How can I find the change of coordinate matrix with non-standard basis?

Question

I have the following question, which seems straight forward but I have no idea how he obtained is answer. How can I gind the coordinate matrix with a non-standard basis?

Let $\mathcal{g} : \mathcal{P}_2 \rightarrow \mathcal{R^3}$ be a linear transformation defined by: $$g(a + bx + cx^2) = \begin{pmatrix} a\\ c\\ b \end{pmatrix}$$

Find $[g]_{\mathcal{C}\leftarrow \mathcal{B}}$ where $\mathcal{B} = \{x, 1, x^2 \}$ and $$\mathcal{C} = \Bigg\{ \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} , \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} \Bigg\} $$ are bases for $\mathcal{P}_2$ and $\mathcal{R}^3$ respectively

My work thus far:

$g(x) = \begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}, g(x^2) = \begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}, g(1) = \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$

First get $g(B)$

$b_{3} = a + b + c$

$b_{2} = b + c$

$b_{1} = c$

Thus:

$c = b_{1}$

$b = -b_1 + b_2$

$a = -b_2 + b_3$

Then we have matrix:

$\begin{pmatrix} 0& 1 & -1 \\ -1 & 0 & 1 \\ 1 & 0 & 0\end{pmatrix}$

Answer 1

Assuming that you meant $\{1,2x,x^2\}$ instead of $\{1+2x+x^2\}$, the algorithm is simple: compute $g$ at each element of $\mathcal{B}$, write the result as a combination of the elements of $\mathcal{C}$, and put the coefficients in the columns. For example, since $$g(1) = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = 1\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}+0\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}+0\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$$we'll have $$[g]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{pmatrix} 1 & ? & ? \\ 0 & ? & ? \\ 0 & ? & ? \end{pmatrix}.$$Do the same for $g(2x)$ and $g(x^2)$ to fill what is missing.

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If $\mathcal{B} = \{x,1,x^2\}$ the idea is the same, just paying attention to the order of the vectors in the basis. Again: $$g(x) = \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}= 0\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}-1\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}+1\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$$and so $$[g]_{\mathcal{C} \leftarrow \mathcal{B}} = \begin{pmatrix} 0 & ? & ? \\ -1 & ? & ? \\ 1 & ? & ? \end{pmatrix}.$$