Matrix right multiplication orbit invariant Let M and N be two square matrices such that they have an equal image.
Show that there exists an invertible matrix P, such that N = MP 
I have managed to prove the converse...but this is tougher because I'm guessing we need to build a linear map which then identifies with a matrix P. 
 A: One construction of such a $P$ is as follows: 
Note that $N^T$ and $M^T$ (where $T$ denotes the transpose) have the same row-space.  Thus, $N^T$ and $M^T$ can be row-reduced to the same form.  In other words, there exist invertible matrices $R_1,R_2$ such that 
$$
R_1N^T = R_2M^T \implies\\
(R_1N^T)^T = (R_2M^T)^T \implies\\
NR_1^T = MR_2^T \implies\\
N = MR_2^T R_1^{-T}.
$$
That is, it suffices to take $P = R_2^T R_1^{-T} = (R_1^{-1}R_2)^T$.

A more direct construction: let $v_1,\dots,v_k$ be a basis for the kernel of $N$, and extend this basis to $v_1,\dots,v_n$, a basis of $\Bbb R^n$. The vectors $Nv_{k+1},\dots,Nv_{n}$ form a basis for the image of $N$. 
Let $w_1,\dots,w_k$ be a basis for the kernel of $M$. There exist vectors $w_{k+1},\dots,w_{n}$ such that $Mw_j = Nv_j$ for $j = k+1,\dots,n$, and the resulting set $w_1,\dots,w_n$ is a basis.
Select a matrix $P$ that satisfies $Pv_j = w_j$ for all $j$.  Note that $P$ is invertible, and $MP v_j = Nv_j$ for all $j$, which means that $N = MP$.
A: $$N=NI=\pmatrix{Ne_1&Ne_2&\cdots&Ne_r\cr}=\pmatrix{Mv_1&Mv_2&\cdots&Mv_r\cr}=MP$$ where $I$ is the identity matrix, $e_1,\dots,e_r$ are the standard basis, $v_1,\dots,v_r$ exist because $M$ and $N$ have the same image, and $P$ is the matrix whose columns are $v_1,\dots,v_r$. 
This is only half an answer, as it doesn't guarantee $P$ is invertible. 
