3
$\begingroup$

We know that every matrix in $\mathrm{GL}(2,\mathbb C)$ is a scalar multiple because of the isomorphism between $\mathrm{GL}(2,\mathbb C)/(\mathbb{C}^*)I_2$.

There are four kinds of products of matrices:

$$\begin{pmatrix} k & 0 \\ 0 & 1 \end{pmatrix},$$ where $k>0$;

$$\begin{pmatrix} \lambda & 0 \\ 0 & 1 \end{pmatrix},$$ where $\lambda = 1$;

$$\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix},$$ where $b$ is inclusive of all complex numbers;

$$\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$

I am a little confused how to utilize these products to prove that every matrix in $\mathrm{GL}(2,\mathbb C)$ is a product of matrices of these four kinds

$\endgroup$
  • $\begingroup$ Elementary row operations? $\endgroup$ – Lord Shark the Unknown Feb 27 '18 at 2:55
  • 1
    $\begingroup$ I added the "matrices" and "linear-algebra" tags to your post. Cheers! $\endgroup$ – Robert Lewis Feb 27 '18 at 3:13
  • $\begingroup$ Should that be $\lambda = -1$? Otherwise the second type is redundant and in fact these 4 types will not be sufficient. $\endgroup$ – Christian Sykes Feb 27 '18 at 4:52
  • $\begingroup$ Also, "scalar multiple" of what? The isomorphism is between what two groups? $\endgroup$ – Christian Sykes Feb 27 '18 at 4:56
  • $\begingroup$ It's true $\mathrm{GL}(2,\mathbb C)/\mathrm(SL)(2,\mathbb C) \cong \mathbb{C}^*$. Do you mean that every matrix in $\mathrm{GL}(2,\mathbb C)$ is a scalar multiple of a matrix in $\mathrm{SL}(2,\mathbb C)$? ($\mathrm{SL}(2,\mathbb C)$ denotes the group of $2\times 2$ complex matrices of determinant 1.) $\endgroup$ – Christian Sykes Feb 27 '18 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy