Question about basis I have to prove that $1, (1-t), (1-t)^2$, and $(1-t)^3$ is a basis of for $P^{3}$ but I am not sure how to start this problem.
Also I have to find the coordinates of $p(t) = 1 +t^3$
Thank you for the help
 A: You need to show that they span and that they are linearly independent.  I would start by showing linear independence.  Start with
$$a + b(1-t) + c(1-t)^2 + d(1-y)^3 = 0.$$
Then by expanding and comparing coefficients prove that $d = c = b = a = 0$.
Now if you know something about dimension then you're done, because the dimension of $P_3$ is $4$.  If not, then you'll have to show that the basis spans, so you'll have to show, given
$$f = a + bt + ct^2 + d^3,$$
how to express $f$ as a linear combination of the basis.  This is a little messy to write out, but shouldn't be hard.
A: The first question follows directly from the change of variable $u = 1-t$.
The same idea can be used for the second question:
$$
1+t^3 = 1 + (1-u)^3 =  1 + (1-3u+3u^2 - u^3) = 2 - 3u + 3u^2 - u^3
$$
A: The operator $\phi:P^3\longrightarrow P^3$ defined by $\phi(P)(t)=P(t-1)$ is an isomorphism of inverse given by $\phi^{-1}(Q)(t)=Q(t+1)$. Therefore, it takes very basis onto a basis. Starting with the canonical basis $\{p_0(t)=1, p_1(t)=t, p_2(t)=t^2, p_3(t)=t^3\}$, we obtain the basis $$\{\phi(p_0)(t)=1, \phi(p_1)(t)=t-1, \phi(p_2)(t)=(t-1)^2, \phi(p_3)(t)=(t-1)^3\}.$$
To find the coordinates of $p(t)=1+t^3$ in the new basis, we look for scalars $a_j$ such that
$$
p(t)=a_0+a_1(t-1)+a_2(t-1)^2+a_3(t-1)^3.
$$
Changing $t$ for $t+1$, this is equivalent to
$$
p(t+1)=a_0+a_1t+a_2t^2+a_3t^3.
$$
Now $p(t+1)=1+(t+1)^3=\ldots$? Develop using the binomial theorem.
