We have two matrixes: $$H_i=I_n-2v_iv_i^T, ||v_i||_2=1,i\in\{1,2\}$$ Moreover $v_2^Tv_1\neq\frac{\sqrt2}{2}$ We need to find algorithm (with $O(n) cost)$ solving following system of equations $$\begin{bmatrix}H_1 & -H_2\\H_2 & H_2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}b_1\\b_2\end{bmatrix}$$ where $x_1,x_2\in\mathbb{R^n}$
I do not fully understand how to do it, so I will write first what I know, and hopefully someone will give me some hint how to continue.$$\begin{cases} H_1x_1-H_2x_2=b_1 \\ H_2x_1+H_1x_2=b_2\end{cases}$$
$H_1,H_2$ are symetrical, so we can transform first equation into $x_1=H_1(b_1-H_2x_2)=H_1b_1-H_1H_2x_2$
We can continue to find $x_2$ in similar way, but I guess at some point we need to use konwledge concerning form of those matrixes, and I am not sure how to do it, so it would be useful to get to know it at this stage. So $$H_1H_2=(I_n-2v_1v_1^T)(I_n-2v_2v_2^T)=I_n-2v_2v_2^T-2v_1v_1^T-4v_1v_1^Tv_2v_2^T$$
Is it correct? To find solution (before construing alogrithm) do I need just to solve in "normally" and substitute $H_i$'s everywhere with what it looks "internally"? Maybe I should wait with multiplication of these matrixes till the end, but, when we have $H_1H_2$ then it is no symetrical anymore (right?) so I cannot use this in further transformations. I would be thankful if someone could give me some advice, how to solve this problem.
Isolate $x_1$ in both equations, $$ x_1=H_1b_1+H_1H_2x_2\\ x_1=H_2b_2-H_2H_1x_2 $$ and subtract to reduce to an equation in $x_2$ only: $$ (H_2H_1+H_1H_2)x_2=H_2b_2-H_1b_1 $$ Then apply the Sherman-Morrison-Woodbury matrix identity for the inverse of a rank-3 or rank-4 modification of the identity matrix. (Actually, it might turn out to be a simple rank-1 modification).