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Casting 2x2 matrix with unit determinant in another form 
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#1
Feb2512, 08:57 AM

P: 27

Dear all,
I have a simple question for you. Any help will be very appreciated. ==Assumptions=== I have a 2x2 matrix, with real entries: A B C D Such matrix has unit determinant, ADBC=1. (For those of you in group theory: the above is a representative of SL(2,R)). ==Question=== The above matrix, due to the unitdeterminant condition, has 3 independent parameters. So, I would like to represent the 2x2 real matrix with unit determinant in a more convenient form: a form which depends on only 3 parameters. For instance, I could try and relate the above matrix to a rotation matrix, that is, one of the independent parameters would be a rotation angle. Any ideas on a convenient "minimal" representation? ========= Thanks a lot! IVL 


#2
Feb2512, 10:42 AM

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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
[tex]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},[/tex] where [itex]\theta \in [0, 2\pi)[/itex], [itex]a > 0[/itex] and [itex]n \in \mathbb R[/itex] are uniquely determined by A. 


#3
Feb2512, 11:05 AM

P: 27

Wow! Thanks Morphism, that is precisely what I was looking for.
Do you have any good reference on the Iwasawa decomposition? Cheers IVL Message to other users: for the sake of completeness, if you think there are other decompositions worth being mentioned, please add them to the thread. Thanks. 


#4
Feb2512, 01:31 PM

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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 socalled 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 [tex]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},[/tex] where [itex]\theta,\psi \in [0,2\pi)[/itex] and [itex]a>0[/itex] are uniquely determined by A. Yet another example is the Bruhat decomposition, which states that [itex]A \in \text{SL}(2,\mathbb R)[/itex] is either uppertriangular, or else can be decomposed as [tex]\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}.[/tex] Note that c is nonzero since we're assuming A is not uppertriangular, so it makes sense to invert it. Also note that b appears to have vanished, but of course it's still there as [itex]b = (ad1)c^{1}[/itex]. 


#5
Feb2512, 02:07 PM

P: 27

Veeery nice!
Thanks a lot! 


#6
Feb2612, 04:31 PM

P: 27

Dear Morphism, and dear all,
thinking more on the above, I have two questions: 1. Doers there exist an AKA decompostion? (where A and K take the same meaning as in the KAK decomposition) 2. I am very interested in the Bruhat decomposition. Can anyone recommend a good reference for that? For some reason, the matrix equations do not appear any more on the blog's page. Is anyone experiencing the same problem? Cheers IVL 


#7
Feb2812, 04:56 PM

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1. No, there can be no such decomposition. If there were, we would be able to write
[tex]\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}.[/tex] 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. 


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