Inner Product and Linear Transformation

J-Wang
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Homework Statement


Let V be a finite-dimensional real inner product space with inner product < , >.

Let L:V->R be a linear map.

Show that there exists a vector u in V such that L(x) = <x,u> for all x in V.


2. The attempt at a solution
It seems really simple but I just can't phrase it properly.

I know since L is linear, I can express as L(x+y) = L(x) + L(y)

I tried to make x and u be expressed as combination of orthogonal basis, and try to work in the fact that inner product will result in ZERO since they are orthogonal.

Any help would be greatly appreciated.
 
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Since V is a finite dimensional vector space, there exist an orthonormal basis \{v_i\} for i from 1 to n. Let a_i= L(v_i). What can you say about <x, v> with v= \sum_{i=1}^n a_iv_i?
 
Does it has something to do with the Projection of x on the basis? :D
 
Are you trying to prove the Reisz-representation theorem?

I would see how this linear functional acted on individual basis elements.
 
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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