Inverse Linear Transformations

# 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)
\begin{align} T_A(T_{A^{-1}}(x)) = AA^{-1}x = Ix = x \\ T_{A^{-1}}(T_A(x)) = A^{-1}Ax = Ix = x \end{align}

## Example 1

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

(2)
\begin{align} w_1 = 2x_1 + 3x_2 \\ w_2 = 6x_1 - 4x_2 \end{align}

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)
\begin{align} w = Ax \\ A^{-1}w = A^{-1}Ax \\ A^{-1}w = x \\ \end{align}

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)
\begin{align} x_1 = \frac{2}{13}w_1 + \frac{3}{26}w_2 \\ x_2 = \frac{3}{13}w_1 - \frac{1}{13}w_2 \end{align}

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