Let $A,B$ be two $m\times n$ matrices, and $$A=BU,\ B=AV$$ for some $n\times n$ matrices $U,V$ . Prove that there exists some invertible $n\times n$ matrix $T$ such that $B=AT$.
It really hinders me. I have knew that the columns of $A,B$ are equivalent, however, I could not find the suitable invertible matrix $T$.
As you already observed, the columns of $A$ span the same space as those of $B$; let's denote the dimension of this space by $k$ ($\le m,n$). Since $A$ maps $\mathbb R^n\to\mathbb R^m$, we have that $n=k+\dim N(A)$.
The product $AV$ has rank $k$, so $V$ has columns that span an at least $k$ dimensional space in $\mathbb R^n/N(A)$. Moreover, we are free to add arbitrary vectors from $N(A)$ to any (or all) of the columns of $V$; the product $AV$ will not be affected. Since we start out with at least dimension $k$ modulo $N(A)$ and may introduce $\dim N(A)=n-k$ additional linearly independent vectors from that space, we can get the dimension of the column space up all the way to $k+n-k=n$.