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 $U$.


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


The vectors are linearly independent since


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?

  • $\begingroup$ $U$ is the set of fourth-degree polynomials that have $6$ as a root. Can you partially factor such a polynomial? $\endgroup$ – amd Dec 22 '19 at 23:50
  • $\begingroup$ ok $(x-6)(ax^3+bx^2+cx+d)$ Is my basis still correct though? $\endgroup$ – user736276 Dec 22 '19 at 23:51
  • 1
    $\begingroup$ 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 $\mathcal P_3$. $\endgroup$ – amd Dec 22 '19 at 23:53
  • $\begingroup$ 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 $U$? $\endgroup$ – Ty Jensen Dec 22 '19 at 23:55
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ If you use this method to show the basis spans $U$ do you still have to show that the vectors are linearly independent? $\endgroup$ – user736276 Dec 23 '19 at 0:07
  • $\begingroup$ @68e1515 Technically yes, one should still show that the polynomials used are linearly independent. Howevere because of the distinct degrees of the polynomial it's fairly easy to deduce that. $\endgroup$ – Anurag A Dec 23 '19 at 0:27

Depending on the method that you used to generate this basis, you should already know that these vectors span it. However, here are a couple of ways to verify this.

$U$ 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.


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