$\newcommand{\rank}{\operatorname{rank}}$We know that $\rank(PA)=\rank(AQ)=\rank(PAQ)=\rank(A)$ where $A\in M_{m\times n}(\mathbb F), P, Q$ are $m\times m, n\times n$ invertible matrices.

mean to say , from the matrix product we can remove the non-singular matrices; rank will not be effected.

After studying this, it came to my mind, then what will happen in the case of $\rank(PAQB)$ where $B\in M_{n\times m}(\mathbb F)$ ?

No idea. The only thing I got is $\rank(PAQB)=\rank(AQB)$. Can we remove $Q$ as well and write $\rank(AQB)=\rank(AB)$?

Please help. In case it has been solved earlier, provide me the link.


No. Worse, the much weaker property of the product being zero is not preserved by insertion of a nonsingular matrix: If we take $$A = B = \pmatrix{0&1\\0&0}, \quad Q = \pmatrix{0&1\\1&0},$$ then $$AQB = A \neq 0$$ but $$AB = 0.$$

  • $\begingroup$ thank you. that cleared my doubt. $\endgroup$ – Anjan3 Jul 23 '15 at 5:20
  • $\begingroup$ You're welcome, I'm glad you found it useful. $\endgroup$ – Travis Jul 23 '15 at 6:04

To give an example in the opposite direction of Travis's answer, consider $$A = B = \left[\begin{array}{cc} 1 & 0\\0 & 0\end{array}\right],\quad Q = \left[\begin{array}{cc} 0 & 1\\1 & 0\end{array}\right]$$

Then, $AB = A$ has rank $1$, but $AQB = 0$ has rank $0$. Geometrically, $A$ fixes $e_1$, and maps $e_2$ to $0$, and $Q$ interchanged $e_1$ and $e_2$, where $e_1 = \left[\begin{array}{c} 1\\0\end{array}\right], e_2 = \left[\begin{array}{c} 0\\1\end{array}\right]$ are the standard basis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.