When working with [[Matrix Transformations]] in a [[Changing Basis Vectors|changed basis]], the key challenge is to ensure that the transformation defined in one coordinate system is correctly applied to vectors expressed in another. This requires translating vectors between coordinate systems before and after applying the transformation.
## Problem Setup
Suppose we have a transformation $T_E$, which performs a $45^\circ$ rotation in the *original coordinate system* (standard basis). The transformation matrix is:
$ T_E=\frac{1}{\sqrt 2}* \begin{bmatrix} 1 & -1 \\ 1 & \phantom-1\end{bmatrix} $
Now, we want to apply this transformation to a vector $r_B$ that is expressed in a *stretched coordinate system* defined by the basis matrix $B$:
$
r_B= \begin{bmatrix}x \\ y\end{bmatrix}, \quad
B=\begin{bmatrix} 3&1 \\ 1&1\end{bmatrix}, \quad
$
## Derivation
To apply the transformation $T_E$ to $r_B$, the following steps are needed:
1. *Translate vector into the original system:* Use the [[Inverse Matrix]] of the basis matrix $B^{-1}$, to convert the stretched coordinate vector $r_B$ into the original coordinate system.
$ r_E = B^{-1} \cdot r_B$
2. *Apply the transformation:* Perform the transformation $T_E$ on the translated vector.
$ r_E^{(\text{trans})} = T_E \cdot r_E $
3. *Translate the transformed vector back into the stretched basis:* Multiply by $B$ to express the result in the stretched coordinate system.
$ r_B^{(\text{trans})} = B \cdot r_E^{(\text{trans})}$
These steps can be combined into a single operation:
$ r_B^{(\text{trans})} = B \cdot T_E \cdot B^{-1} \cdot r_B$
Numerical example:
$
\begin{align}
r_B &=\frac{1}{2} \begin{bmatrix}\phantom-1&-1\\-1&\phantom-3 \end{bmatrix}*\frac{1}{\sqrt 2} \begin{bmatrix} 1 & -1 \\ 1 & \phantom-1\end{bmatrix}*\begin{bmatrix} 3&1 \\ 1&1\end{bmatrix}* \begin{bmatrix} x\\ y\end{bmatrix}\\[12pt]
r_B&=\frac{1}{2\sqrt 2} \begin{bmatrix} -2&-2\\ \phantom-10&\phantom-6\end{bmatrix}\begin{bmatrix} x\\ y\end{bmatrix}
\end{align}
$