How to show that $\operatorname{span} (A)=\mathbb{R}^3$ where $A=\{(t,t^2,t^3):t\in \mathbb{R}\}$?

Question

Let $A=\{(t,t^2,t^3):t\in \mathbb{R}\}$. Show that $\operatorname{span} (A)=\mathbb{R}^3$.

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[My thoughts.]

In order to show $\text{span} (A)=\mathbb{R}^3$ I have to show that every vector in $\mathbb{R}^3$ is a linear combination of vectors in $A$. But I don't know how to do this.

Answer 1

Hint

Take, for example, $t=1$, $t=2$ and $t=3$ and show that $\left(1,1^2,1^3\right)$, $\left(2,2^2,2^3\right)$ and $\left(3,3^2,3^3\right)$ are linearly independent and hence span the (three-dimensional) space $\mathbb{R}^3$.

Answer 2

Alternative approach: the moment curve is not a planar curve, since its torsion is given by

$$ \tau =\frac{((1,2t,3t^2)\times(0,2,6t))\cdot(0,0,6)}{\|(1,2t,3t^2)\times(0,2,6t) \|^2}=\frac{3}{1+9t^2+9t^4}\neq 0. $$

$3\times 3$ real Vandermonde matrices are invertible since the moment curve is non-planar is a nice way to put it. In the classical way, given three distinct real numbers $a,b,c$, the problem of finding a solution to $$ \begin{pmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{pmatrix}\begin{pmatrix}k_1 \\ k_2 \\ k_3\end{pmatrix}=\begin{pmatrix}y_1 \\ y_2 \\ y_3\end{pmatrix}$$ given $y_1,y_2,y_3\in\mathbb{R}$ is equivalent to finding the coefficients of a quadratic polynomial $q$ such that $q(a)=y_1, q(b)=y_2, q(c)=y_3$. A solution always exists, by Lagrange's interpolating polynomials, and such a solution is also unique. It follows that the matrix appearing in the LHS is invertible, hence if $A,B,C\neq 0$ are three distinct points on the moment curve, $\text{Span}(A,B,C)=\mathbb{R}^3$.