A special case of the Iwasawa Decomposition for matrices is the following: 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 am looking at the proof of this in Serge Lang's Undergraduate Algebra, Section 6 Chapter 4 pg 246, however, I need some help in understanding the beginning of the proof:
"Let $e_1,..,e_n$ be the standard vertical unit vectors of $\Bbb{R^n}$. Let $g=(g_{ij})\in{G}$. Then we have $ge_i=\begin{pmatrix} g_{1i} \\ \vdots \\ g_{ni} \end{pmatrix}=g^{(i)}=\sum_{q=1}^n g_{qi}e_q$. There exists an upper triangular matrix $B=(b_{ij})$, so with $b_{ij}=0$ if $i>j$ such that $b_{11}g^{(1)}=e'_1$,……….,$b_{1j}g^{(1)}+b_{2j}g^{(2)}+…+b_{jj}g^{(j)}=e'_j$,………….., $b_{1n}g^{(1)}+b_{2n}g^{(2)}+…+b_{nn}g^{(n)}=e'_n$, such that the diagonal elements are positive, that is $b_{11},…,b_{nn}>0$, and such that the vectors $e'_1,…,e'_n$ are mutually perpendicular unit vectors. Getting such a matrix B is merely applying the usual Gram Schmidt orthogonalisation process, subtracting a linear combination of previous vectors to get orthogonality, and then dividing by the norms to get unit vectors."
My problem here is understanding the matrix $B$ and how it is obtained. Having studied the Gram Schmidt Orthogonalisation process in a linear algebra course, I simply learnt it as a way to construct an orthogonal (orthonormal by diving by norms) basis for a finite dimensional inner product space (in this case $\Bbb{R}$) given any basis. Could someone please explain how Lang then guarantees the existence of this matrix $B$.
Thank you.
ADDITIONAL (carrying on where Lang's proof left off): Let $gB=k\in{K}$. Then $ke_i=e'_i$, so the columns of $k$ are orthonormal, so $k$ is real unitary, and $g=kB^{-1}$. Then $g^{-1}=Bk^{-1}$ and $B=au$, where $a$ is the diagonal matrix with $a_i=b_{ii}$ and $u$ is unipotent, $u=a^{-1}B$.
Best Answer
Gram Schmidt is as follows, depicted visually
such that
$$a_{1} = r_{11} q_{1} \\a_{2} = r_{12}q_{1} + r_{22}q_{2}\\ a_{3}= r_{13}q_{1} +r_{23}q_{2} + r_{33} q_{3} \\ \vdots \\ a_{n} = r_{1n} q_{1} + r_{2n}q_{2} + \cdots + r_{nn} q_{n} $$
giving
$$A = \hat{Q}\hat{R} $$
The matrix $B$ appears to be the matrix $R$.
How it is found
$$ q_{1} = \frac{a_{1}}{r_{11}}\\ q_{2} = \frac{a_{2}-r_{12}q_{1}}{r_{22}}\\q_{3} = \frac{a_{3}-r_{13}q_{1}-r_{23}q_{2}}{r_{33}}\\ \vdots \\ q_{n} = \frac{a_{n} - \sum_{i=1}^{n} r_{in}q_{i}}{r_{nn}}$$
where
$$ |r_{ij} | = q_{i}^{*}a_{j} \\ | r_{jj}| = \| a_{j} = \sum_{i=1}^{j-1} r_{ij}q_{i}\|_{2}$$
seen here
Giving your $B$ from that