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Aug 16, 2013 at 10:27 history closed R W
John Pardon
David White
Andrey Rekalo
Chris Godsil
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Aug 16, 2013 at 3:04 review Close votes
Aug 16, 2013 at 10:27
Aug 12, 2013 at 14:07 comment added Emerton Dear Vishal, This is the Gram--Schmidt process for turning a basis of $\mathbb R^n$ into an orthonormal basis: think of the columns of a matrix in $GL_n(\mathbb R)$ as a basis of $\mathbb R^n$, apply Gram--Schmidt, and then reinterpret it in terms of matrix multiplications. (This is basically what Paul Garrett's answer does.) Regards,
Aug 11, 2013 at 18:32 vote accept Vishal Gupta
Aug 11, 2013 at 18:23 vote accept Vishal Gupta
Aug 11, 2013 at 18:23
Aug 11, 2013 at 17:52 answer added paul garrett timeline score: 6
Aug 11, 2013 at 17:43 comment added Vishal Gupta Could you please give a more elementary argument using only matrices?
Aug 11, 2013 at 17:10 comment added JGordon Sorry about my previous (now deleted) comment, where I completely misread the question. It is, indeed, the special case of the Iwasawa decomposition; the general proof is Proposition 7.31 in Knapp's "Lie groups beyond an introduction", for example. First, we can reduce it to $SL_n$ and $S0_n$, respectively. Now, the idea is to first see the decomposition on the Lie algebra level, and then show that the image of $T\times SO$ has to be open (this uses the Lie algebra decomposition) and closed (this is easy to see since $SO$ is compact) in $SL_n$. This might be an overkill though.
Aug 11, 2013 at 16:52 review Close votes
Aug 11, 2013 at 17:24
Aug 11, 2013 at 16:35 history edited BS.
removed tag functional-analysis
Aug 11, 2013 at 16:06 history asked Vishal Gupta CC BY-SA 3.0