# Linear independence of polynomials from Friedberg

In Page 38 Example 4, in Friedberg's Linear Algebra:

For $k = 0, 1, \ldots, n$ let $$p_k = x^k + x^{k+1} +\cdots + x^n.$$ The set $\{p_0(x), p_1(x), \ldots , p_n(x)\}$ is linearly independent in $P_n(F)$. For if $$a_0 p_0(x) + \cdots a_n p_n(x) = 0$$ for some scalars $a_0, a_1, \ldots, a_n$ then $$a_0 + (a_0 + a_1)x +(a_0+a_1+a_2)x^2+\cdots+ (a_0+\cdots+a_n)x^n = 0.$$

My question is how he arrives at $$a_0 + (a_0 + a_1)x +(a_0+a_1+a_2)x^2+\cdots+ (a_0+\cdots+a_n)x^n = 0.$$ Why is it $(a_0+....+a_n)x^n$ instead of $(a_n)x^n$ since it was $a_n p_n(x)$? It should be $(a_0+\cdots+a_n)x^0$ since $x^0$ is common to each polynomial and then $(a_1+\cdots+ a_n) x$, no?

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## 2 Answers

You have $$p_k = x^k + x^{k + 1} + \cdots + x^n$$ so for $k > 0$ the term $x^0$ does not appear. But $x^n$ always appears. So it's $x^n$ that is common to all of them, not $x^0$.

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Thanks! I somehow missed that. – pati Feb 28 '13 at 7:36

You are taking $n$ as variable, but the variable index is $k$. If you write it out, varying $k$, only $p_0$ has the term $1$ (so only $a_0$ is multiplied by $1$), but all $p_i$ have the term $x^n$. I think if you look at your equations carefully again, you'll see it yourself (visually: in your argument, you make the polynomial shorter from the right; but it is made shorter from the left, when counting from $0$ to $n$).

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Yup got it. thanks! – pati Feb 28 '13 at 7:36