# Inverse Linear Transformations

Definition: If $T_A : \mathbb{R}^n \to \mathbb{R}^n$ is a linear operator that is one-to-one that maps any vector $\vec{x}$ to $\vec{w}$, then the inverse linear operator $T_{A^{-1}} : \mathbb{R}^n \to \mathbb{R}^n$ maps the image $\vec{w}$ to $\vec{x}$. |

We note that $T_A$ is a transformation resulting from multiplication by $A$, thus $T_{A^{-1}}$ is a transformation resulting from multiplication by $A^{-1}$. If $T_A$ is one-to-one, then we know that $A$ is invertible and hence $A^{-1}$ exists. Furthermore it follows that $T_{A^{-1}} : \mathbb{R}^n \to \mathbb{R}^n$ consequentially "reverses" the effects of $T_A$ as seen when we look at the composition $(T_A \circ T_{A^{-1}})(x) = T_A(T_{A^{-1}}(x))$:

(1)## Example 1

**Given that $T_A: \mathbb{R}^n \to \mathbb{R}^n$ is a linear operator defined by the following equations:**

**Show that $T_A$ is one-to-one and define $T_{A^{-1}} : \mathbb{R}^n \to \mathbb{R}^n$.**

We first note that the standard matrix $A = \begin{bmatrix}2 & 3\\ 6 & -4\end{bmatrix}$, and in $w = Ax$ form we get $\begin{bmatrix}w_1\\ w_2 \end{bmatrix}=\begin{bmatrix}2 & 3\\ 6 & -4\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix}$. Since $\det (A) = -26 ≠ 0$, so we know that $T_A$ is also one-to-one. Inverting $A$ we get that $A^{-1} = \begin{bmatrix} \frac{2}{13} & \frac{3}{26}\\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}$, and:

(3)Thus we get that $x = \begin{bmatrix} \frac{2}{13} & \frac{3}{26}\\ \frac{3}{13} & -\frac{1}{13} \end{bmatrix}\begin{bmatrix}w_1\\ w_2 \end{bmatrix} = \begin{bmatrix} x_1\\ x_2\end{bmatrix}$, and the following equations define the image of our inverse transformation:

(4)Thus $T_{A^{-1}} (w_1, w_2) = (\frac{2}{13}w_1 + \frac{3}{26}w_2, \frac{3}{13}w_1 - \frac{1}{13}w_2 )$.