Finding a Set Which Spans a Given Set

Question

Write down a finite set X of elements of V which spans W.

W = $\{p \in \Bbb R_3[x] |$ the sum of the coefficients of p is zero$\}$, $V =\Bbb R_3[x]$

I don't even know where to begin on this one. I've looked around the forum and every similar question references a bunch of stuff I don't know. For reference, this is my first time taking Linear Algebra and doing any sort of work with sets and spaces. I've done the following work:

$p_1 = a_0 + a_1x + a_2x^2 + a_3x^3, a_0 + a_1 + a_2 +a_3 = 0$

That's just one element of the set. Any help would be appreciated, thanks!

Answer 1

The condition given for $p=ax^3+bx^2+cx+d\in W$ is

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

then

$$p=ax^3+bx^2+cx-a-b-c=a(x^3-1)+b(x^2-1)+c(x-1)$$

thus a basis for W is given by $$X=\{x-1,x^2-1,x^3-1\}$$

Answer 2

You can take $X=\{x-1,x^2-1,x^3-1\}$. Obviously, $X\subset W$, and therefore $\langle X\rangle\subset W$. Now, prove that any element of $W$ is a linear combination of elements of $X$ and you're done.

Answer 3

One way to begin:

Observe that if we define the linear functional

$$\;\phi: \Bbb R_3[x]\to\Bbb R\;,\;\;\phi(a_0+a_1x+\ldots+a_3x^3):=\sum_{k=0}^3a_k$$

then we get $\;W=\ker\phi\;$ , which automatically makes $\;W\;$ a hyperspace in $\;\Bbb R_3[x]\;$ since clearly $\;\phi\neq0\;$ , and thus $\;\dim K=\dim\Bbb R_3[x]-1=3\;$ , and you already know you're looking for $\;3\;$ linearly independent vectors in $\;W\;$.

Now, $\;\sum\limits_{k=0}^3a_k=0\implies a_0=-a_1-a_2-a_3\;$ , so you can choose:

$$\begin{cases}\text{First}\,,\;\;a_1=-1\,,\,\,a_2=a_3=0\implies a_0=1\implies 1-x\\{}\\ \text{Second}\,,\;\;a_1=a_3=0,=,a_2=-1\implies a_0=1\implies 1-x^2\\{}\\\text{Third}\;\ldots\end{cases}$$

and I'll leave for you the third one...

Caution: You must be sure you can prove that the three vectors obtained as above are linearly independent.