$A,B$ are $l \times m$ matrices of rank $l.$ Then there is an invertible matrix $P$ such that $AP=B,$

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.

• Are the matrices having same rank congruent ?? If yes, then the problem can be solved easily. – Anik Bhowmick Oct 8 '18 at 3:53
• yes, rank same will imply congruent, then how this can be solved easily ? – user371231 Oct 8 '18 at 3:57
• Then you can transform $A$ into $B$ by multiplying elementary matrices, and elementary matrices are invertible !! :D – Anik Bhowmick Oct 8 '18 at 3:59
• how will I get that form ? remember row operations corresponds to premultiply ! – user371231 Oct 8 '18 at 4:01
• And column operations correspond to post-multiplication !! – Anik Bhowmick Oct 8 '18 at 4:02

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 $$P = (P_A^T)(P_B^T)^{-1}$$.