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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?

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  • $\begingroup$ $1$ because $I_{2\times2}$ is invertible. $\endgroup$
    – user398843
    Commented Mar 8, 2019 at 1:38

1 Answer 1

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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.

  1. My answer

My answer is that you can try four elementary matrices and calculate their product to see if all the entries a、b、c、d can be represented properly. This approach is not as beautiful as the one above, but works in the end.

Remark: Do not forget to prove that three elementary matrices are not enough, you may just come up with a counter example or to give a robust explanation or proof.

By intuition, I think we can make an analogy between this and the relationship between number of variables and equations to have a unique solution.

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  • $\begingroup$ What if $a=0$ in answer #1? $\endgroup$
    – amd
    Commented Jul 22, 2016 at 19:45
  • $\begingroup$ You're right, see my edits. $\endgroup$
    – Mufei Li
    Commented Jul 23, 2016 at 3:14

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