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?



Iwasawa Decomposition for Matrices
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.


