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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.

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  • $\begingroup$ You have two inner products. the $e_i$'s are orthonormal for the old one $\langle\rangle$ but not for the new one $g.$ $\endgroup$ Mar 17, 2023 at 22:28
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    $\begingroup$ What the professor means is that any inner product on a basis orthonormal with respect to that inner product is just the dot product. $\endgroup$ Mar 17, 2023 at 22:30

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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.

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  • $\begingroup$ Thank you very much for what you wrote. This is what makes sense. It is that the basis must be orthonormal with respect to that inner product. As an extension of my question, could kindly please address what the same inner product would look like for a non orthonormal basis? $\endgroup$ Mar 19, 2023 at 0:33
  • $\begingroup$ @HaimSchatz Well if we have a non orthonormal basis $v_1,\dots,v_n$ then we just don't know anything about $\langle v_i,v_j\rangle$ and can't really simplify the double sum which appeared in my answer. So you're just going to be stuck with the double sum. $\endgroup$
    – Wurstcake
    Mar 19, 2023 at 11:21

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