Suppose S is a real inner product space with (x,y) of finite dimension N, and $‖x‖=\sqrt{(x,x)} $
Let $ {(\theta_1,\theta_2,...\theta_N)} $ be an orthonormal basis for S
If X is real, N dimensional, and has an inner product defined as $$ 〈p,q〉=\sum_{i=1}^n p_i q_i$$ and $$ |p|=\sqrt{\sum_{i=1}^n {(p_i)^2} }$$
Define linear operator T: S → X such that $$T(\sum_{i=1}^N {\gamma_i\theta_i)} = [\gamma_1,...\gamma_n]^T $$ How does one prove isometry?
i.e. $ 〈T(x),T(y)〉 =(x,y) $ and $|T(x)| = ‖x‖ $ and the nullspace $ N(T) = {0} $ and therefore injective. Additionally, is T surjective?
You could equivalently define $T(x) = (\langle \theta_1 , x \rangle_S,\cdots, \langle \theta_1 , x \rangle_S)^T$. (Note that $x = \sum_k \langle \theta_k , x \rangle_S \theta_k$.)
$\langle \sum_k \alpha_k \theta_k , \sum_k \beta_k \theta_k \rangle_S = \sum_k \alpha_k \beta_k = \langle \alpha , \beta \rangle_X$.
Note that if the above is true, it follows that $\|T(x) \| = \|x\|$ from which it follows that $\ker T = \{0\}$. Surjectivity follows for a variety of reasons, the most straightforward being that $T$ is linear and $T(\theta_k) = e_k$ (vector of zeros with one in the $k$th position).