product of singular matrix and a full column rank matrix

In $\mathbb{R}^4$, I am given $V,W$ are $4\times 2$ matrices with full column rank i.e $rank(V)=rank(W)=2.$

How can I prove that $EV,AW$ will be also full column rank matrices where $E,A$ are fixed $4\times 4$ matrices with no information about non-singularity? I think the statement is false as I check if $A=E=0_{4\times 4}$ and $V=[e_1,e_2],W=[e_3,e_4]$ then $EV=0_{4\times 2}=AW$

But, I have also an example where $A\ne E\in M_4(\mathbb{R})$ both having rank$=3$, but $(EV)_{4\times 2},(AW)_{4\times 2}$ both having full column rank i.e $2$

Thanks for any suggestion.

• If $E$ and $A$ are at least rank 2, $EV$ and $AW$ will be full column rank. – karakusc Nov 8 '16 at 22:36
• I just realize $rank(AB)\le \min\{rank(A),rank(B)\}$. Thanks – Marso Nov 8 '16 at 22:50
• Yes, in fact I'm not sure about my original statement, but its converse is true by the inequality you have, i.e., if $EV$ and $AW$ are full rank, then $E$ and $A$ are at least rank 2. – karakusc Nov 8 '16 at 22:57
• No, No, I am not given rank of $EV$ and $AW$, I am only given rank of $W,V$, they have full column rank – Marso Nov 8 '16 at 23:04
• Here $V,W:\mathbb{R}^2\to\mathbb{R}^4$ and $A,E:\mathbb{R}^4\to\mathbb{R}^4$ and $EV,AW:\mathbb{R}^2\to\mathbb{R}^4$ – Marso Nov 8 '16 at 23:06