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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

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This is a simple variant of the previous problem you posted. Study the solutions there and solve this one analogously. –  Lepidopterist Mar 17 '13 at 22:20

3 Answers 3

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.

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The first question follows directly from the change of variable $u = 1-t$.

The same idea can be used for the second question: $$ 1+t^3 = 1 + (1-u)^3 = 1 + (1-3u+3u^2 - u^3) = 2 - 3u + 3u^2 - u^3 $$

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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.

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