Bistochastic matrix is a square matrix made of non-negative reals s.t. the sum of elements of any row or column equal 1.
Unitary matrix is a square matrix with complex entries s.t. $UU^*=E$
It's known that if you take a Unitary matrix and take the squared modulus of every entry then you get a bistochastic one. Is the converse true?
Does every bistochastic matrix $A$ have a unitary matrix $U$ s.t. $a_{ij} = |u_{ij}|^2$ ?
If yes, what is the answer if we restrict the set of matrices to orthogonal real ones?
I tried to write all the equations $U$ should satisfy in two-dimensional case. We have $6$ orthogonality relations and $4$ squared-modulus relations. Note that just two orthogonality relation don't hold automatically since we have squared-modulus relations.
I tried to find such matrix for the matrix $\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} $, it lead me to $\frac{1}{\sqrt{2}}(\cos2\pi n / 4 + i \sin 2 \pi n /4)$. We can take any four complex numbers forming a square. The modified construction works for any matrix $\begin{pmatrix} a & 1-a \\ 1-a & a \end{pmatrix} $
Furthermore, we can rotate all the complex numbers and the relations still hold.
When the matrix is $2\times2$, the answer is clearly yes: $\pmatrix{p&1-p\\ 1-p&p}$ is the entrywise square of the real orthogonal matrix $\pmatrix{\cos t&-\sin t\\ \sin t&\cos t}$, where $\cos t=\sqrt{p}$.
When the doubly stochastic matrix is at least $3\times3$, it is not orthostochastic in general. Consider e.g. $$ A=\pmatrix{ \frac12&\frac12&0\\ \frac12&\frac13&\frac16\\ 0&\frac16&\frac56}. $$ If $a_{ij}=|u_{ij}|^2$ for some unitary matrix $U$, then $u_{ij}=\sqrt{a_{ij}}\,\omega_{ij}$ for some $|\omega_{ij}|=1$. Since the first two columns of $U$ are mutually orthogonal, their inner product must be zero. It follows that $$ \frac12\omega_{11}\overline{\omega}_{12}+\frac{1}{\sqrt{6}}\omega_{21}\overline{\omega}_{22}=0 \ \Rightarrow\ \left|\frac{\omega_{21}\overline{\omega}_{22}}{\omega_{11}\overline{\omega}_{12}}\right|=\sqrt{\frac32}, $$ which is a contradiction because all $\omega_{ij}$s have unit moduli.