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Let $l \leq m$ are positive integers and $A$ and $B$ are two $l \times m$ matrices over a field $\mathbb{F}.$ If $\text{rank}(A)=\text{rank}(B)=l$ then I have to show that there is an $m \times m$ invertible matrix $P$ over $\mathbb{F}$ such that $AP = B.$

I have no idea how to start it. I need some help. Thanks.

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  • $\begingroup$ Are the matrices having same rank congruent ?? If yes, then the problem can be solved easily. $\endgroup$ – Anik Bhowmick Oct 8 '18 at 3:53
  • $\begingroup$ yes, rank same will imply congruent, then how this can be solved easily ? $\endgroup$ – user371231 Oct 8 '18 at 3:57
  • $\begingroup$ Then you can transform $A$ into $B$ by multiplying elementary matrices, and elementary matrices are invertible !! :D $\endgroup$ – Anik Bhowmick Oct 8 '18 at 3:59
  • $\begingroup$ how will I get that form ? remember row operations corresponds to premultiply ! $\endgroup$ – user371231 Oct 8 '18 at 4:01
  • $\begingroup$ And column operations correspond to post-multiplication !! $\endgroup$ – Anik Bhowmick Oct 8 '18 at 4:02
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The reduced row echelon form of $A^T$ must be $\displaystyle R = \begin{pmatrix}I_{l\ \times\ l} \\ 0_{m - l\ \times\ l}\end{pmatrix}$, because the columns of $A^T$ are linearly independent. The same is true of $B^T$. Thus there are invertible matrices $P_A$ and $P_B$ such that $$P_A A^T = R \text{ and }P_B B^T = R,$$ so $$AP_A^T = BP_B^T,$$ whence $$A(P_A^T)(P_B^T)^{-1} = B$$ so you can take $P = (P_A^T)(P_B^T)^{-1}$.

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