Iwasawa Decomposition (special case): Let $G=SL_n(\Bbb{R})$, $K=$ real unitary matrices, $U=$ upper triangular matrices with $1$'s on the diagonal (called unipotent), and $A=$ diagonal matrices with positive elements ($0$ everywhere else). Then, the product map $U\times{A}\times{K}\rightarrow{G}$ given by $(u,a,k)\mapsto{uak}$ is a bijection.
Here is the proof by Serge Lang in his book Undergraduate Algebra Section 6 Chapter 4 pg 246 (Read below picture for question(s) please :)
I have a few questions about this proof (understand it roughly as a whole):
1) How does one get that $B=au$, following $g^{-1}=Bk^{-1}$?
2) Why does it follow that A has positive diagonal elements - that is $a_i=b_{ii}>0$? (My guess is that the QR decomposition guarantees this for R and note $B=R$)
3) I can't see any point in the proof where Lang makes a reference to the fact that $g$ has determinant $1$, in fact it seems that $g$ could have any non-zero determinant, hence $g\in{GL_n(\Bbb{R})}$. Why is this not so?
Thank you.
The diagonal entries of $B$ are positive, because they are the lengths of the vectors in Gram-Schmidt. To write $B=au$, I will do it in $2\times 2$ so that you see what happens: $$ B=\begin{bmatrix} b_{11} & b_{12} \\ 0& b_{22}\end{bmatrix} =\begin{bmatrix} b_{11} &0 \\ 0& b_{22} \end{bmatrix} \begin{bmatrix} 1&b_{12}/b_{11}\\ 0&1\end{bmatrix}. $$ As for the determinant, the way the decomposition is phrased it is not restricted to the case $\det g=1$, since $a$ is allowed arbitrary diagonal entries so the determinant can be arbitrary. Note that $\det u=1$, $\det k=\pm1$, so $$\det g=\pm\det a.$$