Multiplication of ACB where A and B are invertible produces a matrix of zeros with an Identity block

Consider m*n matrix C and invertible square matrices A and B with m and n dimentions.

If all three matrices belong to a Field F(all elements of each matrix belongs to F), we can show that there is A and B that:

$ACB = \begin{bmatrix}I_{r} & 0\\0 & 0\end{bmatrix}$

Where $I_{r}$ is and identity block and zeroes represent all the other elements of the result matrix which are zero.

How should I approach proving this expression ? I thought about transforming C to it's row reduced echelon form, but I cannot make much progress from there.

I believe you are referring to the following theorem: if $C$ is a $m\times n$ matrix you can find two matrices A ($m\times m$) and B ($n\times n$) such that

$$ACB=\begin{pmatrix} I_r & 0 \\ 0 &0 \end{pmatrix}.$$

This theorem is true and in fact $r$ is the rank of the matrix C. The proof I've seen of this theorem (existence of the invertible matrices $A$ and $B$) uses induction over $m+n$ (sum of columns and rows of $C$). If $m+n=2$ then $C=(a)$ for some $a\in F$ (and then you can easily find $A,B$).

If $m+n>2$, there are two options

• $a_{ij}=0$ for all $i$ and $j$. Then we can take $A=I_m$ and $B=I_n$ (and r=0).
• $a_{i_0 \ j_0}\neq 0$ for some $i_0,j_0$. Remember that doing elementary changes on our matrix $C$ is the same as multiplying $C$ by an elementary matrix (if you are not familiar with this you can check this article on wikipedia). For exemple, if $C'$ is the $3\times 3$ matrix you obtain after switching rows 1 and 2 of the matrix $C$ then $$C'= \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 &0\\ 0&0&1 \end{pmatrix} C$$ Using this property we can move $a_{i_0 \ j_0}$ to $a_{11}$ by multiplying $C$ with invertible matrices, let's call them $A_0$ and $B_0$. You can also obtain a matrix of the form (multiplying $C$ by invertible matrices $A_1$ and $B_1$):

$$\begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & & \\ \vdots & & C^* \\ 0 & \\ \end{pmatrix}$$ And we have $$\begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & & \\ \vdots & & C^* \\ 0 & \\ \end{pmatrix}=A_1\cdot A_0 \cdot C\cdot B_0\cdot B_1.$$ $A_1\cdot A_0$ is invertible (because both $A_1$ and $A_0$ are invertible) and $B_0\cdot B_1$ is also invertible. $C^*$ is a $(m-1)\times(n-1)$ matrix and due to our induction hypothesis there are invertibles matrices $A^*$ and $B^*$ such that $$A^* C^* B^*=\begin{pmatrix} I_s & 0 \\ 0 &0 \end{pmatrix}.$$

• Thank you. You are right I mistakenly thought I have to prove it for every A and B not finding one instance of it. This is the question. – Vicarious Feb 22 '16 at 20:44

Don't try too hard to prove this theorem because it is incorrect. As a counterexample, take $A$ and $B$ to be identity matrices of $m$M and $n$ dimensions; then $ACB = C$ which is not of the given form.

• I checked the question and updated mine, I have to show that we can find an A and B in field F that works in the expression not that every A and B is correct in the expression. – Vicarious Feb 22 '16 at 20:41