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Let \begin{equation} A = \begin{pmatrix} \lambda & 0\\ 0 & \frac{1}{\lambda}\\ \end{pmatrix} \end{equation} be a matrix in $SL_2(\mathbb{C}$). I am trying to prove that $SL_2(\mathbb{C})$ is generated by the matrices \begin{equation} \begin{pmatrix} 1 & z\\ 0 & 1\\ \end{pmatrix}, \begin{pmatrix} 1 & 0\\ z & 1\\ \end{pmatrix} \end{equation} for $z\in\mathbb{C}$. I have already handled all the cases expect for matrices like $A$. How can I write it as a product of matrices of this type? Any hints are appreciated.

physicist23
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  • Multiplying by those two matrices on the left or right correspond to row or column operations where you add $z$ times a row or column to the remaining row or column. So the question becomes whether you can turn $A$ into the identity matrix using row and column operations (without allowing scalar multiplication.) – Cheerful Parsnip Sep 07 '20 at 00:11
  • So your question is easier and the answer is a special case of the answer given there. – markvs Sep 07 '20 at 00:50
  • @JCAA: I see, I will try if I can apply it to this case – physicist23 Sep 07 '20 at 00:51
  • @JCAA: It worked. – physicist23 Sep 07 '20 at 01:40

2 Answers2

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How about this:

$A = \begin{pmatrix} 1 & -\lambda \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ \frac{1}{\lambda} & 1\\ \end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ -1 & 1\\ \end{pmatrix} \cdot \begin{pmatrix} 1 & 1\\ 0 & 1\\ \end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ \lambda - 1 & 1\\ \end{pmatrix}$

Coriolanus
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Based on the answer in this question provided by JCAA, we can also come up with a way of generating $A$ in terms of four matrices. Following the procedure there yields \begin{equation} A = \begin{pmatrix} 1 & 0\\ \frac{1-\lambda}{\lambda^2} & 1\\ \end{pmatrix} \begin{pmatrix} 1 & \lambda\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ \frac{\lambda-1}{\lambda} & 1\\ \end{pmatrix} \begin{pmatrix} 1 & -1\\ 0 & 1\\ \end{pmatrix}. \end{equation}

physicist23
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