We have that $S=\{1+t^3,3+t-2t^2,-t+3t^2-t^3\}$.
a) Is $S$ linearly dependant in $P_3$?
b) Can the $S$ be a basis for $P_3$?
a) The basis for $P_3$ are $\{1,t,t^2,t^3\}.$ The definition of linearly dependant is that I should be able to write one of the elements as a linear combination of the others. It seems as if I combine any two elements in $S$, I can get the third, since any two of them contain all the basis vectors.
The answer is however that they are linearly independant. Why?
b) Since there are only $3$ elements, they can never be a base in a space of dimension $4$, since $\dim(P_3)=4.$ Is this correct?
Assume $$\alpha(1+t^3)+\beta(3+t-2t^2)+\gamma(-t+3t^2-t^3)=0, \forall t \in \mathbb{R}$$
and you wish to prove $\alpha= \beta =\gamma = 0$.
Plugging in $t = -1$ gives
$$5\gamma = 0 \implies \gamma = 0$$
Now we have $\alpha(1+t^3)+\beta(3+t-2t^2) = 0, \forall t \in \mathbb{R}$.
Plugging in $t = \frac32$ gives
$$\frac{35}{8}\alpha = 0 \implies \alpha = 0$$
Now we have $\beta(3+t-2t^2) = 0, \forall t \in \mathbb{R}$.
Plugging in $t = 1$ gives
$$2\beta = 0 \implies \beta = 0$$
Hence $\alpha = \beta = \gamma = 0$ so $\{1+t^3,3+t-2t^2,-t+3t^2-t^3\}$ is linearly independent.