Section LT  Linear Transformations

From A First Course in Linear Algebra
Version 1.07
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

Here comes a key definition.

Subsection LT: Linear Transformations

Definition LT
Linear Transformation
A linear transformation, $T:U\mapsto V$, is a function that carries elements of the vector space $U$ (called the domain) to the vector space $V$ (called the codomain), and which has two additional properties

1. $T\left(u_1+u_2\right)=T\left(u_1\right)+T\left(u_2\right)$ for all $u_1,u_2\in U$
2. $T\left(\alpha u\right)=\alpha T\left(u\right)$ for all $u\in U$ and all $\alpha \in {ℂ}^{}$

(This definition contains Notation LT.) $△$

The two defining conditions in the definition of a linear transformation should “feel linear,” whatever that means. Conversely, these two conditions could be taken as a exactly what it means to be linear. As every vector space property derives from vector addition and scalar multiplication, so too, every property of a linear transformation derives from these two defining properties. While these conditions may be reminiscent of how we test subspaces, they really are quite different, so do not confuse the two.

Here are two diagrams that convey the essence of the two defining properties of a linear transformation. In each case, begin in the upper left-hand corner, and follow the arrows around the rectangle to the lower-right hand corner, taking two different routes and doing the indicated operations labeled on the arrows. There are two results there. For a linear transformation these two expressions are always equal.

A couple of words about notation. $T$ is the name of the linear transformation, and should be used when we want to discuss the function as a whole. $T\left(u\right)$ is how we talk about the output of the function, it is a vector in the vector space $V$. When we write $T\left(x+y\right)=T\left(x\right)+T\left(y\right)$, the plus sign on the left is the operation of vector addition in the vector space $U$, since $x$ and $y$ are elements of $U$. The plus sign on the right is the operation of vector addition in the vector space $V$, since $T\left(x\right)$ and $T\left(y\right)$ are elements of the vector space $V$. These two instances of vector addition might be wildly different.

Let’s examine several examples and begin to form a catalog of known linear transformations to work with.

Example ALT
A linear transformation
Define $T:{ℂ}^3\mapsto {ℂ}^2$ by describing the output of the function for a generic input with the formula

and check the two defining properties.

So by Definition LT, $T$ is a linear transformation. $⊠$

It can be just as instructive to look at functions that are not linear transformations. Since the defining conditions must be true for all vectors and scalars, it is enough to find just one situation where the properties fail.

Example NLT
Not a linear transformation
Define $S:{ℂ}^3\mapsto {ℂ}^3$ by

This function “looks” linear, but consider

So the second required property fails for the choice of $\alpha =3$ and $x=\left[\begin{array}{c}\hfill 1\hfill \\ \hfill 2\hfill \\ \hfill 3\hfill \end{array}\right]$and by Definition LT, $S$ is not a linear transformation. It is just about as easy to find an example where the first defining property fails (try it!). Notice that it is the “-2” in the third component of the definition of $S$ that prevents the function from being a linear transformation. $⊠$

Example LTPM
Linear transformation, polynomials to matrices
Define a linear transformation $T:P_3\mapsto M_{22}$ by

We verify the two defining conditions of a linear transformations.

So by Definition LT, $T$ is a linear transformation. $⊠$

Example LTPP
Linear transformation, polynomials to polynomials
Define a function $S:P_4\mapsto P_5$ by

Then

So by Definition LT, $S$ is a linear transformation. $⊠$

Linear transformations have many amazing properties, which we will investigate through the next few sections. However, as a taste of things to come, here is a theorem we can prove now and put to use immediately.

Theorem LTTZZ
Linear Transformations Take Zero to Zero
Suppose $T:U\mapsto V$ is a linear transformation. Then $T\left(0\right)=0$. $\square$

Proof   The two zero vectors in the conclusion of the theorem are different. The first is from $U$ while the second is from $V$. We will subscript the zero vectors in this proof to highlight the distinction. Think about your objects. (This proof is contributed by Mark Shoemaker).

Return to Example NLT and compute $S\left(\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \end{array}\right]\right)=\left[\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill -2\hfill \end{array}\right]$ to quickly see again that $S$ is not a linear transformation, while in Example LTPM and compute $S\left(0+0x+0x^2+0x^3\right)=\left[\begin{array}{cc}\hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right]$as an example of Theorem LTTZZ at work.

Subsection MLT: Matrices and Linear Transformations

If you give me a matrix, then I can quickly build you a linear transformation. Always. First a motivating example and then the theorem.

Example LTM
Linear transformation from a matrix
Let

and define a function $P:{ℂ}^4\mapsto {ℂ}^3$ by

So we are using an old friend, the matrix-vector product (Definition MVP) as a way to convert a vector with 4 components into a vector with 3 components. Applying Definition MVP allows us to write the defining formula for $P$ in a slightly different form,

So we recognize the action of the function $P$ as using the components of the vector ($x_1,x_2,x_3,x_4$) as scalars to form the output of $P$ as a linear combination of the four columns of the matrix $A$, which are all members of ${ℂ}^3$, so the result is a vector in ${ℂ}^3$. We can rearrange this expression further, using our definitions of operations in ${ℂ}^3$ (Section VO).

You might recognize this final expression as being similar in style to some previous examples (Example ALT) and some linear transformations defined in the archetypes (Archetype M through Archetype R). But the expression that says the output of this linear transformation is a linear combination of the columns of $A$ is probably the most powerful way of thinking about examples of this type.

Almost forgot — we should verify that $P$ is indeed a linear transformation. This is easy with two matrix properties from Section MM.

So by Definition LT, $P$ is a linear transformation. $⊠$

So the multiplication of a vector by a matrix “transforms” the input vector into an output vector, possibly of a different size, by performing a linear combination. And this transformation happens in a “linear” fashion. This “functional” view of the matrix-vector product is the most important shift you can make right now in how you think about linear algebra. Here’s the theorem, whose proof is very nearly an exact copy of the verification in the last example.

Theorem MBLT
Matrices Build Linear Transformations
Suppose that $A$ is an $m\times n$ matrix. Define a function $T:{ℂ}^n\mapsto {ℂ}^m$ by $T\left(x\right)=Ax$. Then $T$ is a linear transformation. $\square$

Proof  

So by Definition LT, $T$ is a linear transformation. $■$

So Theorem MBLT gives us a rapid way to construct linear transformations. Grab an $m\times n$ matrix $A$, define $T\left(x\right)=Ax$ and Theorem MBLT tells us that $T$ is a linear transformation from ${ℂ}^n$ to ${ℂ}^m$, without any further checking.

We can turn Theorem MBLT around. You give me a linear transformation and I will give you a matrix.

Example MFLT
Matrix from a linear transformation
Define the function $R:{ℂ}^3\mapsto {ℂ}^4$ by

You could verify that $R$ is a linear transformation by applying the definition, but we will instead massage the expression defining a typical output until we recognize the form of a known class of linear transformations.

So if we define the matrix

then $R\left(x\right)=Bx$. By Theorem MBLT, we can easily recognize $R$ as a linear transformation since it has the form described in the hypothesis of the theorem. $⊠$

Example MFLT was not accident. Consider any one of the archetypes where both the domain and codomain are sets of column vectors (Archetype M through Archetype R) and you should be able to mimic the previous example. Here’s the theorem, which is notable since it is our first occasion to use the full power of the defining properties of a linear transformation when our hypothesis includes a linear transformation.

Theorem MLTCV
Matrix of a Linear Transformation, Column Vectors
Suppose that $T:{ℂ}^n\mapsto {ℂ}^m$ is a linear transformation. Then there is an $m\times n$ matrix $A$ such that $T\left(x\right)=Ax$. $\square$

Proof   The conclusion says a certain matrix exists. What better way to prove something exists than to actually build it? So our proof will be constructive, and the procedure that we will use abstractly in the proof can be used concretely in specific examples.

Let $e_1,e_2,e_3,\dots ,e_n$ be the columns of the identity matrix of size $n$, $I_n$ (Definition SUV). Evaluate the linear transformation $T$ with each of these standard unit vectors as an input, and record the result. In other words, define $n$ vectors in ${ℂ}^m$, $A_i$, $1\le i\le n$ by

Then package up these vectors as the columns of a matrix

Does $A$ have the desired properties? First, $A$ is clearly an $m\times n$ matrix. Then

as desired. $■$

So if we were to restrict our study of linear transformations to those where the domain and codomain are both vector spaces of column vectors (Definition VSCV), every matrix leads to a linear transformation of this type (Theorem MBLT), while every such linear transformation leads to a matrix (Theorem MLTCV). So matrices and linear transformations are fundamentally the same. We call the matrix $A$ of Theorem MLTCV the matrix representation of $T$.

We have defined linear transformations for more general vector spaces than just ${ℂ}^m$, can we extend this correspondence between linear transformations and matrices to more general linear transformations (more general domains and codomains)? Yes, and this is the main theme of Chapter R. Stay tuned. For now, let’s illustrate Theorem MLTCV with an example.

Example MOLT
Matrix of a linear transformation
Suppose $S:{ℂ}^3\mapsto {ℂ}^4$ is defined by

Then

so define

and Theorem MLTCV guarantees that $S\left(x\right)=Cx$.

As an illuminating exercise, let $z=\left[\begin{array}{c}\hfill 2\hfill \\ \hfill -3\hfill \\ \hfill 3\hfill \end{array}\right]$ and compute $S\left(z\right)$ two different ways. First, return to the definition of $S$ and evaluate $S\left(z\right)$ directly. Then do the matrix-vector product $Cz$. In both cases you should obtain the vector $S\left(z\right)=\left[\begin{array}{c}\hfill 27\hfill \\ \hfill 2\hfill \\ \hfill 39\hfill \\ \hfill -12\hfill \end{array}\right]$. $⊠$

Subsection LTLC: Linear Transformations and Linear Combinations

It is the interaction between linear transformations and linear combinations that lies at the heart of many of the important theorems of linear algebra. The next theorem distills the essence of this. The proof is not deep, the result is hardly startling, but it will be referenced frequently. We have already passed by one occasion to employ it, in the proof of Theorem MLTCV. Paraphrasing, this theorem says that we can “push” linear transformations “down into” linear combinations, or “pull” linear transformations “up out” of linear combinations. We’ll have opportunities to both push and pull.

Theorem LTLC
Linear Transformations and Linear Combinations
Suppose that $T:U\mapsto V$ is a linear transformation, $u_1,u_2,u_3,\dots ,u_t$ are vectors from $U$ and $a_1,a_2,a_3,\dots ,a_t$ are scalars from ${ℂ}^{}$. Then

Proof  

Our next theorem says, informally, that it is enough to know how a linear transformation behaves for inputs from a basis of the domain, and all other outputs are described by a linear combination of these values. Again, the theorem and its proof are not remarkable, but the insight that goes along with it is fundamental.

Theorem LTDB
Linear Transformation Defined on a Basis
Suppose that $T:U\mapsto V$ is a linear transformation, $B=\left\{u_1,u_2,u_3,\dots ,u_n\right\}$ is a basis for $U$ and $w$ is a vector from $U$. Let $a_1,a_2,a_3,\dots ,a_n$ be the scalars from ${ℂ}^{}$ such that

Then

Proof   For any $w\in U$, Theorem VRRB says there are (unique) scalars such that $w$ is a linear combination of the basis vectors in $B$. The result then follows from a straightforward application of Theorem LTLC to the linear combination. $■$

Example LTDB1
Linear transformation defined on a basis
Suppose you are told that $T:{ℂ}^3\mapsto {ℂ}^2$ is a linear transformation and given the three values,

Because

is a basis for ${ℂ}^3$ (Theorem SUVB), Theorem LTDB says we can compute any output of $T$ with just this information. For example, consider,

so

Doing it again,

so

Any other value of $T$ could be computed in a similar manner. So rather than being given a formula for the outputs of $T$, the requirement that $T$ behave as a linear transformation, along with its values on a handful of vectors (the basis), are just as sufficient as a formula for computing any value of the function. You might notice some parallels between this example and Example MOLT or Theorem MLTCV. $⊠$

Example LTDB2
Linear transformation defined on a basis
Suppose you are told that $R:{ℂ}^3\mapsto {ℂ}^2$ is a linear transformation and given the three values,

You can check that

is a basis for ${ℂ}^3$ (make the vectors the columns of a square matrix and check that the matrix is nonsingular, Theorem CNMB). By Theorem LTDB we can compute any output of $R$ with just this information. However, we have to work just a bit harder to take an input vector and express it as a linear combination of the vectors in $D$. For example, consider,

Then we must first write $y$ as a linear combination of the vectors in $D$ and solve for the unknown scalars, to arrive at

Then Theorem LTDB gives us

Any other value of $R$ could be computed in a similar manner. $⊠$

Here is a third example of a linear transformation defined by its action on a basis, only with more abstract vector spaces involved.

Example LTDB3
Linear transformation defined on a basis
The set $W=\left\{p\left(x\right)\in P_3|p\left(1\right)=0,p\left(3\right)=0\right\}\subseteq P_3$ is a subspace of the vector space of polynomials $P_3$. This subspace has $C=\left\{3-4x+x^2,12-13x+x^3\right\}$ as a basis (check this!). Suppose we define a linear transformation $S:P_3\mapsto M_{22}$ by the values

To illustrate a sample computation of $S$, consider $q\left(x\right)=9-6x-5x^2+2x^3$. Verify that $q\left(x\right)$ is an element of $W$ (does it have roots at $x=1$ and $x=3$?), then find the scalars needed to write it as a linear combination of the basis vectors in $C$. Because

Theorem LTDB gives us

And all the other outputs of $S$ could be computed in the same manner. Every output of $S$ will have a zero in the second row, second column. Can you see why this is so? $⊠$

Subsection PI: Pre-Images

The definition of a function requires that for each input in the domain there is exactly one output in the codomain. However, the correspondence does not have to behave the other way around. A member of the codomain might have many inputs from the domain that create it, or it may have none at all. To formalize our discussion of this aspect of linear transformations, we define the pre-image.

Definition PI
Pre-Image
Suppose that $T:U\mapsto V$ is a linear transformation. For each $v$, define the pre-image of $v$ to be the subset of $U$ given by

In other words, $T^{-1}\left(v\right)$ is the set of all those vectors in the domain $U$ that get “sent” to the vector $v$.

TODO: All preimages form a partition of $U$, an equivalence relation is about. Maybe to exercises.

Example SPIAS
Sample pre-images, Archetype S
Archetype S is the linear transformation defined by

We could compute a pre-image for every element of the codomain $M_{22}$. However, even in a free textbook, we do not have the room to do that, so we will compute just two.

Choose

for no particular reason. What is $T^{-1}\left(v\right)$? Suppose $u=\left[\begin{array}{c}\hfill u_1\hfill \\ \hfill u_2\hfill \\ \hfill u_3\hfill \end{array}\right]\in T^{-1}\left(v\right)$. That $T\left(u\right)=v$ becomes

Using matrix equality (Definition ME), we arrive at a system of four equations in the three unknowns $u_1,u_2,u_3$ with an augmented matrix that we can row-reduce in the hunt for solutions,

We recognize this system as having infinitely many solutions described by the single free variable $u_3$. Eventually obtaining the vector form of the solutions (Theorem VFSLS), we can describe the preimage precisely as,

This last line is merely a suggestive way of describing the set on the previous line. You might create three or four vectors in the preimage, and evaluate $T$ with each. Was the result what you expected? For a hint of things to come, you might try evaluating $T$ with just the lone vector in the spanning set above. What was the result? Now take a look back at Theorem PSPHS. Hmmmm.

OK, let’s compute another preimage, but with a different outcome this time. Choose