Let $A$ be a full rank matrix with the property that columns $1 ,3 , 5 , 7 , ...\quad$ are orthogonal to columns $2 ,4 , 6 , 8, ...$. In a reduced $QR$ factorization $A=\hat{Q}\hat{R}$, what special structure does $\hat{R}$ possess?
Solving this manually leads to following matrix. $$R= \begin{bmatrix} x_{1}&0&y_{1}&0&&&&...\\0&x_{2}&0&y_{2}&0&&&...\\0&0&x_{3}&0&y_{3}&&&...\\0&0&0&x_{4}&0&&&y_{4}\\ &&&...&&&&0\\0&...&&&&&&x_{5}\end{bmatrix} $$ I don't know how to prove this formally. Does any one have an insight on this?
Thanks.
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I added the steps I have done to solve the problem manually:
I’ve tried the Gram-Schmidt orthogonalization as follow:
$ q_n = \frac{a_n- \sum_{i=1}^{n-1}r_{in}q_{i}}{r_{nn}}$
$ r_{ij} = q_i^{*}a_{j} \quad \quad i \neq j $
$ |r_{jj} |= || a_{jj} - \sum_{i=1}^{j-1} q_{i}r_{ij} ||_{2}\quad \quad i = j $
$q_{1} = \frac{a_{1}}{||a_{11}||} \quad q_{2} = \frac{a_{2}}{||a_{22}||}\quad q_{3} = \frac{a_{3}-\frac{(a_{1}.a_{3})a_{1}}{||a_{1}||^{2}}}{|| a_{3}-\frac{(a_{1}.a_{3})a_{1}}{||a_{1}||^{2}}||} $
$r_{11} = ||a_1||\quad$ $r_{22} = ||a_2||\quad$ $r_{12}=0 \quad ....$
$q_{1}$ is a linear combination of $a_{1}$
$q_{2}$ is a linear combination of $a_{2}$
$q_{3}$ is a linear combination of $a_{1},a_{3} $
$q_{4}$ is a linear combination of $a_{2},a_{4} $
...
$$R= \begin{bmatrix} ||a_{1}|| & 0 & \frac{a_1.a_3}{||a_{1}||} & 0 &... \\ 0 & ||a_{2}||& 0 & val\neq0 &0 & ...\\ 0&0&||a_{3}-\frac{a_{1}.a_{3}}{||a_{1}||^{2}}{a_{1}||} & ...\\ .\\.\\.\end{bmatrix} $$
In matrix $R$:
If $i = j$: $r_{ij}\neq0$
If $i > j$ :$r_{ij}=0$
If $i<j$ & both $i,j$ are either odd or even then $r_{ij}\neq0$