# Is there a 4x4 unitary matrix with 0 diagonal and non-zero off-diagonal?

This question is homework from Scott Aaronson's 2017 Quantum Information course.

a) Give an example of a 2x2 unitary matrix where the diagonal entries are 0 but the off-diagonal entries are nonzero.

b) Give an example for a 4x4 unitary matrix.

c) Is it possible to have a 3x3 unitary matrix with this condition? If no, prove it!

(a) is easy.

$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$

(c) is not possible. Since for

$$\begin{pmatrix} 0 & a & b \\ c & 0 & d \\ e & f & 0 \end{pmatrix}$$

to be unitary the inner product $0\cdot a + c \cdot 0 + e \cdot f$ needs to be zero, but then at least one of $e$ or $f$ needs to be 0.

For (b) I assume he means a 4x4 matrix where the diagonal elements are 0 but all other elements are non-zero (not just any unitary matrix).

I tried without luck:

• Building the 4x4 unitary matrix from 2x2 unitary block matrices. The closest I got so far is

$$U = \begin{pmatrix} X & H \\ H & -X \end{pmatrix}$$

with

$$X = \begin{pmatrix} 0 & 1 \\ 1 & -0 \end{pmatrix}$$

and

$$H = \frac{1}{\sqrt 2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$$

Unfortunately $U^\dagger U$ leaves some non-diagonal elements.

$$U^\dagger U = \begin{pmatrix} 2 & 0 & \sqrt 2 & 0 \\ 0 & 2 & 0 & \sqrt 2 \\ \sqrt 2 & 0 & 2 & 0 \\ 0 & \sqrt 2 & 0 & 2 \\ \end{pmatrix}$$

• Starting with an arbitary first column and finding orthogonal column vectors with the desired 0 elements, but this seems tricky.

Any hints?

Example: $$\frac{1}{\sqrt{3}} \begin{bmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & -1 \\ 1 & -1 & 0 & 1 \\ 1 & 1 & -1 & 0 \end{bmatrix}.$$