Change of basis

What to learn

If we want to see vector in transformed coordinates by \(A\) by transformed perspective, we have to calculate the inverse of \(A\).
- (example)
  - In Transformed coordinates by \(A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\), we can find vector \(\vec{v} = \begin{bmatrix} -1 & 2 \end{bmatrix}^T\) in transformed perspective. Vector \(\vec{v}\) that is made by transformed \(-\vec{i} + 2\vec{j}\). So, doing \(A\cdot \vec{v}\) , we can get \(\vec{w} = A\cdot \vec{v} = \begin{bmatrix} -4 & 1 \end{bmatrix}\) in normal perspective.
    - In transformed coordinates by \(A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\), if someone see our vector \(\vec{v} = \begin{bmatrix} 3 & 2 \end{bmatrix}^T\), then \(A^{-1}v\) vector is needed.
How can we do another transformation in transformed coordinates by \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\)?
- Get transformed vector in normal perspective.
  \[\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\]
- apply transformation \(B = \begin{bmatrix} e & f \\ g & h \end{bmatrix}\)
  \[\begin{bmatrix} e & f \\ g & h \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\]
- Inner dot with inverse of A makes it to transformed coordinates.
  \[\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1}\begin{bmatrix} e & f \\ g & h \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\]