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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 $X_1, Y_1 \in C^{r\times r}$.

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Where does $B$ come in? – Robin Chapman May 29 '10 at 15:41
I guess it is between $Q_2$ and $P_2$. – Wadim Zudilin May 29 '10 at 15:45
^_^, sorry. I modified it. – Sunni May 29 '10 at 15:52
up vote 3 down vote accepted

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.

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I assume that y and u are supposed to be the same variable in your answer. Nice proof! – Peter Shor May 31 '10 at 12:05
Yes, I changed that. Thanks! – Anton Jun 2 '10 at 6:58

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