Let $B,C \in \mathcal{M}_{m,n}(\mathbb{C})$. If $B^* B=C^*C$, then there exists a unitary matrix $U \in \mathcal{M}_m(\mathbb{F})$ such that $C=UB$.

Question

I'm trying to prove the SVD theorem and a result like the one mentioned above came up. If I can prove it then I will be done with the SVD theorem proof. Please help.

Answer 1

Since $C^* C=B^* B$, for all vectors $x,y$ we have $⟨ x,B^*B y⟩=⟨ x,C^*C y⟩$, so $$⟨B x,B y⟩=⟨C x,C y⟩.$$ Taking $y=x$, we see that $B x=0\implies C x=0$. So $U:\operatorname{ran} B→\operatorname{ran} C$, $U(B x):=C x$ is well defined, clearly surjective, and the equation above shows that $U$ preserves inner products, so it is unitary. In particular, the ranges of $B$ and $C$ have the equal dimensions, which implies that their orthogonal complements also have equal dimensions, hence we can extend $U$ (in a somewhat arbitrary way) to a unitary $\mathbb{C}^n\to\mathbb{C}^n$.

Answer 2

By polar decomposition there exist partial isometries $U_B$ and $U_C$ in $\mathcal{M}_{m, n}$ such that $$B=U_B|B|,\quad C=U_C|C|, \quad |B|=U_B^*B$$ where $|A|=(A^*A)^{1/2}.$ Since $|B|=|C|$ we get $$C=U_C|C|=U_C|B|=U_CU_B^*B$$ The matrix $U_B^*$ is an isometry from ${\rm Im}B$ onto ${\rm Im}|B|,$ while $U_C$ is an isometry from ${\rm Im}|C|={\rm Im}|B|$ onto ${\rm Im}C.$ Therefore the product $U_CU_B^*$ is an isometry from ${\rm Im}B$ onto ${\rm Im}C.$ Hence it can be extended to a unitary matrix $U$ such that $C=UB.$