0
$\begingroup$

Here is the question I am trying to solve:

Let $V$ be the collection of polynomials with coefficients in $\mathbb Q$ in the variable $x$ of degree at most 5 with $1, x, x^2, \dots , x^5$ as basis. Prove that the following are elements of the dual space of $V$ and express them as a linear combination of the dual basis.

$E : V \to \mathbb Q$ defined by $E(p(x)) = p(3).$

My thoughts:

I was able to prove that it is linear and hence in $V^*.$ but I do not know how to express it as a linear combination of the dual basis? How can I find the dual basis in this case, what are the vectors corresponding to the basis of $V$ that I should use to determine the dual basis? If I answered this question I can complete from there. Could someone clarify this to me please?

Edit:

My question is: How can I write the basis element $x$ for example as a vector, could anyone explain this to me please?

$\endgroup$
7
  • 1
    $\begingroup$ Do you substitute a specific $x$? Because $p(x)$ is a polynomial, not a rational number. $\endgroup$
    – Mark
    Dec 5, 2023 at 0:15
  • $\begingroup$ @Mark I did not get your point. $\endgroup$
    – Intuition
    Dec 5, 2023 at 0:30
  • $\begingroup$ you mean that for example $f_0(1) = 1$?@Mark $\endgroup$
    – Intuition
    Dec 5, 2023 at 0:31
  • 1
    $\begingroup$ You defined $E:V\to\mathbb{Q}$ as $E(p(x))=p(x)$. I'm asking how exactly is that a map from $V$ to $\mathbb{Q}$. $p(x)$ is a polynomial, not an element of $\mathbb{Q}$. $\endgroup$
    – Mark
    Dec 5, 2023 at 0:34
  • $\begingroup$ oh sorry I will edit my question @Mark $\endgroup$
    – Intuition
    Dec 5, 2023 at 0:57

1 Answer 1

2
$\begingroup$

By definition, if your basis is $1,x,x^2,...,x^5$ then its dual basis of $V^*$ consists of the linear functionals $\varphi_0,\varphi_1,...,\varphi_5$ that satisfy:

$\varphi_i(x^j)=\begin{cases} 1 & i=j\\ 0 & i\ne j\\ \end{cases}$

In more detail, the map $\varphi_i:V\to\mathbb{Q}$ (for $0\leq i\leq 5$) is defined by $\varphi_i(\sum\limits_{j=0}^5 a_jx^j)=a_i$, i.e it gives us the $i$-th coefficient in the expansion of the vector as a linear combination of our basis of $V$.

Note: To this point it was the explanation of what is the dual basis. The next part is about solving the problem. If you want to try alone, stop reading here.

Let $f(x)=\sum\limits_{j=0}^5 a_jx^j$ be an element of $V$. Then $\varphi_i(f)=a_i$ for each $i$, and $E(f)=\sum\limits_{j=0}^5a_j3^j$. Therefore:

$E(f)=\sum\limits_{j=0}^5 3^j\varphi_j(f).$

This is true for any $f\in V$, and hence as functions $V\to\mathbb{Q}$ we have $E=\sum\limits_{j=0}^53^j\varphi_j.$ This is the required linear combination.

$\endgroup$
1
  • $\begingroup$ Downvoter, care to explain the reasons? $\endgroup$
    – Mark
    Dec 10, 2023 at 22:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .