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Let $V$ be a real vector space of $\dim 2$ and let $B:=\{u_1,u_2\}$ be a basis. How do you find all the inner products that satisfy $\langle u_1,u_1 \rangle=1$ $\langle u_2,u_2 \rangle=1$?

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    $\begingroup$ It does not mean that the basis $B$ is orthonormal, that would only be the case if $\langle u_1, u_2\rangle$ were zero. $\endgroup$ – Matthew Towers Nov 13 '11 at 16:21
  • $\begingroup$ @mt_: thanks for pointing that out! $\endgroup$ – Freeman Nov 13 '11 at 16:23
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Pretty much anything will do, no? Namely, any inner product is determined by the values $\langle u_1,u_2\rangle,\langle u_1,u_1\rangle,\langle u_2,u_2\rangle$ and these values are independent. So, setting $\langle u_1,u_2\rangle=\alpha$ one sees that the inner product is given by $\langle au_1+bu_2,cu_1+du_2\rangle=ac+\alpha(ad+bu)+bd$

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  • $\begingroup$ I see what you're saying, it does seem like almost anything.. so you're then defining what happens to the basis then taking a linear combination of it to describe what happens to an arbitrary vector? $\endgroup$ – Freeman Nov 13 '11 at 16:12
  • $\begingroup$ Exactly. Without any more information, that's all that can be said. $\endgroup$ – Alex Youcis Nov 13 '11 at 16:13
  • $\begingroup$ Thank you! that certainly makes a lot of sense! :) $\endgroup$ – Freeman Nov 13 '11 at 16:21
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    $\begingroup$ An inner product is usually required to be positive definite. So you must find the values of $\alpha$ such that $\langle x,x\rangle\geq 0$ with equality if and only if $x=0$. $\endgroup$ – Matthew Towers Nov 13 '11 at 16:22
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    $\begingroup$ LHS, you are missing the point. Positive definiteness does not require $\langle u_1, u_2 \rangle \geq 0$. $\endgroup$ – Matthew Towers Nov 13 '11 at 16:47

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