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Show that $P_n(F)$ is generated by $\{1, x, ... x^n \}$.

My Workings

$S = \{1, x, ..., x^n\}$

$W = \{a_nw^n + a_{n - 1}w^{n - 1} + ... + a_0 : a \in F\}$

Let $x \in Span(S)$.

$\therefore x = c_1x^0 + c_2x^1 + ... + c_nx^n$ for some scalars $c_1, c_2, ..., c_n$ in $F$.

We have $S \subseteq W$ where $W$ is a subspace of $P_n(F)$.

$\therefore$ There are $w \in W$ such that $w = c_1x^0 + c_2x^1 + ... + c_nx^n$.

$\therefore Span(S) \subseteq W$

I have two questions:

  1. Is my solution correct? If not, what is the error and what is the correct solution?
  2. Instead of $Span(S) \subseteq W$, I have also seen my textbook describe this as $Span(S) = W$. I am having a hard time understanding how these two are interchangeable; indeed, it would seem like they should not be, since $S$ is only a subset of $W$, which means that $W$ could contain other elements and, therefore, linear combinations that are NOT spanned by $S$. So, again, how does this make sense?

I would greatly appreciate it if people could please take the time to answer these two questions.

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    Do not to denote at the same time the variable of your polynomials *and* a general element of the span of $S$. – 2017-02-14
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    @MarianoSuárez-Álvarez Thanks for the response. Are you referring to the $x$ in $S$ and the $x$ in $W$? I should use different variables? – 2017-02-14
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    If they are referring to different things, then using the same letter to denote both is a bad idea, you surely agree? – 2017-02-14
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    @MarianoSuárez-Álvarez Yes, you are absolutely correct. I should have known better. I will edit it now for clarity. Thank you for the assistance. – 2017-02-14
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    @MarianoSuárez-Álvarez Is that better? – 2017-02-14
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    You should say "$x = c_1x^0 + c_2x^1 + ... + c_nx^n$ for some scalars $c_1, c_2, ..., c_n$ in $F$". Omit statement about vectors in that line. This is because $0 \in F$ so we can decide which $x^i$'s to have and which ones not to have. – 2017-02-14
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    @VikrantDesai You mean we can decide by selecting certain scalars $c_1, c_2, ..., c_n$ to be equal to $0$, which then makes the relevant vector also 0? – 2017-02-14
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    @ThePointer Absolutely Yes. – 2017-02-14
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    @VikrantDesai Interesting. Thank you for clarifying this. I will edit my post. – 2017-02-14
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    @ThePointer You can't say "Let $S \subset W$". Instead, say "We have $S \subset W$. – 2017-02-14
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    @VikrantDesai you're right. I will update the OP. Thank you again for the assistance. – 2017-02-14
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53605/discussion-between-vikrant-desai-and-the-pointer). – 2017-02-14

1 Answers 1

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Vikrant Desai clarified my misunderstanding:

We have established that $Span(S) \subseteq W$.

We also know that $W = \{a_nw^n + a_{n - 1}w^{n - 1} + ... + a_0 : a \in F\}$ and $x = c_1x^0 + c_2x^1 + ... + c_nx^n$ for some scalars $c_1, c_2, ..., c_n$ in $F$ and $x \in span(S)$.

$\therefore$ We also have $W \subseteq span(S)$.

Since $Span(S) \subseteq W$ and $W \subseteq span(S)$, $W = span(S)$