So I am trying to solve this practice problem, and so far I managed to get that the kernel of transformation is zero.
Show that $L$($\vec{x}$)=$A\vec{x}$ from $Im(A^{T})$ to $Im(A)$ is an isomorphism.
Let L(x)=0. Then L(x)=$A$A^{T}y$ for some y. We know that Ker(A)=$Im(A^{T})^{\perp}$. So if $A$A^{T}y$=0, $A^{T}y$ is in $Im(A^{T})^{\perp}$, which is possible only if $A^{T}y$=0, since it is both in the $Im(A^{T})^{\perp}$ and $Im(A^{T})$
I don't want use the fact that dim of these two subspaces (Im($A$) and Im($A^{T}$)) are equal, as I am trying to use this problem as another proof of that. So I need to show that Im($L$)=Im($A$) to complete the proof of the isomorphism. What would be a way of showing that everything in Im($A$) gets hit given this transformation (seems intuitive, but can't show it)? Thanks