# Finding the Vector $v$ for a Given Householder Matrix Transformation of Non-Collinear Vectors $a$ and $b$

Consider a vector $$v$$ in $$\mathbb{R}^{n\times1}$$. The Householder matrix is defined as follows: $$H(v)=I-\dfrac{2vv^T}{v^Tv}.$$ It can be demonstrated that $$H(v)$$ is symmetric and orthonormal. The challenge lies in identifying a vector $$v$$ for two vectors $$a$$ and $$b$$ that are not collinear (i.e., $$a\neq\alpha b$$ for any scalar $$\alpha$$), which satisfies the equation: $$H(v)a=\dfrac{||a||_2}{||b||_2}b.$$

An attempt to solve this problem might involve calculating each entry of the matrices on both sides of the equation. On the left-hand side (LHS), we have: $$\begin{bmatrix} a_1+\dfrac{2v_1}{||v||^2}(-v_1a_1+v_2a_2+\dots+v_na_n)\\ a_2+\dfrac{2v_2}{||v||^2}(+v_1a_1-v_2a_2+\dots+v_na_n)\\ \vdots\\ a_n+\dfrac{2v_n}{||v||^2}(+v_1a_1+v_2a_2+\dots-v_na_n) \end{bmatrix}$$

On the right-hand side (RHS), we obtain: $$\sqrt{\dfrac{a_1^2+a_2^2+\dots+a_n^2}{b_1^2+b_2^2+\dots+b_n^2}}\begin{bmatrix} b_1\\b_2\\\vdots\\b_n \end{bmatrix}$$

However, the determination of $$v$$ remains elusive. Any insights or suggestions would be greatly appreciated.

$$v$$ is one of the bisectors $$\|b\|a\pm\|a\|b$$ between the rays in directions $$a$$ and $$b$$. One of them is the larger one, which gives a more accurate reflection, usually the difference is small, becomes only appreciable if the vectors are almost collinear.
You essentially already got to $$\frac{a}{\|a\|}-cv=\frac{b}{\|b\|}$$, the value of $$c$$ is not really important in this self-normalizing formulation of the reflector.
If the reflectors are used in a QR factorization, the direction on the right side is less important, the signs on the diagonal of the $$R$$ factor can be removed later. That is why I gave two variants above.
Using $$v=\hat b-\hat a$$, where the hatted vectors are normalized, you get $$v^Tv=2(1-\hat a^T\hat b)$$ and $$v^Ta=\|a\|(\hat b^T\hat a-1)$$ so that $$H(v)a=a-2\frac{v\|a\|(\hat a^T\hat b-1)}{2(1-\hat a^T\hat b)} =a+\|a\|(\hat b-\hat a) =\|a\|\hat b=\frac{\|a\|}{\|b\|}b$$
• Sorry I can't find the answer to $v$. I tried both $||b||a+||a||b$ and $v=\frac{a}{||a||}+\frac{b}{||b||}$ and put them in the equation but couldn't prove the equality. How can I prove that without using geometry interpretation? Dec 1, 2023 at 23:10