I've understood what a change of basis matrix is, and how it's structured.
So a change of basis matrix from $B$ to $C$ is the matrix $M$ such that:
$${\begin{bmatrix} &\\ \\ \\\end{bmatrix}}_B = {\begin{bmatrix} &&&&\\ \\ \\\end{bmatrix}}{\begin{bmatrix} &\\ \\ \\\end{bmatrix}}_C$$
or
$$[v]_b = M[v]_c$$
However, my book extends this concept to polynomials. I see no problem if I see the polynomial $1+2t^2+3t^3$ as the vector $\begin{bmatrix} 1\\2\\3\\\end{bmatrix}$ and then I can constructo such matrix. But the exercise asks the following:
The change of basis matrix from the base $B=\{1+t,1-t^2\}$ to the base
$C$ is$$\begin{bmatrix}1&2\\1&-1\end{bmatrix}$$
Find basis $C$
So basically we have this structure:
$${\begin{bmatrix} &\\ \\ \\\end{bmatrix}}_B = \begin{bmatrix}1&2\\1&-1\end{bmatrix}{\begin{bmatrix} &\\ \\ \\\end{bmatrix}}_C$$
where the first column vector is $[b_1]_c$ (vector $b_1$ written in terms of base $C$) and the second is $[b_2]_c$. So in some way I should find the basis $C$, but since there's a $t^2$ term, a '3 dimensional' vector should appear somewhere. How do I proceed?
Best Answer
Let $A=\begin{bmatrix} 1&2\\1&-1\end{bmatrix}$. Then $A^{-1}=\begin{bmatrix}\frac{1}{3}&\frac{2}{3}\\\frac{1}{3}&-\frac{1}{3}\end{bmatrix}$ is the change of basis matrix from C to B, so
$w_1=\frac{1}{3}(1+t)+\frac{1}{3}(1-t^2)=-\frac{1}{3}t^2+\frac{1}{3}t+\frac{2}{3}$ and
$w_2=\frac{2}{3}(1+t)-\frac{1}{3}(1-t^2)=\frac{1}{3}t^2+\frac{2}{3}t+\frac{1}{3}$ are the basis vectors in C.
(Notice that B and C are bases for a 2-dimensional subspace of the 3-dimensional vector space of polynomials of degree at most 2.)