Let $P_3$ be the vector space of polynomials of degree at most three in one variable $t$. Let $p(t)=t^3 +a_2t^2 +a_1t+a_0$ where $a_0,a_1,a_2 \in\mathbb R$ are fixed constants.
Show that $\{p, \frac{dp}{dt}, \frac{d^2p}{dt^2}, \frac{d^3p}{dt^3}\}$ is a basis for $P_3$.
Best Answer
So you have to show that $${\cal B} = \{t^3+a_2t^2+a_1t+a_0, 3t^2+2a_2t+a_1, 6t+2a_2, 6\}$$ ia a basis for $V=P_3$.
Of course you can show it by definition of being a basis, and it's pretty straightforward (try it!). But I'll give a solution a little bit short which uses the following result:
We are going to show that ${\cal B}$ is linearly independent and conclude that $W={\rm span\,}{\cal B}$ is a $4$-dimensional subspace of $V=P_3$, from where we will know that $W=V$ ($P_3$ being a $4$-dimensional vector space).
Write $$\alpha(t^3+a_2t^2+a_1t+a_0)+\beta(3t^2+2a_2t+a_1)+\gamma( 6t+2a_2) + 6\delta = 0,$$ which is just $$\alpha t^3 + (a_2\alpha+3\beta)t^2 + (a_1\alpha + 2a_2\beta + 6\gamma)t + (a_0\alpha + a_1\beta + 2a_2\gamma + 6 \delta) = 0t^3 + 0t^2 + 0 t + 0$$ and it is possible only if $\alpha = \beta = \gamma = \delta = 0$, what is exactly the definition of ${\cal B}$ being linearly independent.
Using the lemma (and the dimensions of ${\rm span\,}{\cal B}$ and $P_3$), we have that ${\rm span\,}{\cal B} = P_3$.
Therefore ${\cal B}$ is a basis for $P_3$.