Matrix inverse

Inverse of a matrix

For a square matrix $A$, if there is a matrix $B$ of the same size such that

\[\begin{aligned} AB &= I &&\text{and} & BA &= I \end{aligned},\]

then $B$ is called an inverse matrix of $A$ (or simply an inverse of $A$). It is important to note that some matrix may not have an inverse at all.

Important non-example. Prove that the matrix

\[A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\]

has no inverse.

If an inverse of a matrix does exist, then it must be unique. This is a nontrivial fact, and we will explore the reasoning in other context. Therefore we say it is the inverse of that matrix.

For a square matrix $A$, we use the notation $A^{-1}$ for the inverse of $A$, if exists. In this case, we say $A$ is invertible (or nonsingular). The following table shows equivalent terms we use:

has an inverse	is invertible	is nonsingular
has no inverse	is not invertible	is singular

It is straightforward to verify the following:

\[\begin{aligned} (A^{-1})^{-1} &= A & (A^\top)^{-1} &= (A^{-1})^\top \\ (AB)^{-1} &= B^{-1} A^{-1} & (A^n)^{-1} &= (A^{-1})^n. \end{aligned}\]

Determine if the following matrices are invertible. If yes, find the inverse matrices.

\[\begin{aligned} A &= \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix} & B &= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} & C &= \begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix} \end{aligned}\]

Exercise. The definition of matrix inverse requires us to check both $AB = I$ and $BA = I$ before we declare $B$ to be the inverse of $A$. Do we have to check both directions? Or will $AB = I$ automatically imply $BA = I$? Explain why.

Inverse matrix and inverse function

For an $n \times n$ matrix $A$, we can consider the linear function $f : \mathbb{R}^n \to \mathbb{R}^n$ given by $f(\mathbf{x}) = A \mathbf{x}$. If $A$ has an inverse $A^{-1}$, then it is easy to verify that $f$ also has an inverse function, and this inverse function is given by $f^{-1}(\mathbf{x}) = A^{-1} \mathbf{x}$.

Inverse matrices and linear systems

Recall that we can always write a linear system as a single equations involving matrix-vector product. In particular, a linear system of $n$ equations in $n$ variables (so called “square system”) can be written as

\[A \mathbf{x} = \mathbf{b}\]

where $A$ is an $n \times n$ matrix collecting all the coefficients, $\mathbf{x}$ contains all the unknowns, and $\mathbf{b} \in \mathbb{R}^n$ represents the constant terms on the right hand side.

In this context, $A$ is invertible if and only if the linear system has a unique solution, which is

\[\mathbf{x} = A^{-1} \mathbf{b}.\]