Section IVLT  Invertible Linear Transformations

One preliminary definition, and then we will have our main definition for this section.

Definition  IDLT
Identity Linear Transformation
The identity linear transformation on the vector space $W$ is defined as

$△$

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.

Definition  IVLT
Invertible Linear Transformations
Suppose that $T:U\to V$ is a linear transformation. If there is a function $S:V\to U$ such that

$\begin{array}{llllllll}\hfill S\circ T& =I_U& \hfill T\circ S& =I_V& \hfill & & \hfill & \end{array}$

then $T$ is invertible. In this case, we call $S$ the inverse of $T$ and write $S=T^{-1}$ . $△$

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.

Example  AIVLT
An invertible linear transformation
Archetype V is the linear transformation

Define the function $S:M_{22}\to P_3$ defined by

Then

$\begin{array}{llll}\hfill \left(T\circ S\right)\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\right)& =T\left(S\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\right)\right)& \hfill & \\ \hfill & =T\left(\left(a-c-d\right)+\left(c+d\right)x+\frac{1}{2}\left(a-b-c-d\right)x^2+c{x}^3\right)& \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill \left(a-c-d\right)+\left(c+d\right)\hfill & \hfill \left(a-c-d\right)-2\left(\frac{1}{2}\left(a-b-c-d\right)\right)\hfill \\ \hfill c\hfill & \hfill \left(c+d\right)-c\hfill \end{array}\right]& \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]& \hfill & \\ \hfill & =I_{M_{22}}\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill d\hfill \end{array}\right]\right)& \hfill & \\ \multicolumn{4}{c}{\text{ And}}\\ \\ \hfill \left(S\circ T\right)\left(a+bx+c{x}^2+d{x}^3\right)& =S\left(T\left(a+bx+c{x}^2+d{x}^3\right)\right)& \hfill & \\ \hfill & =S\left(\left[\begin{array}{cc}\hfill a+b\hfill & \hfill a-2c\hfill \\ \hfill d\hfill & \hfill b-d\hfill \end{array}\right]\right)& \hfill & \\ \hfill & =\left(\left(a+b\right)-d-\left(b-d\right)\right)+\left(d+\left(b-d\right)\right)x& \hfill & \\ \hfill & +\left(\frac{1}{2}\left(\left(a+b\right)-\left(a-2c\right)-d-\left(b-d\right)\right)\right)x^2+\left(d\right)x^3& \hfill & \\ \hfill & =a+bx+c{x}^2+d{x}^3& \hfill & \\ \hfill & =I_{P_3}\left(a+bx+c{x}^2+d{x}^3\right)& \hfill & \end{array}$

For now, understand why these computations show that $T$ is invertible, and that $S=T^{-1}$ . Maybe even be amazed by how $S$ works so perfectly in concert with $T$ ! We will see later just how to arrive at the correct form of $S$ (when it is possible). $⊠$

It can be as instructive to study a linear transformation that is not invertible.

Example  ANILT
A non-invertible linear transformation
Consider the linear transformation $T:{ℂ}^3\to M_{22}$ defined by

Suppose we were to search for an inverse function $S:M_{22}\to {ℂ}^3$ .

First verify that the $2\times 2$ matrix $A=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 3\hfill \\ \hfill 8\hfill & \hfill 2\hfill \end{array}\right]$ is not in the range of $T$ . This will amount to finding an input to $T$ , $\left[\begin{array}{c}\hfill a\hfill \\ \hfill b\hfill \\ \hfill c\hfill \end{array}\right]$ , such that

$\begin{array}{llll}\hfill a-b& =5& \hfill & \\ \hfill 2a+2b+c& =3& \hfill & \\ \hfill 3a+b+c& =8& \hfill & \\ \hfill -2a-6b-2c& =2& \hfill & \end{array}$

As this system of equations is inconsistent, there is no input column vector, and $A\notin \mathcal{ℛ}\left(T\right)$ . How should we define $S\left(A\right)$ ? Note that

So any definition we would provide for $S\left(A\right)$ must then be a column vector that $T$ sends to $A$ and we would have $A\in \mathcal{ℛ}\left(T\right)$ , contrary to the definition of $T$ . This is enough to see that there is no function $S$ that will allow us to conclude that $T$ is invertible, since we cannot provide a consistent definition for $S\left(A\right)$ if we assume $T$ is invertible.

Even though we now know that $T$ is not invertible, let’s not leave this example just yet. Check that

$\begin{array}{llllllll}\hfill T\left(\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 4\hfill \end{array}\right]\right)& =\left[\begin{array}{cc}\hfill 3\hfill & \hfill 2\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]=B& \hfill T\left(\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -3\hfill \\ \hfill 8\hfill \end{array}\right]\right)& =\left[\begin{array}{cc}\hfill 3\hfill & \hfill 2\hfill \\ \hfill 5\hfill & \hfill 2\hfill \end{array}\right]=B& \hfill & & \hfill & \end{array}$

How would we define $S\left(B\right)$ ?

$\begin{array}{llll}\hfill S\left(B\right)& =S\left(T\left(\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 4\hfill \end{array}\right]\right)\right)=\left(S\circ T\right)\left(\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 4\hfill \end{array}\right]\right)=I_{{ℂ}^3}\left(\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 4\hfill \end{array}\right]\right)=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill -2\hfill \\ \hfill 4\hfill \end{array}\right]& \hfill & \\ \multicolumn{4}{c}{\text{ or}}\\ \\ \hfill S\left(B\right)& =S\left(T\left(\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -3\hfill \\ \hfill 8\hfill \end{array}\right]\right)\right)=\left(S\circ T\right)\left(\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -3\hfill \\ \hfill 8\hfill \end{array}\right]\right)=I_{{ℂ}^3}\left(\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -3\hfill \\ \hfill 8\hfill \end{array}\right]\right)=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill -3\hfill \\ \hfill 8\hfill \end{array}\right]& \hfill & \end{array}$

Which definition should we provide for $S\left(B\right)$ ? Both are necessary. But then $S$ is not a function. So we have a second reason to know that there is no function $S$ that will allow us to conclude that $T$ is invertible. It happens that there are infinitely many column vectors that $S$ would have to take to $B$ . Construct the kernel of $T$ ,

Now choose either of the two inputs used above for $T$ and add to it a scalar multiple of the basis vector for the kernel of $T$ . For example,

then verify that $T\left(x\right)=B$ . Practice creating a few more inputs for $T$ that would be sent to $B$ , and see why it is hopeless to think that we could ever provide a reasonable definition for $S\left(B\right)$ ! There is a “whole subspace’s worth” of values that $S\left(B\right)$ would have to take on. $⊠$

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.

Theorem  ILTLT
Inverse of a Linear Transformation is a Linear Transformation
Suppose that $T:U\to V$ is an invertible linear transformation. Then the function $T^{-1}:V\to U$ is a linear transformation. $\square$

Proof   We work through verifying Definition LT for $T^{-1}$ , using the fact that $T$ is a linear transformation to obtain the second equality in each half of the proof. To this end, suppose $x,y\in V$ and $\alpha \in {ℂ}^{}$ .

$\begin{array}{llllllll}\hfill T^{-1}\left(x+y\right)& =T^{-1}\left(T\left(T^{-1}\left(x\right)\right)+T\left(T^{-1}\left(y\right)\right)\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(T\left(T^{-1}\left(x\right)+T^{-1}\left(y\right)\right)\right)& \hfill & \text{Definition LT}& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(x\right)+T^{-1}\left(y\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \multicolumn{8}{c}{\text{ Now check the second defining property of a linear transformation for}{T}^{-1}\text{,}}\\ \\ \hfill T^{-1}\left(\alpha x\right)& =T^{-1}\left(\alpha T\left(T^{-1}\left(x\right)\right)\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(T\left(\alpha T^{-1}\left(x\right)\right)\right)& \hfill & \text{Definition LT}& \hfill & & \hfill & \\ \hfill & =\alpha T^{-1}\left(x\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \end{array}$

$■$

So $T^{-1}$ fulfills the requirements of Definition LT and is therefore a linear transformation. So when $T$ has an inverse, $T^{-1}$ is also a linear transformation. Additionally, $T^{-1}$ is invertible and its inverse is what you might expect.

Theorem  IILT
Inverse of an Invertible Linear Transformation
Suppose that $T:U\to V$ is an invertible linear transformation. Then $T^{-1}$ is an invertible linear transformation and ${\left(T^{-1}\right)}^{-1}=T$ . $\square$

Proof   Because $T$ is invertible, Definition IVLT tells us there is a function $T^{-1}:V\to U$ such that

$\begin{array}{llllllll}\hfill T^{-1}\circ T& =I_U& \hfill T\circ T^{-1}& =I_V& \hfill & & \hfill & \end{array}$

Additionally, Theorem ILTLT tells us that $T^{-1}$ is more than just a function, it is a linear transformation. Now view these two statements as properties of the linear transformation $T^{-1}$ . In light of Definition IVLT, they together say that $T^{-1}$ is invertible (let $T$ play the role of $S$ in the statement of the definition). Furthermore, the inverse of $T^{-1}$ is then $T$ , i.e.  ${\left(T^{-1}\right)}^{-1}=T$ . $■$

We now know what an inverse linear transformation is, but just which linear transformations have inverses? Here is a theorem we have been preparing for all chapter long.

Theorem  ILTIS
Invertible Linear Transformations are Injective and Surjective
Suppose $T:U\to V$ is a linear transformation. Then $T$ is invertible if and only if $T$ is injective and surjective. $\square$

Proof   ( $\Rightarrow$ ) Since $T$ is presumed invertible, we can employ its inverse, $T^{-1}$ ( Definition IVLT). To see that $T$ is injective, suppose $x,y\in U$ and assume that $T\left(x\right)=T\left(y\right)$ ,

$\begin{array}{llllllll}\hfill x& =I_U\left(x\right)& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \\ \hfill & =\left(T^{-1}\circ T\right)\left(x\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(T\left(x\right)\right)& \hfill & \text{Definition LTC}& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(T\left(y\right)\right)& \hfill & \text{Definition ILT}& \hfill & & \hfill & \\ \hfill & =\left(T^{-1}\circ T\right)\left(y\right)& \hfill & \text{Definition LTC}& \hfill & & \hfill & \\ \hfill & =I_U\left(y\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =y& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \\ \multicolumn{8}{c}{\text{ So by }\text{Definition ILT}T\text{ is injective. To check that }T\text{ is surjective, suppose}v\in V\text{. Then }{T}^{-1}\left(v\right)\text{ is a vector in }U\text{. Compute }}\\ \\ \hfill T\left(T^{-1}\left(v\right)\right)& =\left(T\circ T^{-1}\right)\left(v\right)& \hfill & \text{Definition LTC}& \hfill & & \hfill & \\ \hfill & =I_V\left(v\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =v& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \end{array}$

So there is an element from $U$ , when used as an input to $T$ (namely $T^{-1}\left(v\right)$ ) that produces the desired output, $v$ , and hence $T$ is surjective by Definition SLT.

( $\Leftarrow$ ) Now assume that $T$ is both injective and surjective. We will build a function $S:V\to U$ that will establish that $T$ is invertible. To this end, choose any $v\in V$ . Since $T$ is surjective, Theorem RSLT says $\mathcal{ℛ}\left(T\right)=V$ , so we have $v\in \mathcal{ℛ}\left(T\right)$ . Theorem RPI says that the pre-image of $v$ , $T^{-1}\left(v\right)$ , is nonempty. So we can choose a vector from the pre-image of $v$ , say $u$ . In other words, there exists $u\in T^{-1}\left(v\right)$ .

Since $T^{-1}\left(v\right)$ is non-empty, Theorem KPI then says that

However, because $T$ is injective, by Theorem KILT the kernel is trivial, $\mathcal{K}\left(T\right)=\left\{0\right\}$ . So the pre-image is a set with just one element, $T^{-1}\left(v\right)=\left\{u\right\}$ . Now we can define $S$ by $S\left(v\right)=u$ . This is the key to this half of this proof. Normally the preimage of a vector from the codomain might be an empty set, or an infinite set. But surjectivity requires that the preimage not be empty, and then injectivity limits the preimage to a singleton. Since our choice of $v$ was arbitrary, we know that every pre-image for $T$ is a set with a single element. This allows us to construct $S$ as a function. Now that it is defined, verifying that it is the inverse of $T$ will be easy. Here we go.

Choose $u\in U$ . Define $v=T\left(u\right)$ . Then $T^{-1}\left(v\right)=\left\{u\right\}$ , so that $S\left(v\right)=u$ and,

and since our choice of $u$ was arbitrary we have function equality, $S\circ T=I_U$ .

Now choose $v\in V$ . Define $u$ to be the single vector in the set $T^{-1}\left(v\right)$ , in other words, $u=S\left(v\right)$ . Then $T\left(u\right)=v$ , so

and since our choice of $v$ was arbitrary we have function equality, $T\circ S=I_V$ . $■$

When a linear transformation is both injective and surjective, the pre-image of any element of the codomain is a set of size one (a “singleton”). This fact allowed us to construct the inverse linear transformation in one half of the proof of Theorem ILTIS (see Technique C). We can follow this approach to construct the inverse of a specific linear transformation, as the next example shows.

Example  CIVLT
Computing the Inverse of a Linear Transformations
Consider the linear transformation $T:S_{22}\to P_2$ defined by

$\begin{array}{llll}\hfill T\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill \end{array}\right]\right)& =\left(a+b+c\right)+\left(-a+2c\right)x+\left(2a+3b+6c\right)x^2& \hfill & \end{array}$

$T$ is invertible, which you are able to verify, perhaps by determining that the kernel of $T$ is empty and the range of $T$ is all of $P_2$ . This will be easier once we have Theorem RPNDD, which appears later in this section.

By Theorem ILTIS we know $T^{-1}$ exists, and it will be critical shortly to realize that $T^{-1}$ is automatically known to be a linear transformation as well ( Theorem ILTLT). To determine the complete behavior of $T^{-1}:P_2\to S_{22}$ we can simply determine its action on a basis for the domain, $P_2$ . This is the substance of Theorem LTDB, and an excellent example of its application. Choose any basis of $P_2$ , the simpler the better, such as $B=\left\{1,x,x^2\right\}$ . Values of $T^{-1}$ for these three basis elements will be the single elements of their preimages. In turn, we have

$\begin{array}{llllll}\hfill T^{-1}\left(1\right):& & \hfill & & \hfill & \\ \hfill & & \hfill T\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill \end{array}\right]\right)& =1+0x+0x^2& \hfill & & \hfill & \\ \hfill & & \hfill \left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 6\hfill & \hfill 0\hfill \end{array}\right]& \underset{}{\overset{\text{ RREF}}{\to }}\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -6\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 10\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -3\hfill \end{array}\right]& \hfill & & \hfill & \\ \hfill & \text{ (preimage)}& \hfill T^{-1}\left(1\right)& =\left\{\left[\begin{array}{cc}\hfill -6\hfill & \hfill 10\hfill \\ \hfill 10\hfill & \hfill -3\hfill \end{array}\right]\right\}& \hfill & & \hfill & \\ \hfill & \text{ (function)}& \hfill T^{-1}\left(1\right)& =\left[\begin{array}{cc}\hfill -6\hfill & \hfill 10\hfill \\ \hfill 10\hfill & \hfill -3\hfill \end{array}\right]& \hfill & & \hfill & \\ \hfill T^{-1}\left(x\right):& & \hfill & & \hfill & \\ \hfill & & \hfill T\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill \end{array}\right]\right)& =0+1x+0x^2& \hfill & & \hfill & \\ \hfill & & \hfill \left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 6\hfill & \hfill 0\hfill \end{array}\right]& \underset{}{\overset{\text{ RREF}}{\to }}\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill -1\hfill \end{array}\right]& \hfill & & \hfill & \\ \hfill & \text{ (preimage)}& \hfill T^{-1}\left(x\right)& =\left\{\left[\begin{array}{cc}\hfill -3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill -1\hfill \end{array}\right]\right\}& \hfill & & \hfill & \\ \hfill & \text{ (function)}& \hfill T^{-1}\left(x\right)& =\left[\begin{array}{cc}\hfill -3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill -1\hfill \end{array}\right]& \hfill & & \hfill & \\ \hfill T^{-1}\left(x^2\right):& & \hfill & & \hfill & \\ \hfill & & \hfill T\left(\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill \end{array}\right]\right)& =0+0x+1x^2& \hfill & & \hfill & \\ \hfill & & \hfill \left[\begin{array}{cccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 6\hfill & \hfill 1\hfill \end{array}\right]& \underset{}{\overset{\text{ RREF}}{\to }}\left[\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]& \hfill & & \hfill & \\ \hfill & \text{ (preimage)}& \hfill T^{-1}\left(x^2\right)& =\left\{\left[\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -3\hfill & \hfill 1\hfill \end{array}\right]\right\}& \hfill & & \hfill & \\ \hfill & \text{ (function)}& \hfill T^{-1}\left(x^2\right)& =\left[\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -3\hfill & \hfill 1\hfill \end{array}\right]& \hfill & & \hfill & \end{array}$

Theorem LTDB says, informally, “it is enough to know what a linear transformation does to a basis.” Formally, we have the outputs of $T^{-1}$ for a basis, so by Theorem LTDB there is a unique linear transformation with these outputs. So we put this information to work. The key step here is that we can convert any element of $P_2$ into a linear combination of the elements of the basis $B$ (Theorem VRRB). We are after a “formula” for the value of $T^{-1}$ on a generic element of $P_2$ , say $p+qx+r{x}^2$ .

$\begin{array}{llllllll}\hfill T^{-1}\left(p+qx+r{x}^2\right)& =T^{-1}\left(p\left(1\right)+q\left(x\right)+r\left(x^2\right)\right)& \hfill & \text{Theorem VRRB}& \hfill & & \hfill & \\ \hfill & =p{T}^{-1}\left(1\right)+q{T}^{-1}\left(x\right)+r{T}^{-1}\left(x^2\right)& \hfill & \text{Theorem LTLC}& \hfill & & \hfill & \\ \hfill & =p\left[\begin{array}{cc}\hfill -6\hfill & \hfill 10\hfill \\ \hfill 10\hfill & \hfill -3\hfill \end{array}\right]+q\left[\begin{array}{cc}\hfill -3\hfill & \hfill 4\hfill \\ \hfill 4\hfill & \hfill -1\hfill \end{array}\right]+r\left[\begin{array}{cc}\hfill 2\hfill & \hfill -3\hfill \\ \hfill -3\hfill & \hfill 1\hfill \end{array}\right]& \hfill & & \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill -6p-3q+2r\hfill & \hfill 10p+4q-3r\hfill \\ \hfill 10p+4q-3r\hfill & \hfill -3p-q+r\hfill \end{array}\right]& \hfill & & \hfill & \end{array}$

Notice how a linear combination in the domain of $T^{-1}$ has been translated into a linear combination in the codomain of $T^{-1}$ since we know $T^{-1}$ is a linear transformation by Theorem ILTLT.

Also, notice how the augmented matrices used to determine the three pre-images could be combined into one calculation of a matrix in extended echelon form, reminiscent of a procedure we know for computing the inverse of a matrix (see Example CMI). Hmmmm. $⊠$

We will make frequent use of the characterization of invertible linear transformations provided by Theorem ILTIS. The next theorem is a good example of this, and we will use it often, too.

Theorem  CIVLT
Composition of Invertible Linear Transformations
Suppose that $T:U\to V$ and $S:V\to W$ are invertible linear transformations. Then the composition, $\left(S\circ T\right):U\to W$ is an invertible linear transformation. $\square$

Proof   Since $S$ and $T$ are both linear transformations, $S\circ T$ is also a linear transformation by Theorem CLTLT. Since $S$ and $T$ are both invertible, Theorem ILTIS says that $S$ and $T$ are both injective and surjective. Then Theorem CILTI says $S\circ T$ is injective, and Theorem CSLTS says $S\circ T$ is surjective. Now apply the “other half” of Theorem ILTIS and conclude that $S\circ T$ is invertible. $■$

When a composition is invertible, the inverse is easy to construct.

Theorem  ICLT
Inverse of a Composition of Linear Transformations
Suppose that $T:U\to V$ and $S:V\to W$ are invertible linear transformations. Then $S\circ T$ is invertible and ${\left(S\circ T\right)}^{-1}=T^{-1}\circ S^{-1}$ . $\square$

Proof   Compute, for all $w\in W$

$\begin{array}{llllll}\hfill \left(\left(S\circ T\right)\circ \left(T^{-1}\circ S^{-1}\right)\right)\left(w\right)& =S\left(T\left(T^{-1}\left(S^{-1}\left(w\right)\right)\right)\right)& \hfill & & \hfill & \\ \hfill & =S\left(I_V\left(S^{-1}\left(w\right)\right)\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =S\left(S^{-1}\left(w\right)\right)& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \\ \hfill & =w& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =I_W\left(w\right)& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \\ \multicolumn{8}{c}{\text{ so }\left(S\circ T\right)\circ \left(T^{-1}\circ S^{-1}\right)=I_W\text{ and also}}\\ \\ \hfill \left(\left(T^{-1}\circ S^{-1}\right)\circ \left(S\circ T\right)\right)\left(u\right)& =T^{-1}\left(S^{-1}\left(S\left(T\left(u\right)\right)\right)\right)& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(I_V\left(T\left(u\right)\right)\right)& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =T^{-1}\left(T\left(u\right)\right)& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \\ \hfill & =u& \hfill & \text{Definition IVLT}& \hfill & & \hfill & \\ \hfill & =I_U\left(u\right)& \hfill & \text{Definition IDLT}& \hfill & & \hfill & \end{array}$

so $\left(T^{-1}\circ S^{-1}\right)\circ \left(S\circ T\right)=I_U$ . By Definition IVLT, $S\circ T$ is invertible and ${\left(S\circ T\right)}^{-1}=T^{-1}\circ S^{-1}$ . $■$

Notice that this theorem not only establishes what the inverse of $S\circ T$ is, it also duplicates the conclusion of Theorem CIVLT and also establishes the invertibility of $S\circ T$ . But somehow, the proof of Theorem CIVLT is nicer way to get this property.

Does Theorem ICLT remind you of the flavor of any theorem we have seen about matrices? (Hint: Think about getting dressed.) Hmmmm.

A vector space is defined (Definition VS) as a set of objects (“vectors”) endowed with a definition of vector addition ( $+$ ) and a definition of scalar multiplication (written with juxtaposition). Many of our definitions about vector spaces involve linear combinations (Definition LC), such as the span of a set (Definition SS) and linear independence (Definition LI). Other definitions are built up from these ideas, such as bases (Definition B) and dimension (Definition D). The defining properties of a linear transformation require that a function “respect” the operations of the two vector spaces that are the domain and the codomain (Definition LT). Finally, an invertible linear transformation is one that can be “undone” — it has a companion that reverses its effect. In this subsection we are going to begin to roll all these ideas into one.

A vector space has “structure” derived from definitions of the two operations and the requirement that these operations interact in ways that satisfy the ten properties of Definition VS. When two different vector spaces have an invertible linear transformation defined between them, then we can translate questions about linear combinations (spans, linear independence, bases, dimension) from the first vector space to the second. The answers obtained in the second vector space can then be translated back, via the inverse linear transformation, and interpreted in the setting of the first vector space. We say that these invertible linear transformations “preserve structure.” And we say that the two vector spaces are “structurally the same.” The precise term is “isomorphic,” from Greek meaning “of the same form.” Let’s begin to try to understand this important concept.

Definition  IVS
Isomorphic Vector Spaces
Two vector spaces $U$ and $V$ are isomorphic if there exists an invertible linear transformation $T$ with domain $U$ and codomain $V$ , $T:U\to V$ . In this case, we write $U\cong V$ , and the linear transformation $T$ is known as an isomorphism between $U$ and $V$ . $△$

A few comments on this definition. First, be careful with your language ( Technique L). Two vector spaces are isomorphic, or not. It is a yes/no situation and the term only applies to a pair of vector spaces. Any invertible linear transformation can be called an isomorphism, it is a term that applies to functions. Second, a given pair of vector spaces there might be several different isomorphisms between the two vector spaces. But it only takes the existence of one to call the pair isomorphic. Third, $U$ isomorphic to $V$ , or $V$ isomorphic to $U$ ? Doesn’t matter, since the inverse linear transformation will provide the needed isomorphism in the “opposite” direction. Being “isomorphic to” is an equivalence relation on the set of all vector spaces (see Theorem SER for a reminder about equivalence relations).

Example  IVSAV
Isomorphic vector spaces, Archetype V
Archetype V is a linear transformation from $P_3$ to $M_{22}$ ,

Since it is injective and surjective, Theorem ILTIS tells us that it is an invertible linear transformation. By Definition IVS we say $P_3$ and $M_{22}$ are isomorphic.

At a basic level, the term “isomorphic” is nothing more than a codeword for the presence of an invertible linear transformation. However, it is also a description of a powerful idea, and this power only becomes apparent in the course of studying examples and related theorems. In this example, we are led to believe that there is nothing “structurally” different about $P_3$ and $M_{22}$ . In a certain sense they are the same. Not equal, but the same. One is as good as the other. One is just as interesting as the other.

Here is an extremely basic application of this idea. Suppose we want to compute the following linear combination of polynomials in $P_3$ ,

Rather than doing it straight-away (which is very easy), we will apply the transformation $T$ to convert into a linear combination of matrices, and then compute in $M_{22}$ according to the definitions of the vector space operations there (Example VSM),

$\begin{array}{llllll}\hfill & T\left(5\left(2+3x-4x^2+5x^3\right)+\left(-3\right)\left(3-5x+3x^2+x^3\right)\right)& \hfill & & \hfill & \\ \hfill & =5T\left(2+3x-4x^2+5x^3\right)+\left(-3\right)T\left(3-5x+3x^2+x^3\right)& \hfill & \text{Theorem LTLC}& \hfill & & \hfill & \\ \hfill & =5\left[\begin{array}{cc}\hfill 5\hfill & \hfill 10\hfill \\ \hfill 5\hfill & \hfill -2\hfill \end{array}\right]+\left(-3\right)\left[\begin{array}{cc}\hfill -2\hfill & \hfill -3\hfill \\ \hfill 1\hfill & \hfill -6\hfill \end{array}\right]& \hfill & \text{ Definition of }T& \hfill & & \hfill & \\ \hfill & =\left[\begin{array}{cc}\hfill 31\hfill & \hfill 59\hfill \\ \hfill 22\hfill & \hfill 8\hfill \end{array}\right]& \hfill & \text{ Operations in }{M}_{22}& \hfill & & \hfill & \end{array}$

Now we will translate our answer back to $P_3$ by applying $T^{-1}$ , which we found in Example AIVLT,

We compute,

which is, as expected, exactly what we would have computed for the original linear combination had we just used the definitions of the operations in $P_3$ (Example VSP). Notice this is meant only as an illustration and not a suggested route for doing this particular computation. $⊠$

Checking the dimensions of two vector spaces can be a quick way to establish that they are not isomorphic. Here’s the theorem.

Theorem  IVSED
Isomorphic Vector Spaces have Equal Dimension
Suppose $U$ and $V$ are isomorphic vector spaces. Then $dim\left(U\right)=dim\left(V\right)$ . $\square$

Proof   If $U$ and $V$ are isomorphic, there is an invertible linear transformation $T:U\to V$ ( Definition IVS). $T$ is injective by Theorem ILTIS and so by Theorem ILTD, $dim\left(U\right)\le dim\left(V\right)$ . Similarly, $T$ is surjective by Theorem ILTIS and so by Theorem SLTD, $dim\left(U\right)\ge dim\left(V\right)$ . The net effect of these two inequalities is that $dim\left(U\right)=dim\left(V\right)$ . $■$

The contrapositive of Theorem IVSED says that if $U$ and $V$ have different dimensions, then they are not isomorphic. Dimension is the simplest “structural” characteristic that will allow you to distinguish non-isomorphic vector spaces. For example $P_6$ is not isomorphic to $M_{34}$ since their dimensions (7 and 12, respectively) are not equal. With tools developed in Section VR we will be able to establish that the converse of Theorem IVSED is true. Think about that one for a moment.

Just as a matrix has a rank and a nullity, so too do linear transformations. And just like the rank and nullity of a matrix are related (they sum to the number of columns, Theorem RPNC) the rank and nullity of a linear transformation are related. Here are the definitions and theorems, see the Archetypes ( Appendix A) for loads of examples.

Definition  ROLT
Rank Of a Linear Transformation
Suppose that $T:U\to V$ is a linear transformation. Then the rank of $T$ , $r\left(T\right)$ , is the dimension of the range of $T$ ,

(This definition contains Notation ROLT.) $△$

Definition  NOLT
Nullity Of a Linear Transformation
Suppose that $T:U\to V$ is a linear transformation. Then the nullity of $T$ , $n\left(T\right)$ , is the dimension of the kernel of $T$ ,

(This definition contains Notation NOLT.) $△$

Here are two quick theorems.

Theorem  ROSLT
Rank Of a Surjective Linear Transformation
Suppose that $T:U\to V$ is a linear transformation. Then the rank of $T$ is the dimension of $V$ , $r\left(T\right)=dim\left(V\right)$ , if and only if $T$ is surjective. $\square$

Proof   By Theorem RSLT, $T$ is surjective if and only if $\mathcal{ℛ}\left(T\right)=V$ . Applying Definition ROLT, $\mathcal{ℛ}\left(T\right)=V$ if and only if $r\left(T\right)=dim\left(\mathcal{ℛ}\left(T\right)\right)=dim\left(V\right)$ . $■$

Theorem  NOILT
Nullity Of an Injective Linear Transformation
Suppose that $T:U\to V$ is a linear transformation. Then the nullity of $T$ is zero, $n\left(T\right)=0$ , if and only if $T$ is injective. $\square$

Proof   By Theorem KILT, $T$ is injective if and only if $\mathcal{K}\left(T\right)=\left\{0\right\}$ . Applying Definition NOLT, $\mathcal{K}\left(T\right)=\left\{0\right\}$ if and only if $n\left(T\right)=0$ . $■$

Just as injectivity and surjectivity come together in invertible linear transformations, there is a clear relationship between rank and nullity of a linear transformation. If one is big, the other is small.

Theorem  RPNDD
Rank Plus Nullity is Domain Dimension
Suppose that $T:U\to V$ is a linear transformation. Then

$\square$

Proof   Let $r=r\left(T\right)$ and $s=n\left(T\right)$ . Suppose that $R=\left\{v_1,v_2,v_3,\dots ,v_r\right\}\subseteq V$ is a basis of the range of $T$ , $\mathcal{ℛ}\left(T\right)$ , and $S=\left\{u_1,u_2,u_3,\dots ,u_s\right\}\subseteq U$ is a basis of the kernel of $T$ , $\mathcal{K}\left(T\right)$ . Note that $R$ and $S$ are possibly empty, which means that some of the sums in this proof are “empty” and are equal to the zero vector.

Because the elements of $R$ are all in the range of $T$ , each must have a non-empty pre-image by Theorem RPI. Choose vectors $w_i\in U$ , $1\le i\le r$ such that $w_i\in T^{-1}\left(v_i\right)$ . So $T\left(w_i\right)=v_i$ , $1\le i\le r$ . Consider the set

We claim that $B$ is a basis for $U$ .

To establish linear independence for $B$ , begin with a relation of linear dependence on $B$ . So suppose there are scalars $a_1,a_2,a_3,\dots ,a_s$ and $b_1,b_2,b_3,\dots ,b_r$

Then

$\begin{array}{llllllll}\hfill 0& =T\left(0\right)& \hfill & \text{Theorem LTTZZ}& \hfill & & \hfill & \\ \hfill & =T\left(a_1{u}_1+a_2{u}_2+a_3{u}_3+\cdots +a_s{u}_s+\right.& \hfill & & \hfill & \\ \hfill & \left.b_1{w}_1+b_2{w}_2+b_3{w}_3+\cdots +b_r{w}_r\right)& \hfill & \text{Definition LI}& \hfill & & \hfill & \\ \hfill & =a_{1}T\left(u_1\right)+a_{2}T\left(u_2\right)+a_{3}T\left(u_3\right)+\cdots +a_{s}T\left(u_s\right)+& \hfill & & \hfill & \\ \hfill & b_{1}T\left(w_1\right)+b_{2}T\left(w_2\right)+b_{3}T\left(w_3\right)+\cdots +b_{r}T\left(w_r\right)& \hfill & \text{Theorem LTLC}& \hfill & & \hfill & \\ \hfill & =a_{1}0+a_{2}0+a_{3}0+\cdots +a_{s}0+& \hfill & & \hfill & \\ \hfill & b_{1}T\left(w_1\right)+b_{2}T\left(w_2\right)+b_{3}T\left(w_3\right)+\cdots +b_{r}T\left(w_r\right)& \hfill & \text{Definition KLT}& \hfill & & \hfill & \\ \hfill & =0+0+0+\cdots +0+& \hfill & & \hfill & \\ \hfill & b_{1}T\left(w_1\right)+b_{2}T\left(w_2\right)+b_{3}T\left(w_3\right)+\cdots +b_{r}T\left(w_r\right)& \hfill & \text{Theorem ZVSM}& \hfill & & \hfill & \\ \hfill & =b_{1}T\left(w_1\right)+b_{2}T\left(w_2\right)+b_{3}T\left(w_3\right)+\cdots +b_{r}T\left(w_r\right)& \hfill & \text{Property Z}& \hfill & & \hfill & \\ \hfill & =b_1{v}_1+b_2{v}_2+b_3{v}_3+\cdots +b_r{v}_r& \hfill & \text{Definition PI}& \hfill & & \hfill & \end{array}$

This is a relation of linear dependence on $R$ (Definition RLD), and since $R$ is a linearly independent set (Definition LI), we see that $b_1=b_2=b_3=\dots =b_r=0$ . Then the original relation of linear dependence on $B$ becomes

$\begin{array}{llllll}\hfill 0& =a_1{u}_1+a_2{u}_2+a_3{u}_3+\cdots +a_s{u}_s+0w_1+0w_2+\dots +0w_r& \hfill & & \hfill & \\ \hfill & =a_1{u}_1+a_2{u}_2+a_3{u}_3+\cdots +a_s{u}_s+0+0+\dots +0& \hfill & \text{Theorem ZSSM}& \hfill & & \hfill & \\ \hfill & =a_1{u}_1+a_2{u}_2+a_3{u}_3+\cdots +a_s{u}_s& \hfill & \text{Property Z}& \hfill & & \hfill & \end{array}$

But this is again a relation of linear independence (Definition RLD), now on the set $S$ . Since $S$ is linearly independent (Definition LI), we have $a_1=a_2=a_3=\dots =a_r=0$ . Since we now know that all the scalars in the relation of linear dependence on $B$ must be zero, we have established the linear independence of $S$ through Definition LI.

To now establish that $B$ spans $U$ , choose an arbitrary vector $u\in U$ . Then $T\left(u\right)\in R\left(T\right)$ , so there are scalars $c_1,c_2,c_3,\dots ,c_r$ such that

Use the scalars $c_1,c_2,c_3,\dots ,c_r$ to define a vector $y\in U$ ,

Then

$\begin{array}{llllllll}\hfill T\left(u-y\right)& =T\left(u\right)-T\left(y\right)& \hfill & \text{Theorem LTLC}& \hfill & & \hfill & \\ \hfill & =T\left(u\right)-T\left(c_1{w}_1+c_2{w}_2+c_3{w}_3+\cdots +c_r{w}_r\right)& \hfill & \text{ Substitution}& \hfill & & \hfill & \\ \hfill & =T\left(u\right)-\left(c_{1}T\left(w_1\right)+c_{2}T\left(w_2\right)+\cdots +c_{r}T\left(w_r\right)\right)& \hfill & \text{Theorem LTLC}& \hfill & & \hfill & \\ \hfill & =T\left(u\right)-\left(c_1{v}_1+c_2{v}_2+c_3{v}_3+\cdots +c_r{v}_r\right)& \hfill & w_i\in T^{-1}\left(v_i\right)& \hfill & & \hfill & \\ \hfill & =T\left(u\right)-T\left(u\right)& \hfill & \text{ Substitution}& \hfill & & \hfill & \\ \hfill & =0& \hfill & \text{Property AI}& \hfill & & \hfill & \end{array}$

So the vector $u-y$ is sent to the zero vector by $T$ and hence is an element of the kernel of $T$ . As such it can be written as a linear combination of the basis vectors for $\mathcal{K}\left(T\right)$ , the elements of the set $S$ . So there are scalars $d_1,d_2,d_3,\dots ,d_s$ such that

Then

$\begin{array}{llll}\hfill u& =\left(u-y\right)+y& \hfill & \\ \hfill & =d_1{u}_1+d_2{u}_2+d_3{u}_3+\cdots +d_s{u}_s+c_1{w}_1+c_2{w}_2+c_3{w}_3+\cdots +c_r{w}_r& \hfill & \end{array}$

This says that for any vector, $u$ , from $U$ , there exist scalars ( $d_1,d_2,d_3,\dots ,d_s,c_1,c_2,c_3,\dots ,c_r$ ) that form $u$ as a linear combination of the vectors in the set $B$ . In other words, $B$ spans $U$ (Definition SS).

So $B$ is a basis (Definition B) of $U$ with $s+r$ vectors, and thus

as desired. $■$

Theorem RPNC said that the rank and nullity of a matrix sum to the number of columns of the matrix. This result is now an easy consequence of Theorem RPNDD when we consider the linear transformation $T:{ℂ}^n\to {ℂ}^m$ defined with the $m\times n$ matrix $A$ by $T\left(x\right)=Ax$ . The range and kernel of $T$ are identical to the column space and null space of the matrix $A$ ( Exercise ILT.T20, Exercise SLT.T20), so the rank and nullity of the matrix $A$ are identical to the rank and nullity of the linear transformation $T$ . The dimension of the domain of $T$ is the dimension of ${ℂ}^n$ , exactly the number of columns for the matrix $A$ .

This theorem can be especially useful in determining basic properties of linear transformations. For example, suppose that $T:{ℂ}^6\to {ℂ}^6$ is a linear transformation and you are able to quickly establish that the kernel is trivial. Then $n\left(T\right)=0$ . First this means that $T$ is injective by Theorem NOILT. Also, Theorem RPNDD becomes

So the rank of $T$ is equal to the rank of the codomain, and by Theorem ROSLT we know $T$ is surjective. Finally, we know $T$ is invertible by Theorem ILTIS. So from the determination that the kernel is trivial, and consideration of various dimensions, the theorems of this section allow us to conclude the existence of an inverse linear transformation for $T$ .

Similarly, Theorem RPNDD can be used to provide alternative proofs for Theorem ILTD, Theorem SLTD and Theorem IVSED. It would be an interesting exercise to construct these proofs.

It would be instructive to study the archetypes that are linear transformations and see how many of their properties can be deduced just from considering only the dimensions of the domain and codomain. Then add in just knowledge of either the nullity or rank, and so how much more you can learn about the linear transformation. The table preceding all of the archetypes ( Appendix A) could be a good place to start this analysis.