Change of basis matrix for basis with dimension n + 1

Question

I have a basis $A$= $ \{1, x, x^2, ... , x^n\}$ for Pn and another basis $B$ = $ \{1, (x-1), (x-1)^2, . . ., (x-1)^n\}$ for Pn and I need to find the change of basis matrix from A to B. How do I go about this question?

Answer 1

You should have seen that the columns of the change of basis matrix from base A to base B are the coordinates of the vectors of basis A in the basis B. In other word if $\phi_A(y)= \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n\end{pmatrix}$ is the vector that contains the coordinates of the vector $y$ in the basis $A$ (i.e. $y=\sum_{j=1}^n y_j\cdot u_j$ if $A = (u_1,u_2,\ldots,u_n)$) and if the basis B is $(v_1,v_2,\ldots,v_n)$ then the the change of basis from A to B is the matrix $\Phi_{A\rightarrow B} =(\phi_B(u_1) \quad \phi_B(u_2)\quad \ldots \quad \phi_B(u_n)) $ and the change of basis matrix from B to A is $\Phi_{B\rightarrow A} =(\phi_A(v_1) \quad \phi_A(v_2)\quad \ldots \quad \phi_A(v_n)) $. Also you may be aware that $\Phi_{A\rightarrow B} = \Phi_{B\rightarrow A}^{-1}$.

So in your question you have two choices : you either find $\Phi_{B\rightarrow A}$ and inverse it or you have to find the coordinates of the vectors forming base A in the base B. In your case finding $\Phi_{B\rightarrow A}$ is pretty easy (think of Newton's binomial) but maybe finding its inverse will be harder, you can try to find the coordinates of the vectors forming base A in base B is possible for example : $$\phi_B(1)=\begin{pmatrix}1\\0\\0\\0\\ \vdots \\0\end{pmatrix},\phi_B(x)=\begin{pmatrix}1\\1\\0\\0\\ \vdots \\0\end{pmatrix},\phi_B(x^2) = \begin{pmatrix}1\\2\\1\\0\\ \vdots \\0\end{pmatrix},\phi_B(x^3) = \begin{pmatrix}1\\3\\3\\1\\ \vdots \\0\end{pmatrix}$$ We see that the coordinates of $u_j$ in the basis B may be the numbers in the $j^{th}$ row of Pascal's triangle (it's a good exercise to prove it!), normally now you should have an idea on how to go about this question :)