# 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

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

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?

• If $(1,x,x^2,x^3)$ is a basis, $(1+x^3,x+x^3,x^2+x^3,x^3)$ is a basis, too. Commented Jun 2, 2016 at 21:44
• maybe I am misunderstanding what is being asked. $x^2 + x^3$ has an element of deg 2 in it, so isn't that considered a polynomial that has an element of deg 2? i.e cannot be a part of the basis? Commented Jun 2, 2016 at 21:50
• $x^2+x^3$ has a monomial of degree two but is not a polynomial with degree two. Commented Jun 2, 2016 at 21:51
• Then I did misunderstand the concept. Thanks for the clarification Commented Jun 2, 2016 at 21:53

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.

• Is that his site? I have been using that sight as a reference as well. Commented Jul 26, 2016 at 21:39
• @dc3rd I'm not sure if Axler maintains that site, but I did send him an email letting him know that there was a typo. If it turns out he isn't responsible for that content, then I'll amend my answer. The proof is still sound, though. Commented Jul 26, 2016 at 23:18
• The answer above has a link to "Axler's solution". The solutions at that link are not written by me and are not authorized by me. I do not know who wrote and posted those solutions. Math.stackexchange.com is wonderful when it is used appropriately. Students learn best by struggling to discover solutions themselves instead of giving up too quickly and seeking an answer on the web. Commented Jul 28, 2016 at 0:06

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

$$\deg P_j = 3\neq 2$$ 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 $$\mathcal{P}_3(\mathbb{F})$$.