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After reading about inner products as a generalization of the dot product, I was hoping to be able to prove that the dot product is in some sense the unique inner product in Euclidean space (e.g., up to constant scaling).

But it seems that there are a whole bunch of alternative inner products in $\mathbb{R}^2$ with nonzero cross-terms between basis vectors, for example, $\langle (a, b)^\intercal, (x, y)^\intercal \rangle = ax + by + 0.5(ay + bx)$. Unless I've made a mistake, this satisfies symmetry, linearity, and positive-definiteness.

Is there a sense in which the dot product is the canonical inner product on Euclidean space? Or do we just pick it because the implied norm matches our notion of distance?

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    $\begingroup$ Not quite what you're asking - but we do know that any two inner products on a finite-dimensional vector space are equivalent, which means there are positive constants $c, C$ such that $c \langle x, y \rangle_1 \le \langle x, y \rangle_2 \le C \langle x, y \rangle_2$ for all $x,y$. So although the inner product is not unique, at least any two are within a constant scaling factor of each other. (This fact is most useful when studying a topology induced by the inner product - it means the corresponding topology doesn't depend on the choice of inner product.) $\endgroup$ – Daniel Schepler Apr 15 at 17:09
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Any inner product is dot product in some basis. For example, your inner product is standard dot product written in basis $\left(e_1, \frac{1}{2}e_1 + \frac{\sqrt{3}}{2}e_2\right)$.

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  • $\begingroup$ No it's not. $e_2 = (-1,2)/\sqrt{3}$ in that basis, which has euclidean norm 5/3. But $\left<e_2,e_2\right> = 1$. $\endgroup$ – eyeballfrog Apr 15 at 19:01
  • $\begingroup$ Using author's inner product we have $\langle (-1, 2) / \sqrt{3}, (-1, 2) / \sqrt{3}\rangle = 1$. And in general - if we write two vectors in this basis and take inner product as defined in question, we get their standard dot product. $\endgroup$ – mihaild Apr 15 at 19:44
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There is nothing special about the dot product. Yes, it corresponds to the Euclidean norm if you are using an orthonormal basis. But if your basis is not orthonormal then the Euclidean norm will be represented by some other symmetric matrix.

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For an arbitrary inner product $\left<\right>$ on $\mathbb R^n$, there is a positive definite real symmetric matrix $A_{ij} = \left<e_i|e_j\right>$ that defines the transform. Since it is real and symmetric, it is orthogonally diagonalizable. That is, for any inner product on $\mathbb R^n$, there is a set of real numbers $\lambda_j$ and an orthonormal basis $\left|\xi_j\right>$ such that $$ \left<a|b\right> = \sum_j\lambda_j\left<a|\xi_j\right>\left<\xi_j|b\right> $$ Roughly speaking, the inner product resolves $a$ and $b$ into their $\xi_j$ components, then weights the resulting dot product by $\lambda_j$.

In general, this choice of $\left|\xi_j\right>$ will be unique. However, for some inner products, there will be multiple possible choices of $\left|\xi_j\right>$. The Euclidean norm is unique (up to a constant scaling) in that every choice of $\left|\xi_j\right>$ allows the inner product to be written in that form--it is independent of the chosen basis.

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