Basis of polynomials $R[x]<4$ none degree one

Question

Is there a basis of the set of polynomials with degree less than $4$ such that none of the basis polynomials have degree 1?

I tried $\{1,x^2,x^3-x,x^3+x\}$ and showed they are linearly independent but I don't think they span the whole set as in $\{a,b,c,d\in \mathbb{R}:a+(d-c)x+bx^2+(c+d)x^3\}$, the coefficients of $x$ and $x^3$ cannot be chosen freely.

Answer 1

The set $$\{1,x^2, x^3-x, x^3+x\}$$ is a linearly independent set. Let's call the subspace it generates $S$.

We know

1. $S$ is a $4$-dimensional
2. $S$ is a subspace of the space $\mathbb P_4$ of polynomials of degree less than $4$
3. $\mathbb P_4$ is $4$-dimensional

Therefore, $S=\mathbb P_4$.

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If you're unsure of that, you can see that

$$\begin{align}ax^3+bx^2+cx+d &= \frac a2 (x^3-x) + \frac a2 (x^3+x) + bx^2+\frac{c}{2}(x^3+x) - \frac c2 (x^3-x) + d\\ &=\left(\frac{a}{2}+\frac c2\right)(x^3+x) + \left(\frac{a}{2}-\frac c2\right)(x^3-x)+bx^2+d\end{align}$$

Answer 2

Your example is fine. What's the problem? Note that $1$ and $x^2$ are part of your set. Besides$$x^3=\frac12\bigl((x^3-x)+(x^3+x)\bigr)\text{ and }x=\frac12\bigl(-(x^3-x)+(x^3+x)\bigr).$$

Answer 3

A classic basis is given by the Bernstein polynomials. Without the coefficients they are $$ x^3, x^2(1-x), x(1-x)^2, (1-x)^3 $$ The matrix that expresses these polynomials with respect to the canonical basis is triangular with a diagonal of only $1$s, and so is invertible.