# Multiplication of unitary matrices to make symmetric off-diagonal elements zero

Context

Starting with a unitary matrix $U$ of size $m \times m$, I have read of a way to obtain a diagonal matrix by sequentially multiplying $U$ from the right by unitary matrices $V$ of a certain form. Each of these matrices is the identity matrix, except that the elements $V_{ab}$, $V_{ba}$, $V_{aa}$, $V_{bb}$ are replaced by elements of a $2\times2$ unitary matrix, chosen such that the target element $U^{\mathrm{new}}_{ab}$ of the output matrix $U^{\mathrm{new}}=U^{\mathrm{old}}V$ becomes zero. For the diagonalization, the sequential multiplications are performed to make all off-diagonal elements of each row zero, starting with the last. When a row has been completed in this way, the off-diagonal elements of the corresponding column have automatically become zero as well.

Problem

My question is why the off-diagonal elements of the corresponding column become zero if the off-diagonal elements of a row are made zero in this way?

Example

As a $3\times 3$ example, starting with this unitary matrix $$U=\left( \begin{matrix} 0.5000+0.0000i & 0.5000+0.5000i & -0.5000+0.0000i \\ 0.0000-0.5774i & 0.0000+0.5774i & 0.5774+0.0000i \\ 0.0000+0.6455i & 0.3873+0.1291i & 0.5164+0.3873i \\ \end{matrix}\right)$$

the first steps that lead to a diagonalized last row and column would be achieved with the following matrices $$V1=\left( \begin{matrix} 1.0000+0.0000i & 0.0000+0.0000i & 0.0000+0.0000i \\ 0.0000+0.0000i & -0.8452+0.0000i & 0.5345+0.0000i \\ 0.0000+0.0000i & 0.5071-0.1690i & 0.8018-0.2673i \\ \end{matrix}\right)$$ $$V2=\left( \begin{matrix} -0.7638+0.0000i & 0.0000+0.0000i & 0.6455+0.0000i \\ 0.0000+0.0000i & 1.0000+0.0000i & 0.0000+0.0000i \\ 0.2041+0.6124i & 0.0000+0.0000i & 0.2415+0.7246i \\ \end{matrix}\right).$$

The results of the first two multiplication steps are: $$U\times V1=\left( \begin{matrix} 0.5000+0.0000i & -0.6761-0.3381i & -0.1336+0.4009i \\ 0.0000-0.5774i & 0.2928-0.5855i & 0.4629+0.1543i \\ 0.0000+0.6455i & 0.0000+0.0000i & 0.7246+0.2415i \\ \end{matrix}\right)$$ $$U\times V1\times V2=\left( \begin{matrix} -0.6547+0.0000i & -0.6761-0.3381i & 0.0000+0.0000i \\ 0.0000+0.7559i & 0.2928-0.5855i & 0.0000+0.0000i \\ 0.0000+0.0000i & 0.0000+0.0000i & 0.0000+1.0000i \\ \end{matrix}\right),$$ where as expected the last row and column have zeros as off-diagonal entries.

Known property

Since $U^{\mathrm{old}}$ and $V$ are both unitary, so is their product $U^{\mathrm{new}}$. However, based on that alone I do not see why the effect on the columns happens. If anyone knows an explanation, it would be a great help!

The key is that the matrix $U^{\text{new}}$ is again unitary, as you mention.

Let $A$ be a unitary matrix with $m$th row $(0,\ldots,0,a,0,\ldots,0)$.

Since $A$ is unitary, the product of $A$ with its transpose conjugate is the identity, that is, $A \overline{A^t} = I$. The $m$th diagonal element in $A \overline{A^t}$ is:

$$0 + \ldots + 0 + a \cdot \overline{a} + 0 +\ldots + 0 = a \cdot \overline{a} = |a|^2$$

and this has to be one, so $|a|^2 = 1$.

Say the $m$th column is $(r_1,\ldots,r_k,a,r_{k+2},\ldots,r_n)$. Now $\overline{A^t}A=I$ must hold too and the $m$th diagonal element in $\overline{A^t}A$ is:

$$|r_1|^2+ \ldots +|r_k|^2+|a|^2+|r_{k+2}|^2+\ldots+|r_n|^2$$

and this has to be one. Since $|a|^2=1$, this means

$$|r_1|^2+ \ldots +|r_k|^2+|r_{k+2}|^2+\ldots+|r_n|^2 = 0$$

from where all the $r_i$ are zero, that is, the off-diagonal elements of the $m$th column are all zero.