Let $T:U\rightarrow V$ be a linear map and suppose that $rank(T)=dim(U)=dim(V)=n$. Show that the are bases where the matrix is $I_n$ I found this problem that I cannot solve, but I believe is quite interesting. We have to state whether the statement is true or false.
Let $T:U\rightarrow V$ be a linear map and suppose that $rank(T)=dim(U)=dim(V)=n$. Then there are bases: $E$ of $U$ and $F$ of $V$, such that the corresponding matrix $[F,T,E]$ is the indentity matrix $I_n$.
I believe that the claim is false. I was thinking whether to use similar matrices to solve it but I don't really know how to start. Has anyone have any suggestions how to tackle it?

Let $\langle e_1,\cdots,e_n\rangle$ be a basis for $U$. Show that $\langle T(e_1),\cdots,T(e_n)\rangle$ is a basis for $V$. The matrix representing $T$ with respect to these bases is the identity matrix.
