Spanning of the polynomials of degree 2.

Question

Let $p_1(x)=1+2x+x^2$, $p_2(x)=1+x+x^2$, $p_3(x)=1+x^2$. Show that they span the space of polynomials of degree 2.

I'm not sure why if a general linear combination of these three vectors equals a general polynomial of degree 2 then that means they span the space.

Answer 1

It turns out that the vector space generated by $\{p_1,p_2,p_3\}$ is equal to the one generated by $\{q_1,q_2\}$, where $q_1(x)=x$ and $q_2(x)=1+x^2$. Indeed, each element of $\{p_1,p_2,p_3\}$ is a linear combination of $q_1$ and $q_2$ and $q_1=p_1-p_2$ and $q_2=p_3$.

Answer 2

$\lbrace 1, x, x^{2} \rbrace$ is a basis of 2 degree polynomial.

This basis allows changing the coordinate from 2 degree polynomials to 3D Cartesian coordinate $R^{3}$. That is an isomorphism.

The three polynomial in terms of the basis are:

(1, 2, 1), (1, 1, 1), and (1, 0, 1)

Write them as a matrix:

$\begin{bmatrix} 1&2&1\\ 1&1&1\\ 1&0&1\\ \end{bmatrix}$

If the three rows are linearly independent, the three rows form a basis in $R^{3}$.

If that is the case, the 3 polynomial form a basis for 2 degree polynomials.

That is because since we can change the basis from $R^{3}$ back to 2 degree polynomial, any basis in $R^{3}$ corresponds to a basis over 2 degree polynomial.

Answer 3

HINT.-The space of polynomials of degree $2$ has dimension $3$ and a canonical basis is $\{1,x,x^2\}$. You have to prove that $\{p_1,p_2,p_3\}$ is also a basis. There are several ways to do that. One of them is to prove that the matrix (coming from the coefficients of your polynomials) $$\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}$$ is invertible. This is not the case because the determinant

$$\det\begin{pmatrix} 1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}=0$$ This shows that $\{p_1,p_2,p_3\}$ IS NOT a basis of your space.

Answer 4

The set $\{p_1,p_2,p_3\}$ cannot span the space of polynomials of degree 2. We show that $x^2$ for example cannot be projected on this set.$$x^2=(1+x^2)a_1+(1+x+x^2)a_2+(1+2x+x^2)a_3=(1+x^2)(a_1+a_2+a_3)+(a_1+2a_2)x$$which imposes that $a_1+a_2+a_3=1$ since the coefficient of $x^2$ is $a_1+a_2+a_3$ and $a_1+a_2+a_3=0$ since it is the constant value. This leads to a dead-end, and completes our proof.