How to translate between coordinate systems that use different basis vectors

In my previous blogs I talked about Linear Transformations and how can we find the transformation matrix of a linear transformation. In this blog we will try to understand how to translate basis vectors between different coordinate systems with the help of the 2-dimensional Cartesian Coordinate System.

The standard basis vectors in the 2-dimensional system are,
$\hat{i}=\begin{bmatrix} 1\\0 \end{bmatrix}, \hat{j}=\begin{bmatrix} 0\\1 \end{bmatrix}$
Any vector in this 2-dimensional system can be represented in terms of the standard basis vectors, $\hat{i}$ and $\hat{j}$ as,
$\vec{v}= a\hat{i} + b\hat{j}$ where a and b are any scalars such that $\left \{ a, b \in \mathbb{R} \right \}$
The basis vectors of any other 2-dimensional system (lets call it the ‘Inception Coordinate System‘) can be expressed in terms of the standard basis vectors. This is like creating an analogy of the Inception System within the familiar Cartesian Coordinate System.

Lets say the transformed coordinates of the standard basis vectors which replicate the Inception System are given by vectors $\hat{m}$ and $\hat{n}$ such that,
$\hat{m} = \begin{bmatrix} 2\\1 \end{bmatrix}, \hat{n} = \begin{bmatrix} 1\\-1 \end{bmatrix}$
The vectors $\hat{m}$ and $\hat{n}$ are the basis vectors of the Inception System defined in the language of our familiar 2-dimensional Cartesian Coordinate System. Which means that any vector in the Inception System when multiplied with the transformation matrix, $T=\begin{bmatrix} 2 & -1\\ 1 & 1 \end{bmatrix}$ will be mapped to a vector in the 2-dimensional Cartesian Coordinate System.

Now think the other way round. Lets say, $\vec{v} = \begin{bmatrix} 3\\2 \end{bmatrix}$ is a vector in the 2-dimensional Cartesian Coordinate System. This vector can be mapped to the Inception System by the following relationship,
$\vec{v} = T\vec{x}$ where $\vec{x}$ is a vector of the Inception System.
$\begin{bmatrix} 3\\ 2 \end{bmatrix}= \begin{bmatrix} 2 & -1\\ 1 & 1 \end{bmatrix}\vec{x}$
$\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}^{-1}\begin{bmatrix} 3\\ 2 \end{bmatrix}=\vec{x}$
$\begin{bmatrix} 1/3 & 1/3 \\ -1/3 & 2/3 \end{bmatrix}\begin{bmatrix} 3\\ 2 \end{bmatrix}=\vec{x}$
$3\begin{bmatrix} 1/3\\ -1/3 \end{bmatrix}+ 2\begin{bmatrix} 1/3\\ 2/3 \end{bmatrix}=\vec{x}$
$\begin{bmatrix} 1\\ -1 \end{bmatrix}+ \begin{bmatrix} 2/3\\ 4/3 \end{bmatrix}=\vec{x}$
$\begin{bmatrix} 5/3\\ 1/3 \end{bmatrix}=\vec{x}$
We can see that the vector $\vec{v} = \begin{bmatrix} 3\\ 2 \end{bmatrix}$ is represented as the vector $\vec{x} = \begin{bmatrix} 5/3\\ 1/3 \end{bmatrix}$ in the Inception System.