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I was asked to prove that if

$$ T_{n}^{+}(\mathbb{R}) \subseteq M_{n}(\mathbb{R})$$

denotes the set of upper triangular matrices with positive diagonal entries, then prove that the multiplication map

$$ \mu : O_{n}(\mathbb{R}) \times T_{n}^{+}(\mathbb{R}) \rightarrow GL_{n}(\mathbb{R})$$

is a homeomorphism where $O_{n}(\mathbb{R})$ is the set of orthogonal matrices.

Using polar decomposition, I could write

$$ GL_{n}(\mathbb{R}) = O_{n}(\mathbb{R}) \times Pd_{n}(\mathbb{R})$$

the positive definite matrices. Then positive definite matrices are unitarily diagonalizable, but that does not seem to take me towards $T_{n}^{+}(\mathbb{R})$.

Any ideas about the above?

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closed as off-topic by R W, John Pardon, David White, Andrey Rekalo, Chris Godsil Aug 16 '13 at 10:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – R W, John Pardon, Andrey Rekalo
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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. – JGordon Aug 11 '13 at 17:10
Could you please give a more elementary argument using only matrices? – Vishal Gupta Aug 11 '13 at 17:43
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, – Emerton Aug 12 '13 at 14:07
up vote 4 down vote accepted

There is an elementary argument for the Iwasawa decomposition for classical groups, and it is especially simple for $G=GL_n(\mathbb R)$: use the fact that $K=O(n,\mathbb R)$ is transitive on the set of vectors in $\mathbb R^n$ of a given length, by rotating. Thus, given $g\in G$, first left-multiply by an element to make the left-most column of $g$ be of the form $\pmatrix{*\cr 0 \cr 0 \cr ... \cr 0}$. Then, without disturbing that first column, left multiply by an element of $O(n-1,\mathbb R)$ sitting inside the lower-right block of $O(n,\mathbb R)$ to make the second column of (the updated) $g$ of the form $\pmatrix{*\cr*\cr 0\cr ...\cr 0}$. Continuing, these iterated left multiplications by orthogonal matrices make $g$ upper-triangular with positive diagonal entries, giving the Iwasawa decomposition here.

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