Existence of an orthogonal matrix

Question

I would like the prove the following result. Thanks for any help in advance.

Let $v_1$, $v_2$ $\in$ $\mathbb{R^{n}}$. If |$v_1$| = |$v_2$|, then there exists an orthogonal matrix $\Gamma$ such that $\Gamma$$v_1$= $v_2$.

Answer 1

Let $x_1=v_1/|v_1|$, $y_1=v_2/|v_2|$. Then we can extend $x_1$ to an orthonormal basis $x_1,\ldots,x_n$, and $y_1$ to an orthonormal basis $y_n$. Define $\Gamma$ as the linear operator with $\Gamma(x_j)=y_j$. That is $$ \Gamma(\sum_{k=1}^nt_kx_k)=\sum_{k=1}^nt_ky_k. $$ Since both $\{x_j\}$ and $\{y_j\}$ are basis, $\Gamma$ is well-defined. We have $$ (\Gamma x)\cdot(\Gamma x)=\left(\sum_{k=1}^nt_ky_k\right) \cdot\left(\sum_{k=1}^nt_ky_k\right)=\sum_{k=1}^nt_k^2=x\cdot x, $$ so $\Gamma$ is orthogonal. And $$ \Gamma(v_1)=\Gamma(|v_1|\,x_1)=|v_1|\,y_1=|v_2|\,y_1=v_2. $$

Answer 2

For $v_1$ and $v_2$, consider the linear function that reflects $v_2$ onto $v_1$ with respect to the vector between them which is $w=\frac{v_1+v_2}2$:

$$ \Gamma v=2\frac{v.w}{w.w}w-v. $$ Firstly see that: $$ \Gamma v_1=2\frac{v_1.w}{w.w}w-v_1=2\frac{v_1.(v_1+v_2)}{(v_1+v_2)(v_1+v_2)}(v_1+v_2)-v_1\\ =2\frac{v_1.(v_1+v_2)}{(v_1+v_2)(v_1+v_2)}v_2+\frac{(v_2-v_1).(v_1+v_2)}{(v_1+v_2)(v_1+v_2)}v_1=2\frac{|v_1|^2+v_1.v_2}{2|v_1|^2+2v_1.v_2}v_2=v_2. $$ To check orthogonality, see that: $$ \Gamma u.\Gamma v=(2\frac{u.w}{w.w}w-u).(2\frac{v.w}{w.w}w-v)\\ =u.v-2\frac{u.w}{w.w}w.v-2\frac{v.w}{w.w}w.u+4\frac{u.w}{w.w}\frac{u.w}{w.w}w.w=u.v. $$