Given two $m \times n$ Matrices $A, B$ of the same rank . Prove that there are 2 invertible matrices such that $A=VBU$ Given two $m \times n$ Matrices $A, B$ of the same rank. 
Prove that there are two invertible matrices, $V_{m \times m}$ and $U_{n \times n}$ such that  $A=VBU$
I guess it's true because multiplication by invertible matrices is equivalent to elementary operations on matrices, but is there a formal way to prove that?
 A: One way (nonunique)
Do an SVD of both
$A = U_A \Sigma_A V_A^T$, $B = U_B \Sigma_B V_B^T$.
This way $U_A$ and $U_B$ are $m\times m$, $V_A$ and $V_B$ are $n\times n$, and $\Sigma_A$ and $\Sigma_B$ are $m\times n$. 
If they are both rank $r$, then $r \leq \min\{m,n\}$. Then $\Sigma_A$ and $\Sigma_B$ are tall matrices with a diagonal on the upper $n\times n$ block with the first $r$ entries strictly positive.
Define $S_A^m$ and $S_A^n$ as a diagonal $m\times m$  and $n\times n$ matrix (respectively) where the first $r$ entries are $\sqrt{d_i}$ where $d_i$ are the diagonal entries of $\Sigma_A$. Define the remaining diagonal entries as 1. Do almost the same thing with $\Sigma_B$ to define $S_B^m$ and $S_B^n$, but where the first $r$ entries are $1/\sqrt{d_i}$ and $d_i$ are the diagonal nonzero entries of $\Sigma_B$, and the remaining as 1. Then $S_B^m \Sigma_B S_B^n =  \left[\begin{matrix} I & 0 \\ 0 & 0\end{matrix}\right]$ and $S_A^mS_B^m\Sigma_B S_B^nS_A^n = \Sigma_A$.
Then $V =  U_A S_A^m S_B^m U_B^T$, $U = V_B S_B^n S_A^n V_A^T$, and you can show it all works out using the orthogonal properties of the left and right singular vector matrices.
A: Let $rank(A)=rank(B)\triangleq r$ 
$dim(ImA)=r$
d$im(kerA)=n-r $  - according to dimension theorem.
There exist $v_{1},v_{2},\ldots,v_{n-r}\in\mathbb{R}^{n}$  such that: $Av_{i}=o$ for every $i,\ 1\leq i\leq n-r$.
I'll define a $n\times n$ opretor: $V:\mathbb{R}^{n}\longrightarrow\mathbb{R}^{n}$ such that $kerV=sp\{v_{1},\ldots,v_{n-r}\}$.
we'll extend the base of $kerV$
to the base of $\mathbb{R}^{n}=ImV$, thus: $\{w_{1},\ldots\ldots,w_{r}\}$ such that $\{Aw_{i}\}_{i=1}^{r}\neq0$ is spanning $ImA$. 
$r(B)=r \Longrightarrow\ \exists k_{1},\ldots,k_{r}$  such that: 
$u_{1},\ldots,u_{r}$ is a linear indipendent and $\{B(u_{1}),\ldots,B(u_{r})\}$ is spanning $ImB$.
defining $V$: suppose $V(w_{i})=u_{i}$ ,therefore $\{B(u_{1}),\ldots,B(u_{r})\}$ is a linear independent set and $dim(ImB)=r$.
defining $U$: $U(BV(w_{i}))=A(w_{i})$.
I'll extend the base $\{B(u_{1}),\ldots,B(u_{r})\}$ for a base to $\mathbb{R}^{m}$, such that:
$\mathbb{R}^{m}=\{B(u_{1}),\ldots,B(u_{r}),\ z_{r+1},\ldots z_{m}\}$, let $U(z_{i})=0$, for every $i$. 
