I'm not sure how to approach this problem. The examples I've come across on the internet show how to find the change of coordinates matrix from a matrix to another matrix, such as B to C (for example).
I came up with an answer but I'm not sure if it's correct.
I started out with the matrix with those three vectors mentioned above:
3 2 1 0 2 -2 6 -4 3
Then I found the inverse which is the following:
1/21 5/21 1/7 2/7 -1/14 -1/7 2/7 -4/7 -1/7
Then I multiplied by the standard basis of a 3x1 vector:
1/21 5/21 1/7 1 2/7 -1/14 -1/7 * 0 2/7 -4/7 -1/7 0
And came up with the following answer:
1/21 2/7 2/7
Is this correct? Something tells me I'm missing something or perhaps I approached the whole thing incorrectly.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$The matrix which changes coordinates with respect to the basis $B$ to the coordinates with respect to the standard basis $B^{\prime}$ is given by
$P=\begin{pmatrix} 3 & 2 & 1\\ 0 & 2 & -2\\ 6 & -4 & 3\end {pmatrix}$ whose columns are just the vectors in $B$.
To get the matrix which changes coordinates from $B^{\prime}$ to $B$, just take $P^{-1}$.
In general, if $B=\{v_1,\cdots,v_n\}$ and $B^{\prime}=\{u_1,\cdots,u_n\}$ are two bases for a vector space,
the change of coordinate matrix from $B$ to $B^{\prime}$ is given by
$\begin{bmatrix} [v_1]_{B^{\prime}} \;\vert & \cdots &\vert\;\; [v_n]_{B^{\prime}} \end{bmatrix}$.
$\endgroup$