying this historic paper. This paper is about a recursive way of building any $N\times N$ unitary matrix with optical experiments. Its equation 2 shows that by applying in total $N-1$ different $2\times 2$ rotational matrices to a $(N-1)\times (N-1)$ unitary matrix, one can construct a $N\times N$ unitary matrix. Performing this recursively, one can use $N(N-1)/2$ rotational matrices to construct a $N\times N$ unitary matrix starting from an identity matrix.
However, it seems that we just showed that not all unitary matrices $U$ can be constructed in this way...
The answer is no. Let $$\begin{pmatrix}{1\over 2} &{1\over \sqrt{2}} & {1\over 2}\\ -{1\over 2} &{1\over \sqrt{2}}& -{1\over 2}\\ -{1\over \sqrt{2}} & 0 & {1\over \sqrt{2}} \end{pmatrix} $$ The submatrix $$\begin{pmatrix} {1\over \sqrt{2}}& -{1\over 2}\\ 0 & {1\over \sqrt{2}} \end{pmatrix} $$ is not diagonalizable.