# Casting 2x2 matrix with unit determinant in another form

 Sci Advisor HW Helper P: 2,020 There are several such decompositions. Perhaps the most aesthetically pleasing one is the Iwasawa decomposition (also known as the KAN decomposition), which says that any A in SL(2,R) can be written as a product $$A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix} \begin{pmatrix} 1 & n\\ 0 & 1\end{pmatrix},$$ where $\theta \in [0, 2\pi)$, $a > 0$ and $n \in \mathbb R$ are uniquely determined by A.
 Sci Advisor HW Helper P: 2,020 Casting 2x2 matrix with unit determinant in another form For the Iwasawa decomp of SL(2,R), these notes of Keith Conrad are nice. I should mention that what's going on here is part of a bigger picture: the Iwasawa decomposition is really a statement about decomposing certain types of groups (e.g. semisimple Lie groups) of which SL(2,R) is an example. For an example of another decomposition, you can refine the polar decomposition that you learned in linear algebra to obtain the so-called KAK decomposition, which also applies to a broad class of Lie groups. For SL(2,R), this states that any A can be written as $$A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & a^{-1}\end{pmatrix} \begin{pmatrix} \cos\psi & -\sin\psi \\ \sin\psi & \cos\psi \end{pmatrix},$$ where $\theta,\psi \in [0,2\pi)$ and $a>0$ are uniquely determined by A. Yet another example is the Bruhat decomposition, which states that $A \in \text{SL}(2,\mathbb R)$ is either upper-triangular, or else can be decomposed as $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & ac^{-1} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} c & d \\ 0 & c^{-1} \end{pmatrix}.$$ Note that c is nonzero since we're assuming A is not upper-triangular, so it makes sense to invert it. Also note that b appears to have vanished, but of course it's still there as $b = (ad-1)c^{-1}$.
 Sci Advisor HW Helper P: 2,020 1. No, there can be no such decomposition. If there were, we would be able to write $$\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}a & 0 \\ 0 & a^{-1}\end{pmatrix}\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}b & 0 \\ 0 & b^{-1}\end{pmatrix} = \begin{pmatrix}ab\cos\theta & -ab^{-1}\sin\theta \\ a^{-1}b\sin\theta & a^{-1}b^{-1}\cos\theta\end{pmatrix}.$$ Equate the bottom left corners to get $$a^{-1}b\sin\theta = 0 \implies \sin\theta=0.$$ But then the top right corners imply that $$1 = a^{-1}b\sin\theta = 0.$$ 2. There isn't really much to say about the Bruhat decomposition for SL(2,R) besides what I had already mentioned. If you're interested in the Bruhat decomposition for a general semisimple/reductive group, then if you tell me what your representation theory background looks like, I might be able to suggest a suitable reference.