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What will be the basis for the space of cubic polynomials $p$ such that $p(3) = 0$?

I know that: A natural basis for the vector space of cubic polynomials is $p(3)$ is $\langle 1, x, x^2, x^3 \rangle$.

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Look for a polynomial of degree $1$, one of a degree $2$ and one of degree $3$ such that $p(3)=0$. – marwalix Sep 23 '11 at 4:59
But to find p(3) = 0 , how will I apply the natural basis for vector space of cubic polynomials to the question? – Blake Sep 23 '11 at 5:12
Cheap route: The basis for the cubic polynomials such that $P(0)=0$ is just those without a $x^0$ term: $\{x,x^2,x^3\}$. So one can argue the basis for cubic polynomials such that $P(3)=0$ will be $\{(x-3),(x-3)^2,(x-3)^3\}$. – anon Sep 23 '11 at 5:15
I have some nitpicks wrt this question. (1.) You cannot ask for the basis since a vector space might have many (even uncountably many) bases. (2.) A basis is a set, so perhaps you can use braces to represent it, like $\{ 1, x, x^2, x^3 \}$, rather than $\langle \rangle$. (3.) You presumably mean the space of polynomials of degree at most $3$. The space of polynomials of degree exactly $3$ is not a linear space. (I am not sure what the term "cubic" conventionally means.) My apologies if this feels like empty nitpicking, but I think it's good to be correct :-). – Srivatsan Sep 23 '11 at 5:21
@Blake You really need to be speaking of the vector space of polynomials of degree less than or equal to 3. – user38268 Sep 23 '11 at 5:59

One way to arrive at the basis proposed by André is the following: if $p(x)= ax^3 + bx^2 +cx +d$ has $3$ as a root, then

$$ 0 = p(3) = 27a + 9b + 3c + d \ \qquad \Longrightarrow \qquad \ d = -27a - 9b -3c \ . $$

This means that for polynomials of degree less than or equal to $3$ having $3$ as a root, you can choose, for instance, coefficients $a,b,c$ at will, but then you have no other choice for $d$ than $-27a-9b-3c$. So these polynomials look like:

$$ p(x)= ax^3 + bx^2 +cx -27a-9b-3c = a(x^3-27) + b(x^2 - 9) - c(x-3) \ . $$

Remember that you can choose any $a,b,c$? Hence, this last expression means that all polynomials having degree less than or equal to $3$ and $3$ as a root are linear combinations of André's basis:

$$ x^3-27,\ x^2 - 9,\ x-3 \ . $$

So, these ones generate all your polynomials and now you just need to prove that they're linearly independent.

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We want to find a basis for the vector space $V$ of all polynomials $p(x)$ of degree less than or equal to $3$ such that $p(3)=0$. (I will assume that the vector addition is ordinary addition of polynomials, and that the multiplication by scalars is the usual one. Maybe you are expected to prove that $V$ really is a vector space. I leave that part, if it is needed, to you.)

It is easy to see that $V$ is a proper subspace of the space of all polynomials $p(x)$ of degree less than or equal to $3$ (after all, $1$ is not in $V$). In particular, the dimension of $V$ is less than $4$.

We will show that $$\{x-3, x^2-9, x^3-27\}$$ is a basis of $V$.

It is easy to verify that the vectors in our proposed basis are linearly independent (that part is left to you). They all vanish at $x=3$. It follows that any linear combination $a(x-3)+b(x^2-9)+c(x^3-27)$ of them must also vanish at $3$.

There are $3$ vectors in our proposed basis. Thus they generate a subspace of $V$ of dimension $3$. As we observed earlier, $V$ has dimension less than $4$. This shows that our $3$ vectors generate all of $V$.

There are many other choices of basis: any $3$ linearly independent polynomials of degree $\le 3$ that vanish at $3$ will do the job.

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I was asked the question about cubic polynomials specifically. – Blake Sep 23 '11 at 5:31
@Blake Cubic polynomials (i.e., those with degree exactly 3) do not form a linear subspace because they do not have an additive identity (aka zero vector or zero polynomial). The same is true for cubic polynomials such that $p(3)=0$. So asking for a basis becomes meaningless. – Srivatsan Sep 23 '11 at 5:35
You may have been, but the wording used was sloppy. The book or instructor meant polynomials of degree less than or equal to $3$. Technically, a cubic polynomial is a polynomial of shape $ax^3+bx^2+cx+d$ where $a \ne 0$. The cubic polynomials (also the cubic polynomials that vanish at $3$) do not form a vector space. So you can't possibly have been asked to find a basis! (Such sloppy language is common. For example, the basis element $x^2$ that you mentioned in your post is certainly not a cubic polynomial. Don't worry about it, it was degree $\le 3$ that was intended. I am just fussy.) – André Nicolas Sep 23 '11 at 5:41

Take a basis of the space of degree at most $2$ polynomials, and multiply it by $x-3$.

More precisely, a tuple $(p_1,p_2,\dots)$ of polynomials is a basis that fits the bill, if and only if each $p_i$ is divisible by $x-3$, and the tuple of quotients is a basis of the space of degree at most $2$ polynomials.

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The subset of $\mathbb{R}[X]$, $\mathbb{R}_3[X]=\{P\in \mathbb{R}[X], deg(P)\leq 3\}$ is a vectorial subspace of dimension 4 a basis of which is $1$, $x$, $x^2$, $x^3$. $\phi:\mathbb{R}_3[X]\rightarrow \mathbb{R}_3[X], \phi(P)= P(3)$ is a linear form on this subspace. The set we are considering is its kernel and therefore is a subspace of dimension $4-1=3$. The set $\{x-3,(x-3)^2,(x-3)^3\}$ is a free family of three polynomials such that $\phi(P)=0$ and thus the basis we are looking for.

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