Are there any examples of inner products on $\mathbb R$ that is not isometric to the dot product? I cannot come up with an example of an inner product on $\mathbb R$ that is not the dot product......
Are there any examples of inner products on $\mathbb R$ that is not isometric to the dot product?
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linear-algebra
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0Do you mean where $\mathbb R$ is the field or where it is the vector space? – 2017-01-11
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0It is a vector space. – 2017-01-11
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Check out the first proposition here:
https://math.berkeley.edu/~peyam/Math110Sp13/Handouts/Dot%20products.pdf
$\langle x, y \rangle$ is an inner product on $\mathbb R^n$ if and only if $\langle x, y \rangle = x^T A y$, where $A$ is a symmetric matrix whose eigenvalues are strictly positive.
Then for $\mathbb R^1$, we have $\langle x, y \rangle$ is an inner product iff $\langle x, y \rangle = x a y$ where $a > 0$.
Basically, standard multiplication is the only inner product on the vector space $\mathbb R$ up to a change of positive constant $a$. (For the standard inner product on $\mathbb R$, $a=1$.)
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0So they will all be isometric? – 2017-01-11
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0@Adam You'll have to be more specific about what you mean by isometric. – 2017-01-11
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0It just means a map that preserves inner product. – 2017-01-11
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0@Adam Isometries usually preserve distances, which won't happen if you change the constant $a$ ($\langle x,y \rangle_a = xay \ne xby = \langle x,y \rangle_b$ unless $a = b$). However, any two of these inner products will have a nice linear bijection between them. – 2017-01-11
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0@Adam: of course all inner product spaces on $\mathbb{R}$ are isometric. Just take the map that sends $x$ to $x/\sqrt{a}$, which makes sense because $a>0$. – 2017-01-11