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.
I have already proved the surjection of this map myself, however, when viewing Lang's proof (Undergraduate Algebra Section 6 Chapter 4 pg246) for the uniqueness of this mapping I became rather stuck and would like to understand it. Lang states the following:
"For uniqueness of the decomposition, if $g=uak=u'a'k'$, let $u_1=u^{-1}u'$, so using $g^tg$ (what does he mean by 'using'?) you get $a^{2t}u_1^{-1}=u_1a'^2$. These matrices are lower and upper triangular respectively, with diagonals $a^2,a'^2$, so $a=a'$, and finally $u_1=I$, proving uniqueness".
What does Lang mean by 'using' $g^tg$ (I presume he means applying $g^tg$ to both sides, but he is unclear…), and so therefore how does he 'get' $a^{2t}u_1^{-1}=u_1a'^2$.
Thank you
Best Answer
If you write $gg^t$, you get $$ gg^t=uakk^tau^t=ua^2u^t. $$ Similarly, $$ gg^t=u'a'^2u'^t. $$ If you equate both expressions, you get $$ ua^2u^t=u'a'^2u'^t. $$ Multiply by $u^{-1}$ on the left and by $(u'^t)^{-1}$ on the right to get $$ a^2(u'^{-1}u)^t = u^{-1}u' a'^2. $$ So letting $u_1=u^{-1}u'$, $$ a^2u_1^{-t}=u_1a'^2 $$ where $u_1$ is upper triangular.