# Change of basis matrix for polynomials

I've already asked this question here, but theres a misconception about the phrase "change of basis matrix from B to C", and I think the answers were given in the inverse of what's in my book.

So, in my book, the change of basis matrix from $B$ to $C$ is the matrix $M$ such that

$$[ \ \ \ ]_B = M[\ \ \ ]_C$$

where $M$ has the vectors $c_1,c_2$ of the base $C$ written as a linear combination of the basis $B$.

So, the question is:

The change of basis matrix from $B = \{1+t, 1-t^2\}$ to the base $C = \{c_1, c_2\}$ is:

$$\begin{bmatrix}\color{Red}{1} & \color{Blue}{2}\\\color{Red}{1} & \color{Blue}{-1}\end{bmatrix}$$

Find basis $C$.

So what I did was:

$$c_1 = \color{Red}{1}(1+t) + \color{Red}{1}(1-t^2) = 2 + t -t^2\\c_2 = \color{Blue}{2}(1+t) \color{Blue}{-1}(1-t^2) = 1 + 2t + t^2$$

Am I rigth?

• Not quite; $c_1$ and $c_2$ should be multiplied by $\frac13$, and the end of your computation for $c_2$ is wrong. – Bernard Feb 1 '15 at 21:50
• @Bernard why 1/3? – Guerlando OCs Feb 1 '15 at 22:28
• When you compute $M^{-1}$, there's a factor $\frac13$. Actually, $M^{-1}=\frac13M$, – Bernard Feb 1 '15 at 22:31
• @Bernard why I need to compute $M^{-1}$? I've already found $c_1$ and $c_2$. Are you using the definition I have in the exercise? Thank you! – Guerlando OCs Feb 1 '15 at 22:34
• But if $M$ has $c_1$ and $c_2$ expressed with $b_1$ and $b_2$, $M^{-1}$, which is the change of base matrix from $\mathcal C$ to $\mathcal B$ has $b_1$ and $b_2$ expressed with $c_1$ and $c_2$. – Bernard Feb 1 '15 at 22:39