Using Schur-decompostition theorem to prove:
For $A$ $\in$ $\mathbb F^{n \times m}$, there exist unitary matrices $U$, $V$ and a positive diagonal matrix $S$ such that $$A=USV^*$$ This is my mind flow, first I find these equations: $$AA^*=UT_1U^*$$ $$A^*A=VT_2V^*$$ and I could extend $T_1^{n \times n}$ to $S^{n \times m}=[|T_1|^{\frac{1}{2}}\ 0]$, supposing n $\le$ m. Then I could construct $B^{n \times m}$, where $$B=USV^*$$ then I prove $(B-A)$ is both positive semi-definite and negative semi-definite, so they must be equal and $$A=B=USV^*$$
But I have troubles to prove the $(B-A)$ part. Am I on the right way or can you give some hints? Thanks!
So I have done some searches on Wikipedia and find a hint which leads to the answer. Here I just write down the proof of $A^{n \times m}$ when n $\le$ m, the opposite is similar. We have these equalities: $$AA^*=UTU^*$$ $$T=U^*AA^*U$$ So we know $T$ is a positive semi-definite Hermitian matrix and $T$ is also upper triangular according to Schur-decomposition, then $T$ must be positive diagonal. We create: $${V_1}^{m \times n}=A^*UT^{-\frac{1}{2}}$$ $$S^{n \times m}=[T^{\frac{1}{2}} \ \mathbf 0]$$ and we can easily find: $${V_1}^*V_1=I$$ which means the columns of $V_1$ are orthogonal to each others, and it is not hard to find $V_2$ that makes $$V^{m \times m}=[V_1 \ V_2]$$ a unitary matrix. Then we can see: $$USV^*=U[T^{\frac{1}{2}} \ \mathbf 0][V_1 \ V_2]^*=U[T^{\frac{1}{2}}{V_1}^* \ \mathbf 0]$$ $$=UT^{\frac{1}{2}}{V_1}^*=UT^{\frac{1}{2}}T^{-\frac{1}{2}}U^*A=A$$ Here we have shown that $U$, $V$ are both unitary and $S$ is diagonal.