Matrix inverse properties

The polymath Henri Poincaré once said "Mathematics is the art of giving the same name to different things". Now you will learn the connection between invertible matrices, bases and operators as well as their geometry. They are all different ways of talking about the same concept.

These relationships will help you better understand invertible matrices, understand deeper topics, and save you a lot of unnecessary calculations. In the following you'll work with a square matrix $A$ of size $n$.

Basic properties

The following properties show the behavior of the inverse with other common matrix operations. They come in handy when you need to compute the inverse of some matrix in terms of a simpler one.

Suppose you want to find the inverse of the following matrix $B=\begin{pmatrix} 2 & 4 \\ 6 & 2 \end{pmatrix}$. Since all its inputs are even, you can factor the 2 so that

$B=2 \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$. Now, the new matrix $C= \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$ is easier to invert. You can easily compute $C^{-1}=\frac{1}{5}\left(\begin{array}{cc} -1 & 2 \\ 3 & -1 \end{array}\right)$, so thanks to the second property:

$$B^{-1} = \left( 2C \right) ^{-1} = \frac{1}{2} C ^{-1} = \frac{1}{10}\left(\begin{array}{cc} -1 & 2 \\ 3 & -1 \end{array}\right)$$

Systems of linear equations

So far you know that when a matrix is invertible, any system of equations it appears in will have a unique solution. Now, you'll look at other equivalent conditions that are easier to check to find out if the matrix is invertible.

This means that it is enough to guarantee that the homogeneous system $A \mathbf{x}=\mathbf{0}$ has a unique solution (the trivial one) in order to solve every system $A \mathbf{x}=\mathbf{b}$ for each possible $\mathbf{b} \in \mathbb{R}^{n}$. But let's see a geometric interpretation of this result.

The geometry of invertible matrices

The above characterization of invertibility can be paraphrased in terms of the linear transformation $L_A$ associated with $A$. In particular the third point means that for every $\mathbf{b} \in \mathbb{R}^{n}$ there's a unique $\mathbf{x} \in \mathbb{R}^{n}$ such that $L_A(\mathbf{x})=\mathbf{b}$. Well, this signifies, by definition, that $L_A$ is bijective. But this in turn is equivalent to $L_A$ being an invertible function. With this you've just discovered one of the most profound relationships between matrices and linear transformations:

From a geometric point of view, this means that invertible matrices have the best possible behavior in the sense that $L_A$ does not agglomerate vectors (injective) and covers all $\mathbb{R}^{n}$ (surjective). For this reason, you can think of an invertible matrix as a transformation that reassigns the position of all points in space: each point changes position but no pair of points is sent to the same location. You can interpret this as the invertible operators "shaking" the space.

For example, a matrix of size 3 that is not invertible has an associated operator in $\mathbb{R}^3$ that collapses space into a point, a line or a plane.

But the operator of an invertible matrix doesn't collapse the space, it only changes the positions of its points.

A quick test of invertibility

Time to analyze the relationship between invertibility and the columns of the matrix. The key is that the linear system $A \mathbf{x}=\mathbf{0}$ can be rewritten in terms of the columns of $A$ as:

$$A_1 x_1 + A_2 x_2 + \dots + A_n x_n = 0$$Recall that the vectors $A_1, A_2, \dots, A_n$ are linearly independent if and only if $x_1=x_2 = \dots = x_n =0$. Now, notice that this is the same as $0$ being the only possible solution of $A \mathbf{x}=\mathbf{0}$. And you've already seen that this means that $A$ is invertible. You just proved the next result:

Linearly dependent vectors are those that contain redundant information (in the sense that one of them can be obtained through the others). So invertible matrices are those whose columns have the minimum information needed to reconstruct the whole space $\mathbb{R}^{n}$.

Up to this point you're familiar with a lot of properties of invertible matrices. You also know how to calculate them and their geometric interpretation. But first of all, how can you tell if a matrix is invertible? Although the simplest way is to check if the columns are linearly independent, there is a much more direct technique that requires less calculations and is the most widely used:

Let's see this technique in action with the following matrices:

$$\begin{pmatrix} 2 & 1 & 3 \\ 0 & 4 & -1 \\ 1 & -2 & 0 \\ \end{pmatrix} \qquad \begin{pmatrix} 3 & 0 & 1 \\ 2 & 1 & -2 \\ 0 & 4 & 3 \\ \end{pmatrix} \qquad \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{pmatrix}$$

The determinants of the first two matrices are $-17$ and $41$, respectively. Therefore both are invertible. But the determinant of the third is $0$, so it is not invertible. Quick and straightforward.

Conclusion

You just developed a lot of synonyms for invertibility, so let's go over them. The square matrix $A$ is invertible if and only if:

• The linear system $A \mathbf{x}=\mathbf{b}$ has a unique solution for all $\mathbf{b} \in \mathbb{R}^{n}$.

• $L_A$ is invertible

• The columns of $A$ form a basis for $\mathbb{R}^{n}$.

• $\det(A) \neq 0$

Some useful properties of the inverse are:

• $( cA ) ^{-1} = \frac{1}{c} A ^{-1}$ for any non-zero real number $c$.
• $\left( A^T \right ) ^{-1} = \left( A^{-1} \right ) ^{T}$.
• $\left( A^k \right ) ^{-1} = \left( A^{-1} \right ) ^{k}$ for every positive integer $k$.