Show that any $M\in\mathcal{M}_n(\mathbb{C})$ of rank $r$ can be written as $M = A N_r B$

Question

Let $\mathcal{M}_n(\mathbb{C})$ be the set of all $n\times n$ matrices with entries in $\mathbb{C}$ and $\mbox{GL}_n(\mathbb{C})$ denote the set of invertible $n\times n$ matrix in $\mathcal{M}_n(\mathbb{C})$. Show that any $M\in\mathcal{M}_n(\mathbb{C})$ of rank $r$ can be written as $$M=AN_rB,$$ where $A,B\in \mbox{GL}_n(\mathbb{C})$ and

$$N_r = \begin{bmatrix} 0 & I_r \\ 0 & 0 \end{bmatrix}$$

Answer 1

Hint: Can you perform elementary row/column operations on $M$ to transform it into $N_r$? That would mean you can find invertible matrices $P$ and $Q$ such that $PMQ = N_r$

Answer 2

I get that!

Solution: Let $M \in \mathcal{M}_n(\mathbb{C})$ such that $\mbox{rank}(M)=r>0$. Let $R$ be the row-reduced echelon matrix equivalent to $M$. So $R$ could be expressed by $$R=PM,$$ where $P$ is the product of elementary marices used to elementary operation in $M$. Since $R$ is an echelon matrix, we can operate in columns of $R$ until get $N_r$. In this case we say that $N_r$ is "column-equivalent" to $R$ and we have

$$N_r = R Q,$$

where $Q$ is the product of matrices "column-elementary". Via theorem, we have that elementary matrices are invertible, so its products as well, then $P,Q\in GL_n(\mathbb{C})$. Usying the equations above, we obtain

$$N_r=(PM)Q \implies P^{-1}N_r=MQ \implies P^{-1}N_rQ^{-1}=M.$$

Put $A \doteq P^{-1}$ and $B\doteq Q^{-1}$, then $$M=AN_rB,$$ with $A,B\in GL_n(\mathbb{C})$.