Show that $\{x^4+x^3-1,x^3-x,x^2+1\}⊂E$ is a generating set of span($E$).

Question

I'm given the following set of polynomials:

$$E:=\{x^4+x^2+x,x^4+x^3-1,x^3-x,x^2+1\}$$

I know that $E$ is linearly dependent because when $\alpha_1 = -1$, $\alpha_2=1$, $\alpha_3=-1$ and $\alpha_4=1$ then $E=0$. Now I want to show that $\{x^4+x^3-1,x^3-x,x^2+1\}⊂E$ is a generating set of the span($E$) but I'm not sure where to start. Any hints will be helpful!

Answer 1

It suffices to show that the remaining element of $E$ can be written as a linear combination of $x^4+x^3-1$, $x^3-x$ and $x^2+1$. Indeed we have that:

$$ x^4+x^2+x = (x^4+x^3-1)-(x^3-x)+(x^2+1). $$

Then, by the definition of the spanning set we have that: $$ span \{x^4+x^3-1,x^3-x,x^2+1\}=span(E). $$

Note: Of course that doesn't imply that the generating subset is necessarily linearly independent. In this case though it actually is and it can be easily proved.