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Let $V = \mathbb{R}[x]_n$ be the real vector space of all real polynomials of degree $\leq n$.

$$\alpha(f,g,) = \int^1_0 xf(x)g(x)\,dx$$

I need to prove that this is a symmetric bilinear form on $V$. Could someone please point me in the right direction?

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So what is a symmetric bilinear form? Does $\alpha$ satisfy the axioms of a symmetric bilinear form? –  Vectk Nov 29 '12 at 8:30
    
if it's a symmetric matrix then it is symmetric? but I don't know how to prove that –  user51130 Nov 29 '12 at 8:35
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First, show that $\alpha$ is a bilinear form. Then, show that for any $f,g \in \mathbb{R}[x]_n$, $\alpha(f, g) = \alpha(g,f)$. –  Vectk Nov 29 '12 at 8:40
    
I don't know how to prove that it is bilinear. I know that it has to satisfy conditions such as B(u + v, w) = B(u, w) + B(v, w) but I don't know how to apply that to this case –  user51130 Nov 29 '12 at 8:42
    
Here's a start. For any $f,h,g \in \mathbb{R}[x]_n$, $\alpha(f + h, g) = \int^1_0 x(f(x) + h(x))g(x)dx = \int^1_0 (xf(x)g(x) + xh(x)g(x))dx$. Can you go on from there? –  Vectk Nov 29 '12 at 8:51
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