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$$V = \{ f \in P_n(F) \: : \: f(a) = 0 \} .$$

I'm trying to find a basis for the vector space V but I have no idea how to approach it. At first I thought it might be $$(0,x-x,x^2-x^2,...,x^n-x^n)$$ But this subset isn't linearly independent, since $$x-x = x^2 - x^2$$ Come to think of it, the basis can't possibly be more than one element: the zero polynomial, since all the polynomials evaluate to zero. I'm probably just misunderstanding this somehow.

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    The given functions $x^n - x^n$ vanish everywhere, whereas you need functions which vanish at $a$, not everywhere, right? Therefore, consider $x^n-a^n$ instead. Or, look at it this way: If a polynomial $p$ vanishes at $a$, then $p(x) = (x-a)q(x)$ for some polynomial $q$. Hence, another suitable choice could be $(x-a), x(x-a), x^2(x-a)$ etc. – 2017-02-22
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    shouldn't $f(a) = 0$ for all a? because for $f(x) = x^n - a^n$, $f(x) \not= 0$ for all $x \not= a$. – 2017-02-22
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    One second: Your $a$ is not a fixed quantity? *If it is for all values of $a$*, then you are correct. Otherwise,*if it is for one value of $a$*, then you can take the hint I have given. You should clarify this, but my feeling is that $a$ is a fixed number (like $1,\pi,0.2$ etc). However, you could be correct, but this should be clarified independent of the question. – 2017-02-22
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    The exact wording is: "Let $a$ be an element of $F$". Does that mean the same basis has to be true for any arbitrary element $a$ in $F$? – 2017-02-22
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    Aha, so $a$ is *one* element of $F$. Depending upon $a$, the basis will change. For example, if $a=1$ , then the basis will be different from $a=2$. The basis is definitely dependent upon the choice of $a$. It's good you clarified this, thank you for that. (Also $+1$ for active participation in your own question). – 2017-02-22
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    That makes a lot sense now! Thanks. – 2017-02-22
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    You are welcome! By the way, either verify (or get verified) the fact that the elements I have given indeed do form a basis of the given vector space. – 2017-02-22
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    One question though, is $(x−a),x(x−a),x^2(x−a)...x^n(x-a)$ linearly independent? isn't $(x-a) = c_1(x-a)$ for $c_1 = x,x^1,x^2...x^n$ – 2017-02-22
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    Good question. $c_1$ has to be a number, not another polynomial. You cannot write $c_1(x-a) = x^n(x-a)$ for any value of $n$ and any constant $c$. This can be easily shown. – 2017-02-22

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Essentially, you're looking for a basis for the vector space of all $n^{\text{th}}$ order polynomials over $\mathbb{F}$ which have a zero at $a$. Consider a general $g\in P_n(\mathbb{F})$ written $$ g(x)=b_0x^n+b_1x^{n-1}+b_2x^{n-2}+\cdots+b_{n-2}x^2+b_{n-1}x+b_n. $$ If $g\in V$, then $$ g(a)=b_0a^n+b_1a^{n-1}+\cdots+b_{n-1}a+b_n=0. $$ Notice that $$ b_n=-b_0a^n-b_1a^{n-1}-\cdots-b_{n-1}a. $$ You're getting some information on the basis for this space now. We know that $b_n$ will just be a linear combination of the other coefficients, so $V$ does not require a constant in its basis. $V$ has at most $n$ elements in its basis then, instead of the $n+1$ basis elements that general vector spaces of $n^{\text{th}}$ order polynomials have. Now think about factoring $(x-a)$ from $g(x)$. This will hopefully lead you to writing down the basis for $V$.

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    But aren't the other terms also linear combination of other terms? $b_0a^n = -b_1a^{n-1}+...+b_{n-1}a+b_n$ – 2017-02-22
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    I think you're misunderstanding the question. A concrete example would be $V=\{f\in P_2(\mathbb{R}):f(2)=0\}$. Here you're looking for polynomials of at most order $2$ with a zero at $2$. This could be, for example, $(x-2)^2$, $x^2-4$, or $x-2$, etc. In this case, you want to find a basis which represents every possible polynomial of at most order 2 which has a zero at 2. – 2017-02-22