# Trouble with problem involving finding a basis for the set

Having a lot of trouble with these problems.Can someone show me how to show the set of vectors I have found spans $$U$$?

Problem Let $$U=\{p \in \mathcal{P}_4(F):p(6)=0\}$$.

(a).Find a basis of $$U$$.

Attempt:

(a). Using methods from this forum to help me I found a basis to be

$$(x-6),(x^2-36),(x^3-216),(x^4-1296)$$

The vectors are linearly independent since

$$a(x-6)+b(x^2-36)+c(x^3-216)+d(x^4-1296)=0$$

Thus $$6a-36b-216c-1296d+ax+bx^2+cx^3+dx^4=0$$

$$\implies a=b=c=d=0$$

Now I cannot figure out how to show that the set spans $$U$$ any help?

How am I supposed to know what a generic vector in $$U$$ is so I can write the generic vector as a linear combination of my basis?

• $U$ is the set of fourth-degree polynomials that have $6$ as a root. Can you partially factor such a polynomial? – amd Dec 22 '19 at 23:50
• ok $(x-6)(ax^3+bx^2+cx+d)$ Is my basis still correct though? – user736276 Dec 22 '19 at 23:51
• Try to write this as a linear combination of your basis vectors. You should be able to factor $(x-6)$ out of both sides, reducing the problem to showing that the remaining factors form a basis of $\mathcal P_3$. – amd Dec 22 '19 at 23:53
• Can you not use the fact that each 'vector' is linearly independent and since each vector is not the zero vector, by definition the set spans $U$? – Ty Jensen Dec 22 '19 at 23:55
• How did you come up with this basis in the first place? Depending on the method, that would’ve already shown that these vectors span the space. – amd Dec 23 '19 at 0:01

Let $$p(x)=ax^4+bx^3+cx^2+dx+e \in U$$. Then $$p(6)=0 \implies 1296a+216b+36c+6d+e=0$$. Thus $$e=-(1296a+216b+36c+6d)$$. So \begin{align*} p(x) & =ax^4+bx^3+cx^2+dx-(1296a+216b+36c+6d)\\ &=a(x^4-1296)+b(x^3-216)+c(x^2-36)+d(x-6). \end{align*} Thus any $$p(x) \in U$$ is a linear combination of the vectors in your basis.
• If you use this method to show the basis spans $U$ do you still have to show that the vectors are linearly independent? – user736276 Dec 23 '19 at 0:07
$$U$$ is the set of polynomials annihilated by the “evaluate at $$6$$” functional. By a variant of the Rank-Nullity theorem, this implies that $$U$$ is four-dimensional. All four polynomials in your basis clearly vanish at $$x=6$$ and they’re obviously linearly independent, so they must be a basis of $$U$$.
Alternatively, $$U$$ is the set of polynomials of degree at most four that have $$6$$ as a root. Every such polynomial is of the form $$(x-6)(ax^3+bx^3+cx+d)$$. Factoring $$x-6$$ out of your proposed basis vectors, the problem reduces to showing that $$\{1,x+6,x^2+\cdots,x^3+\cdots\}$$ is a basis of $$\mathcal P_3(F)$$, but these four polynomials are obviously linearly independent, so you’re done. Notice that I didn’t bother to work out all of the quotients of division by $$x-6$$. It’s enough to know that the four polynomials are all of different degree to determine that they’re independent.