Vector's coordinates Consider the following two bases for $\Bbb R^3$:
$$B_1 =\left\{\begin{bmatrix}2 \\4 \\ -2 \end{bmatrix} , \begin{bmatrix}5 \\0 \\ 1 \end{bmatrix} , \begin{bmatrix}-1 \\6 \\ 4 \end{bmatrix}\right\}$$
$$B_2 =\left\{\begin{bmatrix}3 \\1 \\ 7 \end{bmatrix} , \begin{bmatrix}1 \\0 \\ 3 \end{bmatrix} , \begin{bmatrix}-1 \\3 \\ 1 \end{bmatrix}\right\}$$
If a vector’s coordinates in $B_1$ are $x_1=\begin{bmatrix}1 \\1 \\ 1 \end{bmatrix}$, what are the point’s coordinates
in $B_2$?
 A: Hints:
What effect does the transformation $A = \begin{bmatrix}2&5&-1\\4&0&6\\-2&1&4\end{bmatrix}$ have on an arbitrary vector $x$?

 It takes a vector currently represented with respect to the basis $B_1$ and represents it instead with respect to the standard basis.

What effect does $A^{-1}$ have then?

 It takes a vector currently represented in the standard basis and represents it with respect to the basis $B_1$.

What about the matrix $C=\begin{bmatrix}3&1&-1\\1&0&3\\7&3&1\end{bmatrix}$ and its inverse $C^{-1}$?
Can you think of a way of combining these matrices or their inverses to have the desired effect?

 What does $A^{-1}C$ do to a vector?  What does $C^{-1}A$ do to a vector?

A: What JMoravitz said is correct. Another way of saying the same thing is like this (in case he wasn't clear enough):
Your basis $B_1$ and $B_2$ can be thought of as a transformation involving matrix multiplication on $x_1$, where $y$ is a vector in $B_1$ as $y=T_1x_1$. i.e.
$$
y=\begin{bmatrix}2 & 5 & -1\\ 4 & 0 & 6 \\ -2 & 1 &4\end{bmatrix}\begin{bmatrix}
   1 \\ 1 \\ 1
\end{bmatrix}
$$
We want to find vector coordinates $x_2$ such that $T_2x_2=y$, i.e. we want $x_2$ where:
$$
T_2x_2 = \begin{bmatrix}3 & 1 & -1 \\ 1 &0 &3\\ 3 &2 &1\end{bmatrix}\begin{bmatrix}x_{2,1}\\ x_{2,2}\\ x_{2,3}\end{bmatrix}=y
$$
which is really just $T_2x_2=T_1x_1$ This tells us that to find $x_2$ we need to first "find" the vector from the basis of $B_1$ then "locate it" in basis $B_2$, i.e. $ x_2=T_2^{-1}T_1x_1$.
