Say $T$ is an operator on a finite dimensional real inner product space $V$ which has an orthonormal basis $v_1,v_2,…,v_n$, $T$ preserves the norm of the sum of any two orthonormal basis vectors $v_i,v_j$, i.e. $$||T(v_i+v_j)||=||v_i+v_j||,$$where $1\leq i,j\leq n.$ How do you prove T is an isometry?
I tried expanding the norm and played around with the adjoint, but it didn't seem to work.
Best Answer
Counter-example: Let $V=\mathbb R^{2}, v_1=(1,0), v_2=(0,1)$ and $T(x,y)=(\frac {x+y} {\sqrt 2},0)$. Then $\|T(v_1+v_2)\|=\|v_1+v_2\|$ but $T$ is not an isometry.
If we allow $i=j$ (assuming that $T$ is a linear map) we see that $\|Tv_i\|=\|v_i\|$ and expanding $\|T(v_i+v_j)\|^{2}$ as well as $\|v_i+v_j\|^{2}$ we see that $\langle Tv_i, Tv_j \rangle =0$ for $i \neq j$. Now we can check that $\|T(\sum a_iv_i)\|^{2}=\|\sum a_iv_i\|^{2}$ for all $a_1,a_2,...,a_n$.