A simple question in matrix theory, but I was puzzled:

Question

If matrices $AA^\dagger=BB^\dagger$, then there exists some unitary matrix V, s.t. A=BV. Is it true? If it is, how to prove it?(Here, $\dagger$ denotes transpose conjugate)

Answer 1

let $A^*$ denote the transpose conjugate of $A$.

We assume that $A,B$ are complex $n\times n$ matrices. We use the SVD decomposition, cf.

http://en.wikipedia.org/wiki/Singular_value_decomposition

Thus $A=U\Sigma V^*$ where the columns of $U$ are eigenvectors of $AA^*$. Since $AA^*$=$BB^*$, these matrices have same eigenvectors and $A,B$ have same singular values. Therefore, we deduce that one of the SVD of $B$ is in the form $B=U\Sigma {V_1}^*$. Finally $A=BW$ where $W=V_1V^*$ is unitary.

EDIT: @Frank, we have the same result if $A,B$ are both $n\times m$ rectangular matrices. By the same reasoning, from a SVD of $A$, we construct a SVD of $B$. In particular, beware, your equality $U_1\Sigma \Sigma^*{U_1}^*=U_2\Sigma \Sigma^*{U_2}^*$ does not imply that $U_1=U_2$ ; indeed, if we have multiple singular values, then there exists an infinity of choices for the eigenvectors of $AA^*$. Moreover, if we choose $U_2=U_1$, then possibly, there exists an infinity of choices for $V_2$. It is equivalent to say that there exists an infinity of choices for $W$ in $A=BW$ ; for example, this happens if $\dim(\ker(B))\geq 2$.