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.
better to demand that only unit vectors will be allowed. Then the $v$ is just $v = a + b,$ or half of that which is the midpoint of the segment between them. Make some drawings and try examples in a plane, what is required is just $a,b$ be the same length and $a+b \neq 0$