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I know that in $\mathbb{R}^n$ (and in general $F^n$) you can define an Inner Product in the following way:

Let $X,Y \in \mathbb{R}^n$. Let $C \in \mathbb{R}^n$, and let all the components of $C$ be positive.

Then: $\langle 'X,Y \rangle = \sum_{i=1}^n c_ix_iy_i$ is an Inner Product.

I was wondering if this is the only type of Inner Product in $\mathbb{R}^n$? Can you define an inner product in $\mathbb{R}^n$ that doesn't take this form? Thanks.

glS
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user146485
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    Short answer: Yes. Long answer: No, not really. You can define an inner product by choosing an arbitrary basis and declaring them to be orthonormal (extending the inner product bilinearly), when expressed in the standard basis this won't usually be of the form you are suggesting. But it turns out that every inner product can be defined this way. For example the one you defined is given by taking as a basis the standard basis vectors scaled down by $\sqrt{c_i}$. – Nate Apr 28 '14 at 23:11
  • you need to watch out for generalizing your observation to arbitrary fields $F$. Your summation then yields an element of $F$, while for the definition to give an inner product, the answer has to be a real or a complex number. You also may want to wrap things between dollar signs so that things look like $\mathbb R^n$ rather than R^n. – Ittay Weiss Apr 29 '14 at 00:46

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There are inner products on $\mathbb{R}^n$ that do not take the form you suggestion. In the inner product $$ \langle x, y \rangle = \sum_i c_i x_i y_i$$ each component in $x$ is matched with exactly one component in $y$. But why can't $x_i$ be matched with some $y_j$ for $i \neq j$? Indeed, one can actually arrange for this. In general

$$ \langle x, y \rangle = x^T A y, $$ where $A$ is a symmetric positive definite matrix, classifies all inner products on $\mathbb{R}^n$. Your example arises as a special case where $A$ is a diagonal matrix, $$ A = \begin{bmatrix} c_1 & & \\ & \ddots & \\ & & c_n \end{bmatrix}. $$

Christopher A. Wong
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