For any nonzero vector $v,\:w \in \mathbb{R}^n$, construct an orthogonal matrix $A$ such that $Av=\frac{\|v\|}{\|w\|}w$.
Here is something that came to my mind.
Let $v=\begin{bmatrix}
v_1\\
v_2\end{bmatrix},\:w=\begin{bmatrix}
w_1\\
w_2\end{bmatrix}\in\mathbb{R}^2$.
As, $A$ is an orthogonal matrix, its columns are orthogonal, and so it's similar to a matrix in the form of $\begin{bmatrix}
a&0\\
0&b
\end{bmatrix}$. Thus,
$$\begin{bmatrix}
a&0\\
0&b
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2\end{bmatrix}=\frac{\|v\|}{\|w\|}\begin{bmatrix}
w_1\\
w_n
\end{bmatrix}.
$$
Hence, $a=\frac{\|v\|}{\|w\|}\frac{w_1}{v_1} $ and $b=\frac{\|v\|}{\|w\|}\frac{w_2}{v_2}.$
I'm wondering if my argument is valid and I can do the same for $\mathbb{R}^n.$
Best Answer
Suppose that $v,w$ are linearly indepedent. All vectors of $\mathbf R^n$ are to be treated as column vectors.
1) Perform Gram--Schmidt process on the set $\{v,w\}$ to obtain an orthonormal basis $(v_1,v_2)$ for the linear span $L=\langle v,w\rangle$ such that $$ v_1 = \frac 1{|v|}v. $$
2) Perform Gram--Schmidt on the set $\{w,v\}$ to obtain an orthonormal basis $(w_1,w_2)$ for $L$ such that $$ w_1=\frac 1{|w|}w. $$
3) Extend the linear independent set $\{v_1,v_2\}$ to a basis $\mathcal B$ of $\mathbf R^n,$ say by adding to it some of the unit vectors $e_i$ from the standard basis for $\mathbf R^n.$ Perform Gram--Schmidt on $\mathcal B,$ followed by normalizing of all vectors to obtain an orthnormal basis $$ (v_1,v_2,v_3,\ldots,v_n) $$ for $\mathbf R^n.$
4) Consider an orthonormal basis $$ (w_1,w_2,w_3,\ldots,w_n)=(w_1,w_2,v_3,\ldots,v_n) $$ for $\mathbf R^n$ which is an extension of the linear independent set $\{w_1,w_2\}.$
5) Form the $(n\times n)$ matrices $V,W$ whose columns are the vectors $v_1,v_2,v_3,\ldots,v_n$ and $w_1,w_2,\ldots,w_n,$ respectively: $$ V=(v_1 \, v_2 \, \ldots v_n) $$ and $$ W=(w_1 \, w_2 \, \ldots w_n). $$ Now both matrices $V,W$ are orthogonal, since each of them takes the standard (orthonormal) basis $(e_1,e_2,\ldots,e_n)$ to an orthonormal basis. Say, $$ V(e_1\, e_2 \, \ldots e_n)=(v_1 \, v_2 \, \ldots v_n) $$
6) Finally, the matrix $$ A=W V^{-1} $$ takes the orthonormal basis $(v_1,v_2,\ldots,v_n)$ to the orthonormal basis $(w_1,w_2,\ldots,w_n),$ and hence orthogonal. In particular, $$ A v_1=w_1, $$ whence $$ Av=A |v| v_1=|v| w_1=\frac{|v|}{|w|}w, $$ as required.