Is the following claim true?


**Claim** Let $A, B\in C^{n\times n}$ with $rank(A)=rank(B)=r$. Then there exist nonsingular matrices $P\_1, P\_2, Q\_1, Q\_2$ such that


$$ Q\_1AP\_1=Q\_2BP\_2=\left(\begin{array}{cc}I\_{r}&0\\\
0&0
\end{array}\right)$$ and
$$Q\_1Q\_2^{-1}=\left(\begin{array}{cc}X\_1&0\\\
X\_2&X\_3
\end{array}\right), \qquad
P\_1^{-1}P\_2=\left(\begin{array}{cc}Y\_1&Y\_2\\\
0&Y\_3
\end{array}\right),$$ where $X\_1, Y\_1 \in C^{r\times r}$.



A question on matrix decomposition.
The Gauss-algorithm tells you that you can find matrices $Q\_i$ and $P\_i$ satisfying the first line. For the second line you ask, how can one modify, say, the $Q\_i$ to make the second line come true? One thing you certainly can do, is changing $Q\_1$ by 
$$\left(\begin{array}{cc} 1&x\\\ 0&1\end{array}\right)Q\_1$$ 
and replace $Q\_2$ by $\left(\begin{array}{cc} 1&-y\\\ 0&1\end{array}\right)Q\_2$. The matrix $A=\left(\begin{matrix} a&b\\\ c&d\end{matrix}\right)=Q\_1Q\_2^{-1}$ then changes to 
$$\left(\begin{array}{cc} \ast&ay+b+x(cy+d)\\\ \ast&\ast\end{array}\right).$$
Now $A$ being invertible, the matrix $(c\ d)$ has full rank. From this it is easy to see that there exists $y$ such that $(cy+d)$ is invertible.
Hence one can find $x$ such that $ay+bx(cy+d)=0$.
The $P$'s are treated similary.


