Unitary diagonalization of a square matrix whose entries are all the same

Question

Let

$$M = \left(\begin{matrix} a & a &... & a\\a & a & ... & a\\\vdots & \vdots & \ddots & \vdots\\a & a & ... &a\end{matrix}\right)$$

For unitary diagonalization: $U^{\dagger}MU =D$. What are $U$ and $D$?

My generalization from examples tell me that

$$D = \left(\begin{matrix} Na & 0 &... & 0\\0 & 0 & ... & 0\\\vdots & \vdots & ... & \vdots\\0 & 0 & ... &0\end{matrix}\right)$$

where $N$ is the dimension of $M$. $Na$ is the only nonzero eigenvalue of $M$! Then for $U$,

$$U = \left(\begin{matrix} \frac{1}{\sqrt{N}} & 0 &... & 0\\\frac{1}{\sqrt{N}} & 0 & ... & 0\\\vdots & \vdots & ... & \vdots\\\frac{1}{\sqrt{N}} & 0 & ... &0\end{matrix}\right)$$

But how to calculate the rest of the columns to make it unitary?

Answer 1

The nicest matrix to use here (in my opinion) is the DFT matrix. Note that it's a unitary matrix whose first column is $\frac 1{\sqrt{N}}(1,\dots,1)$, which is all we really need here.

Answer 2

Couple of things you should ask yourself and probably that will solve the problem for you:

1. Why can this matrix be diagonalized by a unitary matrix?
2. What is the rank of this matrix or how many linearly independent vectors (rows or columns) are there? How is rank related to the number of non-zero eigenvalues for a diagonalizable matrix?
3. Now you have a eigenvalue $Na$ and zero is a multiple eigenvalue which repeats $N-1$ times. Now that you have found eigenvector corresponding to the non-zero eigenvalue, what are remaining eigenvectors corresponding to?