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

Question

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!

Answer 1

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.