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Q: Suppose $p_0, \dots , p_m$ are polynomeals in $P_m(F)$ s.t. $p_j(2)=0$ for all $j$. Prove that $p_0, \dots , p_m$ is not linearly independant in $P_m(F)$.

Edit: $P_m(F)$ is the vector space of all polynomials of degree at most m in a Field

My Proof

Suppose that $p_0, \dots , p_m$ is linearly independant. Now, consider the set $z, p_0, \dots , p_m$ where this family spans $P_m(F)$.

Note: $1,x, \dots , x^m$ spans $P_m(F)$ and has $m+1$ elements. So, no linearly independant set in $P_m(F)$ has more than $m+1$ elements

So, $z, p_0, \dots , p_m$ is linearly dependent (has $m+2$ elements) So, for $\beta, \alpha_j \in F$ (not all $=0$), $\beta z + \alpha_0 p_0 + \dots +\alpha_m p_m = 0$

if $\beta = 0$, then $\alpha_0 p_0 + \dots +\alpha_m p_m = 0$ and since it is assumed that $p_0, \dots , p_m$ is linearly independant, then this forces all $\alpha_j = 0$, which contradicts the fact that $z, p_0, \dots , p_m$ is linearly dependant. Hence, $\beta \neq 0$.

Then, $z = (-1/\beta)(\alpha_0 p_0 + \dots +\alpha_m p_m)$ which is in the $span\{p_0 , \dots , p_m\}$, so $z \in span\{p_0, \dots , p_m\}$. So, $\exists \alpha_j \in F$ such that $z=\alpha_0 p_0 (z) + \dots + \alpha_m p_m (z)$. But, $p_j(2)=0$ for $\forall j$, and $2 \neq 0 + \dots + 0 = 0$, so contradiction. SO, $p_0, \dots , p_m$ is not linearly independent.

Is this proof sufficient? Are there any holes or inconsistencies that I should be worried about?

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  • $\begingroup$ What is $P_m(F)$? $\endgroup$ – Lord Shark the Unknown May 10 '17 at 15:41
  • $\begingroup$ The vector space of polynomials of degree m in a Field. Thank you, I should clarify $\endgroup$ – Pascal May 10 '17 at 15:42
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    $\begingroup$ Some remarks: a) I suppose that you are in characteristic $0$. In characteristic $2$, you have $2=0$. b) If you suppose that $p_0,\cdots,p_m$ are linearly independants, as they are $m+1$ such polynomials this is a basis for $P_m(F)$. So you can write directly that $z$ (the variable) is a linear combination of the $p_j$ c) You can also write $p_j(z)=(z-2)q_j(z)$. Then the degree of the polynomials $q_j$ are $\leq m-1$, they are in $P_{m-1}(F)$. This family of $m+1$ elements of this vector space,( that is of dimension $m$) are hence linearly dependant, and this imply the same for the $p_j$. $\endgroup$ – Kelenner May 10 '17 at 16:43
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The map $\varepsilon:P_m(F)\to F$ given $\varepsilon(p)=p(2)$ is a surjective linear transformation. Therefore, its kernel has dimension $(m+1)-1=m$ and so any $m+1$ elements in the kernel must be linearly dependent.

Here is another way to express the same idea:

If $p_0, \dots , p_m$ were linearly independent, then they'd form a basis for $P_m(F)$ and every element $p \in P_m(F)$ would be a linear combination of them. But then $p(2)=0$, which is certainly not the case for the constant polynomial $p=1$.

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