Here is how I would do it:
- First remember the change of basis matrix $P_\mathcal B^{\mathcal B'}$ from an
old base $\mathcal B$ of a vector space $ E $ to a new base $\mathcal B\,' $ has as column-vectors the coordinates of the newbase in the oldbase. It is the matrix of the identity map from $ (E, \mathcal B\,') $ to $ (E, \mathcal B) $.
- The change of basis matrix the other way is just the inverse matrix of the previous one: $$P_{\mathcal B'}^{\mathcal B}=\bigl(P_\mathcal B^{\mathcal B'}\bigr)^{-1}$$
- It allows to express the
old coordinates $X$ of a vector from the new coordinates $X'$: $$X=P_\mathcal B^{\mathcal B'}X'$$
- You can compose the change of basis matrix:
$$ P_{\mathcal B}^{\mathcal B''} = P_{\mathcal B}^{\mathcal B’}P_{\mathcal B’}^{\mathcal B''}$$
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 $b_1=\begin{bmatrix} 1 \\-\tfrac 13\end{bmatrix}$, $\quad b_2=\begin{bmatrix} 0 \\\tfrac 23\end{bmatrix}$.
I assume you want to compute the change-of-basis-matrix from basis B1 to basis B2.
The first step is to write the new basis vectors as a linear combination of the old basis vectors:
$$(1,0,1)=1\cdot(0,0,1)+1\cdot(1,0,0)+0\cdot(0,1,0)$$
$$(1,1,0)=0\cdot(0,0,1)+1\cdot(1,0,0)+1\cdot(0,1,0)$$
$$(0,1,1)=1\cdot(0,0,1)+0\cdot(1,0,0)+1\cdot(0,1,0)$$
The first column in the change-of-basis-matrix is the coordinate of the first new basis vector with respect to the old basis vectors. Repeat this for the second and third column. The you get the following change-of-basis matrix T:
\begin{equation}
\begin{bmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix}
\end{equation}
The general formula to transform new coordinates into old coordinates and vice versa:
coordinate of a point with respect to old basis = $T$ multiplied by coordinate in new basis
coordinate of a point with respect to new basis = $T^{−1}$ multiplied by coordinate in old basis
Remark: T translates from the new basis to the old basis!
Best Answer
First, make sure you understand what it means to write a vector $v$ in the basis $B$. In the standard basis, $S$, the vector $(1,2,3)_S$ is the linear combination $$1\cdot \left[\begin{array}{c}1\\0\\0\end{array}\right] + 2\cdot \left[\begin{array}{c}0\\1\\0\end{array}\right] + 3\cdot \left[\begin{array}{c}0\\0\\1\end{array}\right] $$ which is the same as the matrix multiplication problem: $$ \left[ \begin{array}{ccc} 1&0&0\\0&1&0\\0&0&1 \end{array}\right] \left[\begin{array}{c}1\\2\\3\end{array}\right].$$
In the basis $B$, the vector $(1,2,3)_B$ is the linear combination $$1\cdot \left[\begin{array}{c}3\\0\\6\end{array}\right] + 2\cdot \left[\begin{array}{c}2\\2\\-4\end{array}\right] + 3\cdot \left[\begin{array}{c}1\\-2\\3\end{array}\right] = \left[ \begin{array}{ccc} 3&0&6\\2&2&-4\\1&-2&3 \end{array}\right] \left[\begin{array}{c}1\\2\\3\end{array}\right].$$
In general, a the matrix $T$ of a basis can be used to change a vector $v_T$ in the basis to the standard basis $S$ via $T\cdot v_T = v_S$.
(This agrees with the fact that $I\cdot v_S = v_S$.)
In this case, a vector represented in both $S$ and $B$ would satsify $B\cdot v_B = I \cdot v_S$ and $v_B = B^{-1} \cdot v_S$.
This shows how $B$ and $B^{-1}$ are the matrices to go back and forth from $B$ to the standard basis.