# Set of Polynomials divided by single linear factor is linearly independent

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.

• Are we to assume each $a_i$ is unique? This is trivially not true otherwise... – rnrstopstraffic Jun 26 '17 at 3:19
• @mrstopstraffic Yes I just specified that. Sorry for the confusion. – rubikscube09 Jun 26 '17 at 3:20

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.
• Why is your second equality true? Why is $b_k \prod_{i \neq k} (a_k-a_i)$ equal to $\sum_i b_ig_i(a_k)$? – Nick Jun 26 '17 at 3:23
• 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. – Aweygan Jun 26 '17 at 3:25
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).