# Singular value decomposition of 2x2 matrix with unit norm entries

I've got a question and I would appreciate if one could help me to understanding it.

I have a 2x2 complex matrix $F$. The absolute value of the entries of $F$ are equal to one, i.e., $|F(m,n)|=1$. I find the singular value decomposition of this matrix as

$[U S V] = svd(F)$.

I see that the absolute value of the entries of $U$ and $V$ are equal to $0.5$, i.e., $|U(m,n)|^2=0.5$ and $|V(m,n)|^2=0.5$.

Could one explain why the absolute values of $U$ and $V$ are $0.5$?

It is not true in general, with $A=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ then $U (\sqrt{2}I ) (-I)$, with $U=-{1 \over \sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ is a singular value decomposition of $A$ but the entries of $V=-I$ are zero or minus one.