# For nonzero $v, w\in\Bbb{R}^n$, prove a linear isometry exists s.t. $Tv=\|v\|e_1$, and $Tw\in\operatorname{span}\{e_1, e_2\}.$

The problem is this: Given nonzero vectors $$v, w \in \mathbb{R}^n$$ ($$n \geqslant 2$$), prove that there is a linear isometry $$T : \mathbb{R}^n \rightarrow \mathbb{R}^n$$ s.t. $$Tv = \|v\|e_1$$, and $$Tw \in \operatorname{span}\{e_1, e_2\}.$$

I have not done any problems quite like this before, and I am mostly looking for help regarding where to start with the problem? I am not sure if I should try to find an explicit isometry $$T$$, or if I should just try to prove that such an isometry must exist (whether by contradiction or directly). Any help would be greatly appreciated, at any rate. Thank you!

Suppose $$v \neq 0$$. Assume also that $$v$$ and $$w$$ are linearly independent. Let $$v_1=\frac v {\|v\|}$$. Use Gram-Schmidt process to construct an orthonormal basis $$x_1,x_2,...,x_n$$ such that $$x_1=v_1$$ and $$x_2 \in span \{v,w\}$$. Then define $$T( \sum\limits_{k=1}^{n} a_kx_k)=\sum\limits_{k=1}^{n} a_ke_k$$. Then $$T$$ is an isometry and $$Tv=\|v\|Tv_1=\|v\|Tx_1==\|v\|e_1$$ and $$Tw \in span \{v,w\}$$.
The case $$v=0$$ is is simpler and I will let you handle that case. The case when $$v$$ and $$W$$ are linearly dependent is also easier.