linear algebra - Writing a matrix as a product of elementary matrices.

So if I have a matrix and I put it into RREF and keep track of the row operations, I can then write it as a product of elementary matrices. An elementary matrix is achieved when you take an identity matrix and perform one row operation on it. So if we have a matrix like $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, one elementary matrix could look like $\begin{bmatrix}1&0\\-1&1\end{bmatrix}$ for the row operation $r_2 - r_1$ or $\begin{bmatrix}1&0\\0&1/2\end{bmatrix}$ for the row operation $\dfrac{r_2}{2}$. So if you put a matrix into reduced row echelon form then the row operations that you did can form a bunch of elementary matrices which you can put together as a product of the original matrix. So if a have a $2\times{2}$ matrix, what is the most elementary matrices that can be used. What would that look like?

Answer

Let's assume that nonzero entries in our matrices are invertible.

If $a \ne 0$, then a $2\times 2$ matrix with $a$ in the upper corner can be written as a product of 4 matrices that are elementary in the sense described:

$$\left( \begin{array}{cc} 1 & 0 \\ \frac{c}{a} & 1 \end{array}\right) \left( \begin{array}{cc} a & 0 \\ 0 & 1 \end{array}\right) \left( \begin{array}{cc} 1 & 0 \\ 0 & d-\frac{bc}{a} \end{array}\right) \left( \begin{array}{cc} 1 & \frac{b}{a} \\ 0 & 1 \end{array}\right) = \left( \begin{array}{cc} a & b \\ c & d \end{array}\right)$$

Notice that when $a=1$, three elementary matrices suffice.

If $a=0$ but $c\ne 0$, then
$$\left( \begin{array}{cc} 1 & \frac{1}{c} \\ 0 & 1\end{array}\right) \left( \begin{array}{cc} 0 & b \\ c & d\end{array}\right)= \left( \begin{array}{cc} 1 & * \\ c & d\end{array}\right)$$
Since $\left( \begin{array}{cc} 1 & * \\ c & d\end{array}\right)$ can be written as a product of 3 elementary matrices, $\left( \begin{array}{cc} 0 & b \\ c & d\end{array}\right)$ can again be written as the product of 4. A similar argument holds when $a=0$ but $b \ne 0$.

I'll leave the case $a=b=c=0$ to the reader.