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 $\langle u_1,u_1 \rangle=1$ $\langle u_2,u_2 \rangle=1$?
How to find all the inner products that satisfy $\langle u_1,u_1 \rangle=1$ $\langle u_2,u_2 \rangle=1$? where $B:=\{u_1,u_2\}$ is a basis
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linear-algebra
inner-product-space
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1It does not mean that the basis $B$ is orthonormal, that would only be the case if $\langle u_1, u_2\rangle$ were zero. – 2011-11-13
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0@mt_: thanks for pointing that out! – 2011-11-13
1 Answers
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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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0I 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? – 2011-11-13
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0Exactly. Without any more information, that's all that can be said. – 2011-11-13
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0Thank you! that certainly makes a lot of sense! :) – 2011-11-13
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0An 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 $x=0$. – 2011-11-13
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0Indeed, so $\alpha>0$ as $u_1,u_2 \neq0$ – 2011-11-13
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0And moreover, any $\alpha$ >=1 will satisfy the condition. And, on the contrary, if less than(<)1, then it cannot be defined. So it corresponds to the pre-image of the logarithmic function of the group of positive real numbers with multiplication.(This is for fun only, which does not imply something deep deliberately.) – 2011-11-13
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0LHS, you are missing the point. Positive definiteness does *not* require $\langle u_1, u_2 \rangle \geq 0$. – 2011-11-13