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Suppose that $p_0, p_1, ..., p_m$ are polynomials in $P_m(F)$ such that $p_i(2)=0$ for each $i$. Prove that the set of vectors $p_0, ..., p_m$ is linearly dependent.

Note that $P_m(F)$ denotes the vector space of all polynomials of degree less than or equal to $m$.

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  • $\begingroup$ Are the $m+1$ polynomials $0,0,0,\ldots,0$ linearly independent? $\endgroup$ – Julien Feb 18 '13 at 20:29
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Suppose that $p_0, p_1, ..., p_m$ are polynomials in $P_m(F)$ such that $p_i(2)=0$ for each $i$. Prove that the set of vectors $p_0, ..., p_m$ is linearly dependent.

If each $p_i(t)$ evaluates to $0$ at $t = 2$, why can you conclude that the $p_i$ must therefore be linearly dependent?:

For intuition: Try to construct linearly independent polynomials, say for $P_2(F)$, such that for each, $p_i(2) = 0$; and see why these polynomials cannot, in fact, be linearly independent.


Suppose there are $m+1$ linearly independent polynomials such that $p_i(2) = 0 \forall p_i$. Then $p_0 = 0$. Hence, e.g., 1 would necessarily be linearly independent of the $m+1$ $p_i$, also in $P_m(F)$. That would give a basis of $P_m(F)$ that has $m+2$ linear independent vectors. Impossible!

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  • $\begingroup$ You miss one $0$ to strictly fit the assumptions. Not that it matters much... $\endgroup$ – Julien Feb 18 '13 at 20:35
  • $\begingroup$ I think I corrected now...thanks, @julien $\endgroup$ – Namaste Feb 18 '13 at 20:39
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If $p_0,\dots,p_m$ were linearly independent and $p_i(2)=0$ for each $i$, then also $p_0,\dots,p_m,1$ would be linearly independent (see below for details). Hence you would have a base of $P_m$ with $m+2$ elements...

some more detail

Suppose: $$ \lambda_0 p_0 + \dots + \lambda_m p_m + \lambda_{m+1} 1 = 0 $$ then evaluating each polinomial in $t=2$ we get $$ 0 + \dots + 0 + \lambda_{m+1} = 0 $$ hence $\lambda_{m+1}=0$. So $$ \lambda_0 p_0 + \dots + \lambda_m p_m = 0 $$ which, by contradiction, we suppose that implies $\lambda_0 = 0, \dots, \lambda_m = 0$. Hence if $p_0,\dots,p_m$ are linearly independent also $p_0, \dots, p_m, 1$ would be linearly independent.

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  • $\begingroup$ @julien: I don't understand your comment... I'm suggesting a proof by contradiction. To prove that $p_0,...,p_m,1$ are linearly independent you need to know that $p_i(2)=0$ for all $i$. $\endgroup$ – Emanuele Paolini Feb 18 '13 at 21:09
  • $\begingroup$ @julien: $p_0,...,p_m$ are supposed to satisfy the condition $p_i(2)=0$ see the details I have added. $\endgroup$ – Emanuele Paolini Feb 18 '13 at 21:15
  • $\begingroup$ Ah! Thanks for the clarification. Fun approach. $\endgroup$ – Julien Feb 18 '13 at 21:17

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