## Question about proving polynomial basis linear independence. |
davidLN |
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Say I have a basis {x^2 - 1, x^2 + x, 5} If I set up ax^2 + bx + c = d(x^2 - 1) + e(x^2) + f(5) and I solve for a, b, and c. If a, b, and c are not zero, does that mean I have proved that the components are linearly independent? |

yulduzkhonbruin: May 5, 2015, 6:52 p.m. |
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Actually, I think if you prove that d, e, and f are zero while a=b=c=0 which means you found only one trivial way/solution that is zero. In that case, it is linearly independent. In other words, when you solve equations d+e=0 (from (d+e)x^2), and 5f-d=0 while e=0, you get e and f equal to zero. That means it has only trivial solution, that is why it is linearly independent. |

Noeland: May 7, 2015, 10:18 a.m. |
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a polynomial $c_{0}x^{n}+c_{1}x^{n-1}+c_{2}x^{n-2}....c_{n-1}x+c_{n}$ with real coefficient could not have more than n roots, thus if $c_{0}x^{n}+c_{1}x^{n-1}+c_{2}x^{n-2}....c_{n-1}x+c_{n}=0$, all the coefficient must be zero. |