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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.

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$v$ 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 $$ H(v)a=a-2\frac{v\|a\|(\hat a^T\hat b-1)}{2(1-\hat a^T\hat b)} =a+\|a\|(\hat b-\hat a) =\|a\|\hat b=\frac{\|a\|}{\|b\|}b $$

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  • $\begingroup$ 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? $\endgroup$ Dec 1, 2023 at 23:10

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