# Can there be multiple inner product defintions for $\mathbb{C}^n$?

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?

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

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.
• You make it sound like there are inner products on $\mathbb C^n$ that are outside of this class. Feb 9, 2020 at 11:25
• 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? Feb 9, 2020 at 11:30
• @PeeyushKushwaha We have $S=C^\dagger C$ (or $S=(C^{-1})^\dagger C^{-1}$, depending on which way $C$ converts). Feb 9, 2020 at 11:53
• @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 $S = (C^{-1})^\dagger C^{-1}$. Feb 9, 2020 at 13:17