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I have seen a proof that the set $\{1, x, ... , x^{n-1}\}$ form a basis for $P_{n}(x)$ as given below:

To show that $\{1, x, ... , x^{n-1}\}$ is a basis for $P_{n}(x)$, let $a_{0},a_{1}, ... ,a_{n-1}$ be scalars such that $a_{0} + a_{1}x + ... + a_{n-1}x^{n-1} = 0$. Now setting $x = 0$ yields $a_{0} = 0$.

I have a doubt in the following claim:

In a similar way by factoring $x$ each time, we see that $a_{0}=a_{1}= ... =a_{n-1} = 0$.

Consider $a_{1}x + ... + a_{n-1}x^{n-1} = 0$,

$x(a_{1} + ... + a_{n-1}x^{n-2}) = 0$.

I cannot set $x = 0$ here right. Then how to prove $a_{1} = 0$.

I just want to know whether the claim is right or wrong?

Vinod
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1 Answers1

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There you obtain a product of two polynomials, and you know that the polynomial x is not the zero polynomial. That means the polynomial in parantheses must be the zero polynomial (if $pq=0$, then at least one of $p,q$ must be zero). Rinse and repeat.

Alp Uzman
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