Proving existence of matrix N such that M = MNM

Question

Doing practice problems for my qualifying exam and am a bit stumped by the following:

Let $n \ge 1$, $F$ be a field, and let $M$ be a matrix in $M_n(F)$. Show that there exists an $N \in M_n(F)$ such that $M = MNM$.

Here is my attempt at a solution:

We first define the vector space $V = F^n$ on which the matrices in $M_n(F)$ act. Now take $M \in M_n(F)$ and by the rank-nullity theorem we have that $$ker(M) \oplus Im(M) = V$$ Now, first note that if $M$ is invertible, then there exists $N = M^{-1}$ such that $NM = I$ and moreover $$M = MNM$$ giving us the result. We now assume that $M$ is not invertible (ker$(M) \neq 0$). We have by the First isomorphism theorem that $$V / ker(M) \cong Im(M)$$ I want to use the above fact with the property that ideals of fields are trivial to give us that $Im(M) = 0$ and therefore $M$ is just the zero map or something to this effect?

Answer 1

Many times, if you need to define a map, you can think about what it does on a basis. We also know that if $M$ were invertible then we could use $M^{-1}$ so maybe we need to get some sort of inverse $M' : \operatorname{im}(M) \to F^n$.

In terms of a basis, I will let $e_1, \dots, e_n$ be a basis for $F^n$ such that $e_{m + 1}, \dots, e_n$ is a basis for $\ker M$ and $f_i = Me_i$ for $i = 1, \dots, m$ is a basis for $\operatorname{im} M$.

The $M'$ I was talking about above is given by $M'f_i = e_i$. If you play around with this a bit, you should be able to construct $N$ (complete $f_1, \dots, f_m$ to a basis for $F^n$ and define $Nf_i = e_i$).

In terms of matrices, factoring $M$ through the projection $P : F^n \to \operatorname{span}\{e_1,\dots,e_m\}$ gives a rank factorization $M = QP$ where $Qe_i = Me_i$ and $P$ is $m \times n$ and $Q$ is $n \times m$. This was talked about in the comments.