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The Iwasawa decomposition of $\text{SL}(2,\mathbf R)$ is

$$ \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} r & 0 \\ 0 & 1/r \end{bmatrix} \begin{bmatrix}1&x\\ 0 & 1 \end{bmatrix} $$

where $\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \in SO(2)$, where $\begin{bmatrix} r & 0 \\ 0 & 1/r \end{bmatrix}$ is a squeeze matrix and where $\begin{bmatrix}1&x\\ 0 & 1 \end{bmatrix}$ is a shear matrix.

I wish to extend this to $\text{GL}(2,\mathbf R)$. Can I multiply with

$$ \begin{bmatrix} \sqrt{a} & 0 \\ 0 & \sqrt{a} \end{bmatrix} $$

Perhaps this gives $\text{GL}^+(2,\mathbf R)$? If so, I am also okay with that.

Finally, I was wondering what the lie algebra of $\begin{bmatrix} r & 0 \\ 0 & 1/r \end{bmatrix}$ is. And also for $\begin{bmatrix}1&x\\ 0 & 1 \end{bmatrix}$.

Anon21
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1 Answers1

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The extension you are trying for is called $QR$-decomposition: For a matrix $A$, QR-decomposition means, $A=QR$ where $Q$ is a $2 \times 2$ orthogonal matrix (rotation matrix) and $R$ is a $2 \times 2 $ upper triangular matrix with possibly different non-zero elements along the diagonal. This $QR$ decomposition is obtained by Gram-Schmidt orthogonalization. So the matrix $[r,0;0,1/r]$ must be replaced by $[r_1,0;0,r_2]$ (arbitrary diagonal matrix). The last matrix $[1,x;0,1]$ remains the same.

Good luck !

Balaji sb
  • 4,357
  • Would you know the lie algebra of the last matrix [1,x;0,1]? – Anon21 Mar 22 '23 at 13:39
  • Dang. I randomly guessed the matrix of the lie algebra to be x[0,1;0,0] on the first try: $\exp x\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & x \ 0 & 1 \end{bmatrix}$ – Anon21 Mar 22 '23 at 14:08
  • The lie algebra for the other matrix is simply $\exp \left( r_1 \begin{bmatrix} 1 & 0 \ 0 &0 \ \end{bmatrix} + r_2 \begin{bmatrix} 0 & 0\ 0& 1 \end{bmatrix} \right)$ – Anon21 Mar 22 '23 at 14:25
  • However, I cannot figure out the lie algebra of this matrix $[r,0;0,1/r]$ – Anon21 Mar 22 '23 at 14:26