The matrix of basis change from basis $B = \{b_1,b_2\}$ to the basis $C = \{(1,1),(0,2)\}$ is
$$ M = \begin{bmatrix} 1 & 0 \\2&3\end{bmatrix}$$
Find basis $B$.
Well, here's what I did:
Since $M$ changes from $B$ to $C$, then it Works like this:
$${\begin{bmatrix}& \\\\\\\end{bmatrix}}_C = \begin{bmatrix} 1 & 0 \\2&3\end{bmatrix}{\begin{bmatrix}a_1\\a_2\end{bmatrix}}_B$$
In other words, a vector with coordinates in $C$ is written as the matrix $M$ multiplied by the vector written in $B$.
$${\begin{bmatrix}& \\\\\\\end{bmatrix}}_C = a_1\begin{bmatrix} 1 \\2\end{bmatrix}+a_2\begin{bmatrix} 0 \\3\end{bmatrix}$$
where $\begin{bmatrix} 1 \\2\end{bmatrix}$and $\begin{bmatrix} 0 \\3\end{bmatrix}$ are the vectors $b_1,b_2$ written in $C$, so:
$$\vec{b_1} = 1(1,1) + 2(0,2) = (1,5)\\\vec{b_2} = 0(1,1) + 3(0,2) = (0,6)$$
but this is not the answer.
Could somebody tell me what I'm doing wrong? (instead os just solving in another way)
Best Answer
Here is how I would do it:
old base
$\mathcal B$ of a vector space $ E $ to anew base
$\mathcal B\,' $ has as column-vectors the coordinates of thenewbase
in theoldbase
. It is the matrix of theidentity map
from $ (E, \mathcal B\,') $ to $ (E, \mathcal B) $.old coordinates
$X$ of a vector from thenew coordinates
$X'$: $$X=P_\mathcal B^{\mathcal B'}X'$$Denote the canonical base as $\mathrm{Can}$. Determining $\mathcal B$ is the same as determining $P_{\mathrm{Can}}^{\mathcal B}$. By the composition formula, we have: $$P_{\mathrm{Can}}^{\mathcal B}=P_{\mathrm{Can}}^{\mathcal C}P_{\mathcal C}^{\mathcal B}=P_{\mathrm{Can}}^{\mathcal C}\bigl(P_{\mathcal B }^{\mathcal C}\bigr)^{-1}.$$
Now $ P_{\mathrm{Can}}^{\mathcal C}= \begin{bmatrix} 1 & 0 \\1&2\end{bmatrix} $, and we compute with the pivot method that $$\bigl(P_{\mathcal B }^{\mathcal C}\bigr)^{-1}= \begin{bmatrix} 1 & 0 \\2&3\vphantom{\tfrac 13}\end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 \\-\tfrac23& \tfrac 13\end{bmatrix}$$ so that $$P_{\mathrm{Can}}^{\mathcal B}=\begin{bmatrix} 1 & 0 \\-\tfrac 13& \tfrac 23\end{bmatrix}$$
Hence $b_1=\begin{bmatrix} 1 \\-\tfrac 13\end{bmatrix}$, $\quad b_2=\begin{bmatrix} 0 \\\tfrac 23\end{bmatrix}$.