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Let $f(t)$ be a polynomial composed of linear factors $(t-a_i)$ for $i \in [n]$, i.e $f(t) = (t-a_1)\cdots (t-a_n) $. Let $g_k(t)$ be given by: $$g_k(t) = \frac{f(t)}{(t-a_k)}$$ for $k \in[n]$. Prove that the set of polynomials $\{g_k(t) \mid 1 \leq k \leq n \}$ is linearly independent.

Forgot to specify, but each $a_i$ is distinct.

I am not sure how to proceed here. I was thinking of using induction but am not entirely sure how to work in the inductive hypothesis.

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  • $\begingroup$ Are we to assume each $a_i$ is unique? This is trivially not true otherwise... $\endgroup$ – rnrstopstraffic Jun 26 '17 at 3:19
  • $\begingroup$ @mrstopstraffic Yes I just specified that. Sorry for the confusion. $\endgroup$ – rubikscube09 Jun 26 '17 at 3:20
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Suppose there are constants $b_1,\ldots,b_n$, not all zero, such that $$b_1g_1(t)+\cdots+b_ng_n(t)=0$$ for all $t$. But if $b_k\neq0$, then $$b_k\prod_{i\neq k}(a_k-a_i)=b_kg_k(a_k)=b_1g_1(a_k)+\cdots+b_ng_n(a_k)=0,$$ and so $a_i=a_k$ for some $i\neq k$. Thus, if all the $a_i$ are distinct, then $g_1,\ldots,g_n$ are linearly independent.

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  • $\begingroup$ Why is your second equality true? Why is $b_k \prod_{i \neq k} (a_k-a_i)$ equal to $\sum_i b_ig_i(a_k)$? $\endgroup$ – Nick Jun 26 '17 at 3:23
  • $\begingroup$ I suppose the ordering was weird. We have $b_kg_k(a_k)=b_1g_1(a_k)+\cdots+b_ng_n(a_k)$ because all other summands are zero. $\endgroup$ – Aweygan Jun 26 '17 at 3:25
  • $\begingroup$ I see now. Thanks! $\endgroup$ – Nick Jun 26 '17 at 3:26
  • $\begingroup$ You're welcome. Glad to help! $\endgroup$ – Aweygan Jun 26 '17 at 3:27
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Consider the linear map from your space of polynomials to $\Bbb R^n$, defined by the fact that component $i$ of the result is given by evaluating your polynomial in $x=a_i$. Then the image of $g_j$ is a vector with exactly one nonzero coordinate, which is in position $j$. These images are then clearly linearly independent, which implies that the $g_j$ are so as well (since linear maps preserve linear dependence relations).

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