Change of basis

As we saw in Intuitive understanding of multiplying a matrix, matrix multiplication can be seen as change of basis, preserving a vector’s coefficients.

In this point of view, a single vector can be represented differently according to what basis you choose.

If a vectorspace \(\mathbb{A}\) chooses \(\begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix}\) and \(\begin{bmatrix} a_{12} \\ a_{22} \end{bmatrix}\) as it’s basis,

\(\begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix}_{\mathbb{R}} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}_{\mathbb{A}}\) and \(\begin{bmatrix} a_{12} \\ a_{22} \end{bmatrix}_{\mathbb{R}} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}_{\mathbb{A}}\)

In \(\mathbb{R}\), \(90^\circ\) rotation of \(\begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix}_{\mathbb{R}}=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}_{\mathbb{R}} \begin{bmatrix} a_{11} \\ a_{21} \end{bmatrix}_{\mathbb{R}}\)

In \(\mathbb{A}\), \(90^\circ\) rotation of \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}_{\mathbb{A}} =\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}^{-1} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}_{\mathbb{A}}\)

So, if you want to perform linear transformation \(P_{\mathbb{R}}\) to \(\begin{bmatrix} x \\ y \end{bmatrix}_{\mathbb{A}}\), you can do

\[A^{-1}PA \begin{bmatrix} x \\ y \end{bmatrix}\]