$T:P_n(F) \rightarrow F$ PROOF OUTLINE

Question

I'd like some heavy critique if you don't mind. See here for more details.

Let $S=\{f \in P_n(F) : f(1)=0\}$. Clearly, the polynomial $f(x)=0 \in S$ because $f(c)=0$ for any choice of $c\in F$. To demonstrate closure under addition and multiplication consider the fact that $cf(1)+g(1)=c\cdot 0+0=(cf+g)(1)=0$ for $f,g\in S$. To find dimension of $P_n(F)$, consider the linear transformation $T:P_n \rightarrow F$ defined by $T(f(x))=f(1)$, namely evaluating a polynomial at $1$ is a linear transformation from $P_n$ to $F$: \begin{eqnarray} T(f(x)+cg(x))=T(f(x))+cT(g(x))=f(1)+cg(1) \end{eqnarray} Thus, $S$ is the null space of $T$, and from Lemma 1 and Lemma 2 (don't worry about these) we see that \begin{eqnarray} \dim R(T) + \dim N(T) = \dim P_n(F) \leadsto 1 + \dim N(T) = n+1 \leadsto \dim N(T) = n \end{eqnarray}

We all are guessing that $,P_n(F),$ is the linear (vector) space of all the polynomials of degree not more than $,n,$ over some field $,F,$ ...Yet I find it odd that in two cases to show linearity you choose to show $,cf+g;,;;c\in F;,;f,g\in P_n(F);$ , instead of $,cf+dg,$...why? — DonAntonio, Apr 18 '13 at 02:23
@DonAntonio I was just using the general idea expounded on page 35 of Hoffmam's "Linear Algebra," namely Theorem 1 on that page. — Trancot, Apr 18 '13 at 03:45
Indeed so, @Trancot. Never saw it that way and it looks odd to me. — DonAntonio, Apr 18 '13 at 03:56
So, @DonAntonio, what's your story? You don't have to answer, just curious. — Trancot, Apr 18 '13 at 03:59
Don't worry, I won't, in particular because I didn't understand your question. — DonAntonio, Apr 18 '13 at 04:03

score 1 · Answer 1 · answered Jun 15 '13 at 13:54

Yes, your proof is correct. But the first part is superfluous. Define $T: P_n(F)\to F$ by $f\mapsto f(1)$. In your question you checked that $T$ is linear. Then as you noted, $S$ is the null space (sometimes also called kernel) of $T$. As the null space of a linear map it is thus a subspace of $P_n(F)$.

score 0 · Answer 2 · answered Apr 18 '13 at 17:48

0

Yes, your proof is correct. Alternatively, note that $\{1 - x, 1 - x^2, \dots , 1 - x^n\} \subseteq S$ is linearly independent. Since it's clear that $S \ne P_n(F)$, it follows that $\dim S = n$.

answered Apr 18 '13 at 17:48

Ink

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$T:P_n(F) \rightarrow F$ PROOF OUTLINE

2 Answers2