Proving Orthonormal Basis & Uniqueness of Inner Product | Linear Algebra HW Help

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Homework Statement


Prove that any basis of R^n is an orthonormal basis with respect to some inner product. Is the inner product uniquely determined?


Homework Equations


I am not sure where to begin. Should I just define an arbitrary basis for a arbitrary R^n? I mean I think I understand the question about the inner product being uniquely determined but I am not sure where to begin.


The Attempt at a Solution


See above.
 
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How do you define inner products in R^n? A familiar question: how and when do symmetric matrices induce inner products on R^n?

Uniquely determined means that there is no other inner product that has those properties. That is, given a basis there is only one inner product that makes the basis an orthonormal set. If you don't know what I meant by symmetric matrices you can just play around with scaling inner products by positive reals.
 
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Do you mean in (x^T)Kx or in notation <x,x> or in formula? I'm not going to lie I'm a bit confused with what you're asking.
 
Do you mean by bilinearity, symmetry, and positivity?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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