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Let $p_0, p_1, \ldots, p_m $ polynomials in $\mathbb{P}_m$ with the property that $p_j(1)=0\ \forall j$. Show that the polynomial $p_0, p_1, \ldots,p_m $ are linearly dependent.

My approach for this problem is the following.

Consider $$ \sum_{i=0}^m\alpha_ip_i(x)=0\ \ \forall x. \tag{1} $$

Put $x=1 $ in $(1)$ $ \implies \sum_{i=0}^m\alpha_ip_i(1)=0 \implies \alpha_i $ can be nonzero also. Hence the given set of polynomial is linearly dependent.

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  • $\begingroup$ What is $\mathbb{P}_{m}$? $\endgroup$ – preferred_anon Sep 23 '15 at 15:57
  • $\begingroup$ Nice problem (+1) but also a nice exercise is: Is the converse of this theorem true as well? $\endgroup$ – imranfat Sep 23 '15 at 15:57
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    $\begingroup$ Unfortunately, you haven't shown anything - you've just shown that there's one value of $x$ for which non-zero $\alpha_i$ produce a zero result. You need to show the result for all $x$, though - that is, you need to show that the relation holds as an equation in polynomials. Plugging in any particular value of $x$ by itself won't be enough to do this. $\endgroup$ – Steven Stadnicki Sep 23 '15 at 16:00
  • $\begingroup$ @Steven Stadnicki So, how will I show that the set is LD. $\endgroup$ – Sachchidanand Prasad Sep 23 '15 at 16:01
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Your proof is wrong : you are asked to exhibit concrete $\alpha_j$, and instead of that you assume that the $\alpha_j$ exist. So you’re basically reasoning backwards.

For a correct solution, notice that there are polynomials $q_0,\ldots,q_m$ in ${\mathbb P}_{m-1}$ such that $p_j=(x-1)q_j$ for every $j$. Since the dimension of ${\mathbb P}_{m-1}$ is $m$, the $q_j$ must be linearly dependent and hence the $p_j$ also.

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You have $m+1$ polynomials and you know that $1$ is already a root. So $$P_j(x)=(x-1)P'_j(x)$$ for each $j$ and $P'_j(x)$ is a polynomial of degree less or equal to $m-1$. Now there are $m +1$ of them and dimension can be maximum $m$.

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