An inverse operator

Imagine you are developing some 3D graphics software. How can you transform the scene, for example, rotate it? It turns out that most of the transformations in computer graphics are, in fact, linear operators. And so you can ask the question: is it possible to return to the previous scene (without just saving it somewhere)? Can you apply another linear operator that will "cancel" the effect of the first one? That's where the concept of an inverse operator arises.

What is an inverse operator?

Consider a vector space $V$ and a linear operator $\mathcal{A} \colon V \to V$. The linear operator $\mathcal{A}$ is said to be invertible if, for any vector $y \in V$, there exists a unique vector $x \in V$ such that $\mathcal{A}(x)=y$.

Notice that a linear operator is invertible if and only if there exists a linear operator $\mathcal{B} \colon V \to V$ such that

$\mathcal{AB} = \mathcal{BA} = \mathcal{I}.$The linear operator $\mathcal{B}$ is called an inverse operator and is denoted by $\mathcal{A}^{-1}$.

For example, the unit operator has the inverse operator and $\mathcal{I}^{-1} = \mathcal{I}$.

Properties of an inverse operator

An inverse operator has the following properties.

1) If linear operators $\mathcal{A} \colon V \to V$ and $\mathcal{B} \colon V \to V$ are invertible, then the product $\mathcal{A} \mathcal{B}$ is invertible and $(\mathcal{A} \mathcal{B})^{-1} = \mathcal{B}^{-1} \mathcal{A}^{-1}.$

2) If a linear operator $\mathcal{A}$ is invertible and a scalar $\lambda \ne 0$, then $\lambda \mathcal{A}$ is invertible and $(\lambda \mathcal{A})^{-1} = \lambda^{-1}\mathcal{A}^{-1}.$

3) If a linear operator $\mathcal{A}$ is invertible, then its inverse operator is also invertible, and $(\mathcal{A}^{-1})^{-1} = \mathcal{A}.$

4) A linear operator $\mathcal{A} \colon V \to V$ is invertible if and only if $\ker \mathcal{A} = \{0\}$ and $\text{Im} \ \mathcal{A} = V$.

5) If a linear operator $\mathcal{A} \colon V \to V$ is invertible and $\dim V < \infty$, then its inverse operator has a matrix inverse to the matrix of the operator $\mathcal{A}$ in the same basis.

Example

Consider a linear operator in a vector space $\mathbb{R}^3$:

$\mathcal{A}x = \begin{pmatrix} 2x_1 +2x_2-3x_3 \\ -x_2 + 2x_3 \\ -x_1-x_2+x_3 \end{pmatrix}$where $x = (x_1, x_2, x_3)$. Is this linear operator invertible?

The operator $\mathcal{A}$ can be rewritten as follows:

$\mathcal{A}x = \begin{pmatrix} 2 & 2 & -3 \\ 0 & -1 & 2 \\ -1 & -1 & 1 \end{pmatrix}\begin{pmatrix} x_1\\x_2\\x_3\end{pmatrix}.$So, the operator $\mathcal{A}$ has a matrix

$A= \begin{pmatrix} 2 & 2 & -3 \\ 0 & -1 & 2 \\ -1 & -1 & 1 \end{pmatrix}.$We need to find the matrix inverse to this matrix $A$. It will be the matrix of the inverse operator $\mathcal{A}^{-1}$.

To find the inverse matrix $A^{-1}$, we write the extended matrix: matrix $A$ will be on its left side and the identity matrix on the right side. After reducing the matrix from the left to the unit using elementary transformations, the correct inverse matrix will be on the right side.

$\left(\begin{array}{ccc|ccc} 2& 2 & -3 &1 & 0 & 0\\ 0 & -1 & 2 & 0 & 1 & 0\\ -1 & -1 & 1&0 & 0 & 1 \end{array}\right) \overset{\frac{1}{2}r_1, \ -r_2, \ -r_3}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1& 1 & -3/2 &1/2 & 0 & 0\\ 0 & 1 & -2 & 0 & -1 & 0\\ 1 & 1& -1&0 & 0 & -1 \end{array}\right) \overset{r_3-r_1}{\longrightarrow} \\ \longrightarrow \left(\begin{array}{ccc|ccc} 1& 1 & -3/2 &1/2 & 0 & 0\\ 0 & 1 & -2 & 0 & -1 & 0\\ 0 & 0& 1/2&-1/2 & 0 & -1 \end{array}\right) \overset{r_1-r_2, \ 2r_3}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1& 0 & 1/2 &1/2 & 1 & 0\\ 0 & 1 & -2 & 0 & -1 & 0\\ 0 & 0& 1&-1 & 0 & -2 \end{array}\right) \overset{r_2 + 2r_3}{\longrightarrow} \\ \longrightarrow \left(\begin{array}{ccc|ccc} 1& 0 & 1/2 &1/2 & 1 & 0\\ 0 & 1 & 0 & -2 & -1 & -4\\ 0 & 0& 1&-1 & 0 & -2 \end{array}\right) \overset{r_1 - \frac{1}{2}r_3}{\longrightarrow} \left(\begin{array}{ccc|ccc} 1& 0 & 0 &1& 1 & 1\\ 0 & 1 & 0 & -2 & -1 & -4\\ 0 & 0& 1&-1 & 0 & -2 \end{array}\right) .$Hence,

$A^{-1} = \begin{pmatrix} 1 & 1 & 1\\ -2 & -1 & -4 \\ -1 & 0 & -2 \end{pmatrix}$and we find an inverse operator:

$\mathcal{A}^{-1}x = \begin{pmatrix} x_1 + x_2 + x_3 \\ -2x_1 - x_2 - 4x_3\\ -x_1 - 2x_3 \end{pmatrix}.$

Conclusion

In this topic we have learned about an inverse operator to a given one. A product (in other words, a composition) of an original operator and its inverse is a unit operator (and their order does not matter). We considered some properties of an inverse operator and studied an example of how we could find it.