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I learnt in a introductory linear algebra class that an inner product is defined as the dot product, i.e. $\langle v, w\rangle = \Sigma_i (v_i \cdot w_i)$

I recently learnt that inner product is actually a generalization of dot product, and an inner product satisfies certain conditions (e.g. as listed on MathWorld).

One inner product for $\mathbb{C}^n$ is given by

$\langle v,w\rangle = v^\dagger w$, where $v^\dagger$ is the transpose of the complex conjugate, i.e. $v^\dagger = \bar{v}^T$

Are there other inner products? I'm having a hard time seeing how something else would be useful. e.g. I can define norm using the inner product definition above, so it is useful, how would some other inner product be useful?

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  • $\begingroup$ any positive operator in a Hilbert space describe an inner product, so yes, there are infinitely many distinct inner products for any Hilbert space $\endgroup$
    – Masacroso
    Commented Feb 9, 2020 at 11:19

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Any inner product on $\Bbb C^n$ is of the form $$ \langle v, w\rangle = v^\dagger Sw $$ for some positive-definite Hermitian matrix $S$. It's what you get when changing basis without changing the (geometric) inner product. In other words, if you want to preserve the dot product through a basis change, even though all vectors get represented by new components, then this is the new algebraic form of the same product.

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    $\begingroup$ You make it sound like there are inner products on $\mathbb C^n$ that are outside of this class. $\endgroup$
    – Christoph
    Commented Feb 9, 2020 at 11:25
  • $\begingroup$ This looks interesting. If $C$ is the basis-change matrix between two bases $A$ and $B$ (i.e. for a vector $v$ in basis $A$, $Cv$ is its representation in $B$), how is $C$ related to $S$ in your answer above? $\endgroup$
    – Peeyush Kushwaha
    Commented Feb 9, 2020 at 11:30
  • $\begingroup$ @PeeyushKushwaha We have $S=C^\dagger C$ (or $S=(C^{-1})^\dagger C^{-1}$, depending on which way $C$ converts). $\endgroup$
    – Arthur
    Commented Feb 9, 2020 at 11:53
  • $\begingroup$ @Christoph Yes, I did. Or at least I made it sound like I wasn't certain. I fixed that now. $\endgroup$
    – Arthur
    Commented Feb 9, 2020 at 12:03
  • $\begingroup$ @PeeyushKushwaha If we have the standard dot product on $A$, and we want the same (geometric) product on $B$ in spite of the basis change, then it must be $$\langle v, w\rangle = v_A^\dagger w_A = (C^{-1}v_B)^\dagger (C^{-1}w_B) = v_B^\dagger (C^{-1})^\dagger C^{-1}w_B$$So in the basis $B$, the inner product is given by $S = (C^{-1})^\dagger C^{-1}$. $\endgroup$
    – Arthur
    Commented Feb 9, 2020 at 13:17

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