# Representation of an invertible square matrix as a product of elementary matrices.

Suppose $A$ is an invertible $2 \times 2$ matrix. What is the smallest integer $n$ such that $A$ is a product of $n$ elementary matrices?

My guess is that at most 4 elementary matrices are sufficient assuming the four entries in $A$ are linearly independent of each other. I'm stuck then. Could you please give me some tips on whether my guess is correct and how to prove it?

• $1$ because $I_{2\times2}$ is invertible. – user398843 Mar 8 '19 at 1:38

After a discussion with a friend of mine, we both figure out the answer in two different ways.

1. Answer the question in the forward logic, credit to Yiwei Shi.

For any 2-by-2 invertible matrix in the following form, a、b、c、d $\in$ R, ad - bc $\neq 0$, a $\neq 0$. $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 \\ \frac{c}{a} & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & d- \frac{bc}{a} \\ \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & \frac{b}{a} \\ 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$ Note that ad - bc $\neq 0$ is required to be satisfied in the second matrix, that's why we want to ensure the invertibility of the matrix.

If a = 0, then in a slightly different way:

$$\begin{bmatrix} 1 & 0 \\ 0 & c \\ \end{bmatrix} \begin{bmatrix} b & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ \frac{d}{c} & 1 \\ \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 0 & b \\ c & d \\ \end{bmatrix}$$

Thanks to @amd, I missed the second case at first.

• What if $a=0$ in answer #1? – amd Jul 22 '16 at 19:45