One preliminary definition, and then we will have our main definition for this section.
The identity linear transformation on the vector space $W$ is defined as \begin{equation*} \ltdefn{I_W}{W}{W},\quad\quad \lt{I_W}{\vect{w}}=\vect{w} \end{equation*}
Informally, $I_W$ is the “do-nothing” function. You should check that $I_W$ is really a linear transformation, as claimed, and then compute its kernel and range to see that it is both injective and surjective. All of these facts should be straightforward to verify (Exercise IVLT.T05). With this in hand we can make our main definition.
Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. If there is a function $\ltdefn{S}{V}{U}$ such that \begin{align*} \compose{S}{T}&=I_U & \compose{T}{S}&=I_V \end{align*} then $T$ is invertible. In this case, we call $S$ the inverse of $T$ and write $S=\ltinverse{T}$.
Informally, a linear transformation $T$ is invertible if there is a companion linear transformation, $S$, which “undoes” the action of $T$. When the two linear transformations are applied consecutively (composition), in either order, the result is to have no real effect. It is entirely analogous to squaring a positive number and then taking its (positive) square root.
Here is an example of a linear transformation that is invertible. As usual at the beginning of a section, do not be concerned with where $S$ came from, just understand how it illustrates Definition IVLT.
It can be as instructive to study a linear transformation that is not invertible.
In Example ANILT you may have noticed that $T$ is not surjective, since the matrix $A$ was not in the range of $T$. And $T$ is not injective since there are two different input column vectors that $T$ sends to the matrix $B$. Linear transformations $T$ that are not surjective lead to putative inverse functions $S$ that are undefined on inputs outside of the range of $T$. Linear transformations $T$ that are not injective lead to putative inverse functions $S$ that are multiply-defined on each of their inputs. We will formalize these ideas in Theorem ILTIS.
But first notice in Definition IVLT that we only require the inverse (when it exists) to be a function. When it does exist, it too is a linear transformation.
Suppose that $\ltdefn{T}{U}{V}$ is an invertible linear transformation. Then the function $\ltdefn{\ltinverse{T}}{V}{U}$ is a linear transformation.
So when $T$ has an inverse, $\ltinverse{T}$ is also a linear transformation. Furthermore, $\ltinverse{T}$ is an invertible linear transformation and its inverse is what you might expect.
Suppose that $\ltdefn{T}{U}{V}$ is an invertible linear transformation. Then $\ltinverse{T}$ is an invertible linear transformation and $\ltinverse{\left(\ltinverse{T}\right)}=T$.