# Proving existence of matrix N such that M = MNM

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?

• I don't think what you want to show can be true. There are matrices over fields, such as $$\begin{pmatrix}1&0\\0&0\end{pmatrix},$$which are not invertible but are also not the zero map. You're definitely on the right track, though -- I'd try thinking more about using the kernel to construct $N$. Commented Jan 4, 2023 at 5:21
• A matrix-theoretical proof is probably simpler here: let $M=UV^T$ be a rank factorisation, $X$ be a right inverse of $V^T$ and $Y^T$ be a left inverse of $U$. Then $N=XY^T$ will do the job. Commented Jan 4, 2023 at 5:45
• @user1551 can we safely assume that the left and right factorization matrices are invertible? Commented Jan 4, 2023 at 15:28
• Yes, because $U$ has full column rank and $V^T$ has full row rank. Commented Jan 4, 2023 at 18:36

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.