Prove or disprove: there exists a basis $p_0, p_1, p_2, p_3 \in P_3(F)$ such that none of the polynomials $p_0, p_1, p_2, p_3$ has degree 2

Question

Prove or disprove: there exists a basis $p_0, p_1, p_2, p_3 \in P_3(F)$ such that none of the polynomials $p_0, p_1, p_2, p_3$ has degree 2

This is a repeat of Does there exist a basis $(p_0,p_1,p_2,p_3)\in P_3(\Bbb F)$ such that none of the polynomials $p_0,p_1,p_2,p_3$ has degree $2$?

But I just have a question in regards to the supposed basis vectors.

My conclusion was that it could not occur because in order to characterize all of the polynomials of degree 3, you will need a polynomial of degree 2.

But the solution said otherwise, particularly how are $x^2 + x^3, x^2$ going to be basis vectors. Do these not have a polynomial of degreee 2? Which is what we are trying to show cannot occur?

Answer 1

Suppose we have a list of vectors $(v_1,v_2,v_3,v_4)$, and that this list forms a basis for the space $V$. Our task is to prove that $$(v_1+v_4,v_2+v_4,v_3+v_4,v_4)$$ also forms a basis of $V$.

Our first step is to prove that $(v_1+v_4,v_2+v_4,v_3+v_4,v_4)$ is linearly independent. To do so, we'll examine the following: $$a(v_1+v_4)+b(v_2+v_4)+c(v_3+v_4)+dv_4=\ av_1+bv_2+cv_3+(a+b+c+d)v_4 =\ 0$$ $$\iff$$ $$a=b=c=d=0$$

$(v_1+v_4,v_2+v_4,v_3+v_4,v_4)$ only produces the zero vector when the coefficients are all zero, so this list is linearly independent.

Now we can show that all of the original elements that formed the basis of $V$ can be represented as a linear combination of potential basis. This isn't as hard as it might sound. For instance $$v_1=(v_1+v_4)-v_4$$ $$v_2=(v_2+v_4) - v_4$$ $$v_3=(v_3+v_4)-v_4$$ $$v_4=v_4$$

Et voila! New basis proven. If you substitute $v_1 = 1$, $v_2=x$, $v_3=x^2$, and $v_4=x^3$, then you will have essentially proven that you can represent $\mathscr{P}_3(\mathbb{F})$ with a group of polynomials where the 2nd degree is not represented.

Answer 2

$p_0:=x^3$
$p_1:=x^2+x^3$
$p_2:=x+x^2+x^3$
$p_3:=1+x+x^2+x^3$

$\deg P_j = 3\neq 2$ for any $j$.

$p_1-p_0=x^2$
$p_2-p_1=x$
$p_3-p_2=1$

So, $p_0,p_1,p_2,p_3$ is a basis of $\mathcal{P}_3(\mathbb{F})$.