Inner Product, Orthonormal basis, and Kronecker Delta

Question

My professor says that any inner product on an orthonormal basis is equal to the dot product, but I am so confused as to what this can mean.

Let $C = \{e_j\}_{j=1}^n$ be an orthonormal basis of an inner product space $V$. Suppose an inner product is given $g(u,v) = \left<u, Av\right>$ where $A:V \to V$ is a linear operator that is symmetric and positive definite.

It would seem to me that $$g(e_j, e_k) = \left<e_j, \sum_{l=1}^n A_{lk} e_l \right> = \sum_{l=1}^n A_{lk}\left< e_j, e_l \right> = \sum_{l=1}^n A_{lk}\delta_{jl}= A_{jk}$$ which may or may not itself be equal to the kronecker delta $\delta_{ij}$.

In this inner product the basis vectors are not orthonormal because $A_{ij}$ is not a kronecker delta, but I had assumed them to be initially. This is one way of communicating what I am confused about.

Answer 1

You have defined a new inner product. I think what was meant (as eyeballfrog mentioned in the comments), is that the inner product of the space $V$ is the dot product when the basis is orthonormal with respect to that inner product.

To see this we take $v,w\in V$ and write them out as $v=a_1e_1+\dots+a_ne_n$, $w=b_1e_1+\dots+b_ne_n$. Then by taking the inner product, we get:

$$\langle v,w\rangle=\langle\sum_{i=1}^n a_ie_i,w\rangle=\sum_{i=1}^na_i\langle e_i,w\rangle=\sum_{i=1}^na_i\langle e_i,\sum_{j=1}^nb_je_j\rangle=\sum_{i=1}^n\sum_{j=1}^na_ib_j\langle e_i,e_j\rangle=\\\sum_{i=1}^n\sum_{j=1}^na_ib_j\delta_{ij}=\sum_{i=1}^na_ib_i$$

which is precisely the dot product.