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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 $\mathbb{R}^n.$

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    $\begingroup$ $A=\frac{\|v\|}{\|w\|}w$ makes no sense. The RHS is a vector. $\endgroup$ – Rodrigo de Azevedo Jan 25 '17 at 20:22
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    $\begingroup$ I edited the equation in the problem statement to the most reasonable fix. If this is incorrect, please undo my edit. $\endgroup$ – Aweygan Jan 25 '17 at 20:26
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    $\begingroup$ Also, your argument assumes both $v_1$ and $v_2$ are nonzero, which isn't a valid assumption. $\endgroup$ – Aweygan Jan 25 '17 at 20:37
  • $\begingroup$ @Aweygan Sorry, I missed $v$. $\endgroup$ – Parisina Jan 25 '17 at 20:44
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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.

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  • $\begingroup$ Thanks! But what if $w, \:v$ aren't linearly independent? $\endgroup$ – Parisina Jan 27 '17 at 5:40
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    $\begingroup$ :) Then they must be proporitional, and, since orthogonal matrices must preserve length, we would have that $v= \pm w.$ So either the identity matrix $I$ takes $v$ to $w,$ or $-I.$ $\endgroup$ – Olod Jan 27 '17 at 11:39

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