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Let $V$ and $W$ be isomorphic vector spaces. "If $\langle\vec u, \vec v\rangle$ is an inner product in $V$, then it is also an inner product in $W$". Does such a relationship exist?

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    $\begingroup$ It can't, unless $V = W$. Inner products on $V$ and on $W$ have different domains if $V\neq W$. $\endgroup$ – Daniel Fischer Aug 18 '15 at 13:36
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    $\begingroup$ @DanielFischer is right. I assume you meant does $V$'s inner product push forward to an inner product in $W$? $\endgroup$ – Zach Stone Aug 18 '15 at 13:39
  • $\begingroup$ @ZachStone I assume inner product returns a real number. For example dot product in $\mathbb R^3$ applied to second degree polynomials. $\endgroup$ – user137035 Aug 18 '15 at 13:45
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    $\begingroup$ Proper notation is $\langle\vec u, \vec v\rangle$, not $<\vec u, \vec v>$. I edited accordingly. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 18 '15 at 13:49
  • $\begingroup$ @DanielFischer This means that we can't apply dot product in $\mathbb R^3$ to the vector space of all second degree polynomials, right? Such an operation would be similar. $\endgroup$ – user137035 Aug 18 '15 at 13:50
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You can't apply the inner-product on $V$ to elements of $W$ without explicitly providing some sort of identification between $V$ and $W$ - that is, a linear isomorphism. If $(V, \left< \cdot, \cdot \right>_V)$ is an inner-product space over $\mathbb{F}$, and $W$ is a vector space over $\mathbb{F}$, and you are given an isomorphism $\phi \colon V \rightarrow W$, you can use the isomorphism to define an inner-product on $W$ by the formula:

$$ \left< w_1, w_1 \right>_W := \left< \phi^{-1}(w_1), \phi^{-1}(w_2) \right>_V. $$

The inner product $\left< \cdot, \cdot \right>_W$ is called the pushfoward of $\left< \cdot, \cdot \right>_V$ using $\phi$ and depends strongly on $\phi$. Different $\phi$'s will result in different inner products on $W$. You can think of $\phi$ as providing an identification of vectors in $W$ with vectors in $V$ and using this identification, you transfer the inner product on $V$ to an inner product on $W$.

For example, if $V = \mathbb{R}^n$ with the standard inner product and $W = \mathbb{R}_{n-1}[x]$ and $\phi(a_0, \ldots, a_{n-1}) = a_0 + a_1x + \ldots + a_{n-1}x^{n-1}$, then pushfoward inner-product on $W$ is

$$ \left< a_0 + \ldots + a_{n-1}x^{n-1}, b_0 + \ldots + b_{n-1}x^{n-1} \right>_W = \left< (a_0, \ldots, a_{n-1}), (b_0, \ldots, b_{n-1}) \right>_V = \sum_{i=0}^{n-1} a_i b_i. $$

Note that by definition, the map $\phi \colon (V, \left< \cdot, \cdot \right>_V) \rightarrow (W, \left< \cdot, \cdot \right>_W)$ becomes an isometry of inner-product spaces (that's how $\left< \cdot, \cdot \right>_W$ was defined in the first place!) and using this map, you can translate questions about inner-products of vectors in $W$ to questions about inner-products of vectors in $V$ and vice-versa.

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  • $\begingroup$ I didn't know pushforward, but this seems what I have been thinking. Thank you for the answer. $\endgroup$ – user137035 Aug 18 '15 at 14:30
  • $\begingroup$ "pushfoward" is just a name (because you take an inner product on $V$ and using $\phi \colon V \rightarrow W$, push it foward to an inner product on $W$) for the construction I described in the answer. You're welcome! $\endgroup$ – levap Aug 18 '15 at 14:31

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