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SectionLTLinear Transformations

Early in Chapter VS we prefaced the definition of a vector space with the comment that it was “one of the two most important definitions in the entire course.” Here comes the other. Any capsule summary of linear algebra would have to describe the subject as the interplay of linear transformations and vector spaces. Here we go.

SubsectionLTLinear Transformations

DefinitionLTLinear Transformation

A linear transformation, \(\ltdefn{T}{U}{V}\text{,}\) 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. \(\lteval{T}{\vect{u}_1+\vect{u}_2}=\lteval{T}{\vect{u}_1}+\lteval{T}{\vect{u}_2}\) for all \(\vect{u}_1,\,\vect{u}_2\in U\text{.}\)
  2. \(\lteval{T}{\alpha\vect{u}}=\alpha\lteval{T}{\vect{u}}\) for all \(\vect{u}\in U\) and all \(\alpha\in\complexes\text{.}\)

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 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.

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Figure7.2Definition of Linear Transformation, Additive

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Figure7.3Definition of Linear Transformation, Multiplicative

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. \(\lteval{T}{\vect{u}}\) is how we talk about the output of the function, it is a vector in the vector space \(V\text{.}\) When we write \(\lteval{T}{\vect{x}+\vect{y}}=\lteval{T}{\vect{x}}+\lteval{T}{\vect{y}}\text{,}\) the plus sign on the left is the operation of vector addition in the vector space \(U\text{,}\) since \(\vect{x}\) and \(\vect{y}\) are elements of \(U\text{.}\) The plus sign on the right is the operation of vector addition in the vector space \(V\text{,}\) since \(\lteval{T}{\vect{x}}\) and \(\lteval{T}{\vect{y}}\) are elements of the vector space \(V\text{.}\) These two instances of vector addition might be wildly different.

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

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.

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.


Return to Example NLT and compute \(\lteval{S}{\colvector{0\\0\\0}}=\colvector{0\\0\\-2}\) to quickly see again that \(S\) is not a linear transformation, while in Example LTPM compute \begin{align*} \lteval{S}{0+0x+0x^2+0x^3}&=\begin{bmatrix}0&0\\0&0\end{bmatrix} \end{align*} as an example of Theorem LTTZZ at work.

SubsectionLTCLinear Transformation Cartoons

Throughout this chapter, and Chapter R, we will include drawings of linear transformations. We will call them “cartoons,” not because they are humorous, but because they will only expose a portion of the truth. A Bugs Bunny cartoon might give us some insights on human nature, but the rules of physics and biology are routinely (and grossly) violated. So it will be with our linear transformation cartoons. Here is our first, followed by a guide to help you understand how these are meant to describe fundamental truths about linear transformations, while simultaneously violating other truths.

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Figure7.10General Linear Transformation

Here we picture a linear transformation \(\ltdefn{T}{U}{V}\text{,}\) where this information will be consistently displayed along the bottom edge. The ovals are meant to represent the vector spaces, in this case \(U\text{,}\) the domain, on the left and \(V\text{,}\) the codomain, on the right. Of course, vector spaces are typically infinite sets, so you will have to imagine that characteristic of these sets. A small dot inside of an oval will represent a vector within that vector space, sometimes with a name, sometimes not (in this case every vector has a name). The sizes of the ovals are meant to be proportional to the dimensions of the vector spaces. However, when we make no assumptions about the dimensions, we will draw the ovals as the same size, as we have done here (which is not meant to suggest that the dimensions have to be equal).

To convey that the linear transformation associates a certain input with a certain output, we will draw an arrow from the input to the output. So, for example, in this cartoon we suggest that \(\lteval{T}{\vect{x}}=\vect{y}\text{.}\) Nothing in the definition of a linear transformation prevents two different inputs being sent to the same output and we see this in \(\lteval{T}{\vect{u}}=\vect{v}=\lteval{T}{\vect{w}}\text{.}\) Similarly, an output may not have any input being sent its way, as illustrated by no arrow pointing at \(\vect{t}\text{.}\) In this cartoon, we have captured the essence of our one general theorem about linear transformations, Theorem LTTZZ, \(\lteval{T}{\zerovector_U}=\zerovector_V\text{.}\) On occasion we might include this basic fact when it is relevant, at other times maybe not. Note that the definition of a linear transformation requires that it be a function, so every element of the domain should be associated with some element of the codomain. This will be reflected by never having an element of the domain without an arrow originating there.

These cartoons are of course no substitute for careful definitions and proofs, but they can be a handy way to think about the various properties we will be studying.

SubsectionMLTMatrices 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.

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 is the theorem, whose proof is very nearly an exact copy of the verification in the last example.


So Theorem MBLT gives us a rapid way to construct linear transformations. Grab an \(m\times n\) matrix \(A\text{,}\) define \(\lteval{T}{\vect{x}}=A\vect{x}\) and Theorem MBLT tells us that \(T\) is a linear transformation from \(\complex{n}\) to \(\complex{m}\text{,}\) 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 was not an 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 is 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.


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 \(T\text{.}\)

We have defined linear transformations for more general vector spaces than just \(\complex{m}\text{.}\) 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 us illustrate Theorem MLTCV with an example.

SubsectionLTLCLinear 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 will have opportunities to both push and pull.


Some authors, especially in more advanced texts, take the conclusion of Theorem LTLC as the defining condition of a linear transformation. This has the appeal of being a single condition, rather than the two-part condition of Definition LT. (See Exercise LT.T20).

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


You might recall facts from analytic geometry, such as “any two points determine a line” and “any three non-collinear points determine a parabola.” Theorem LTDB has much of the same feel. By specifying the \(n\) outputs for inputs from a basis, an entire linear transformation is determined. The analogy is not perfect, but the style of these facts are not very dissimilar from Theorem LTDB.

Notice that the statement of Theorem LTDB asserts the existence of a linear transformation with certain properties, while the proof shows us exactly how to define the desired linear transformation. The next two examples show how to compute values of linear transformations that we create this way.

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

Informally, we can describe Theorem LTDB by saying “it is enough to know what a linear transformation does to a basis (of the domain).”


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. An output from the codomain could have many different inputs from the domain which the transformation sends to that output, or there could be no inputs at all which the transformation sends to that output. To formalize our discussion of this aspect of linear transformations, we define the pre-image.


Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. For each \(\vect{v}\text{,}\) define the pre-image of \(\vect{v}\) to be the subset of \(U\) given by \begin{equation*} \preimage{T}{\vect{v}}=\setparts{\vect{u}\in U}{\lteval{T}{\vect{u}}=\vect{v}}\text{.} \end{equation*}

In other words, \(\preimage{T}{\vect{v}}\) is the set of all those vectors in the domain \(U\) that get “sent” to the vector \(\vect{v}\text{.}\)

The preimage is just a set, it is almost never a subspace of \(U\) (you might think about just when \(\preimage{T}{\vect{v}}\) is a subspace, see Exercise ILT.T10). We will describe its properties going forward, and it will be central to the main ideas of this chapter.

SubsectionNLTFONew Linear Transformations From Old

We can combine linear transformations in natural ways to create new linear transformations. So we will define these combinations and then prove that the results really are still linear transformations. First the sum of two linear transformations.

DefinitionLTALinear Transformation Addition

Suppose that \(\ltdefn{T}{U}{V}\) and \(\ltdefn{S}{U}{V}\) are two linear transformations with the same domain and codomain. Then their sum is the function \(\ltdefn{T+S}{U}{V}\) whose outputs are defined by \begin{equation*} \lteval{(T+S)}{\vect{u}}=\lteval{T}{\vect{u}}+\lteval{S}{\vect{u}}\text{.} \end{equation*}

Notice that the first plus sign in the definition is the operation being defined, while the second one is the vector addition in \(V\text{.}\) (Vector addition in \(U\) will appear just now in the proof that \(T+S\) is a linear transformation.) Definition LTA only provides a function. It would be nice to know that when the constituents (\(T\text{,}\) \(S\)) are linear transformations, then so too is \(T+S\text{.}\)

DefinitionLTSMLinear Transformation Scalar Multiplication

Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation and \(\alpha\in\complexes\text{.}\) Then the scalar multiple is the function \(\ltdefn{\alpha T}{U}{V}\) whose outputs are defined by \begin{equation*} \lteval{(\alpha T)}{\vect{u}}=\alpha\lteval{T}{\vect{u}}\text{.} \end{equation*}

Given that \(T\) is a linear transformation, it would be nice to know that \(\alpha T\) is also a linear transformation.


Now, let us imagine we have two vector spaces, \(U\) and \(V\text{,}\) and we collect every possible linear transformation from \(U\) to \(V\) into one big set, and call it \(\vslt{U}{V}\text{.}\) Definition LTA and Definition LTSM tell us how we can “add” and “scalar multiply” two elements of \(\vslt{U}{V}\text{.}\) Theorem SLTLT and Theorem MLTLT tell us that if we do these operations, then the resulting functions are linear transformations that are also in \(\vslt{U}{V}\text{.}\) Hmmmm, sounds like a vector space to me! A set of objects, an addition and a scalar multiplication. Why not?

DefinitionLTCLinear Transformation Composition

Suppose that \(\ltdefn{T}{U}{V}\) and \(\ltdefn{S}{V}{W}\) are linear transformations. Then the composition of \(S\) and \(T\) is the function \(\ltdefn{(\compose{S}{T})}{U}{W}\) whose outputs are defined by \begin{equation*} \lteval{(\compose{S}{T})}{\vect{u}}=\lteval{S}{\lteval{T}{\vect{u}}}\text{.} \end{equation*}

Given that \(T\) and \(S\) are linear transformations, it would be nice to know that \(\compose{S}{T}\) is also a linear transformation.


Here is an interesting exercise that will presage an important result later. In Example STLT compute (via Theorem MLTCV) the matrix of \(T\text{,}\) \(S\) and \(T+S\text{.}\) Do you see a relationship between these three matrices?

In Example SMLT compute (via Theorem MLTCV) the matrix of \(T\) and \(2T\text{.}\) Do you see a relationship between these two matrices?

Here is the tough one. In Example CTLT compute (via Theorem MLTCV) the matrix of \(T\text{,}\) \(S\) and \(\compose{S}{T}\text{.}\) Do you see a relationship between these three matrices???

SubsectionReading Questions


Is the function below a linear transformation? Why or why not? \begin{equation*} \ltdefn{T}{\complex{3}}{\complex{2}},\quad\lteval{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{3x_1-x_2+x_3\\8x_2-6}\text{.} \end{equation*}


Determine the matrix representation of the linear transformation \(S\) below. \begin{equation*} \ltdefn{S}{\complex{2}}{\complex{3}},\quad\lteval{S}{\colvector{x_1\\x_2}}=\colvector{3x_1+5x_2\\8x_1-3x_2\\-4x_1}\text{.} \end{equation*}


Theorem LTLC has a fairly simple proof. Yet the result itself is very powerful. Comment on why we might say this.



The archetypes below are all linear transformations whose domains and codomains are vector spaces of column vectors (Definition VSCV). For each one, compute the matrix representation described in the proof of Theorem MLTCV.

Archetype M, Archetype N, Archetype O, Archetype P, Archetype Q, Archetype R


Find the matrix representation of \(\ltdefn{T}{\complex{3}}{\complex{4}}\text{,}\) \(\lteval{T}{\colvector{x\\y\\z}} = \colvector{3x + 2y + z\\ x + y + z \\ x - 3y \\2x + 3y + z }\text{.}\)


Let \(\vect{w}=\colvector{-3\\1\\4}\text{.}\) Referring to Example MOLT, compute \(\lteval{S}{\vect{w}}\) two different ways. First use the definition of \(S\text{,}\) then compute the matrix-vector product \(C\vect{w}\) (Definition MVP).


Define the linear transformation \begin{equation*} \ltdefn{T}{\complex{3}}{\complex{2}},\quad\lteval{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3}\text{.} \end{equation*} Verify that \(T\) is a linear transformation.


Verify that the function below is a linear transformation. \begin{equation*} \ltdefn{T}{P_2}{\complex{2}},\quad \lteval{T}{a+bx+cx^2}=\colvector{2a-b\\b+c}\text{.} \end{equation*}


Define the linear transformation \begin{equation*} \ltdefn{T}{\complex{3}}{\complex{2}},\quad\lteval{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3}\text{.} \end{equation*} Compute the preimages, \(\preimage{T}{\colvector{2\\3}}\) and \(\preimage{T}{\colvector{4\\-8}}\text{.}\)


For the linear transformation \(S\) compute the pre-images. \begin{gather*} \ltdefn{S}{\complex{3}}{\complex{3}},\quad \lteval{S}{\colvector{a\\b\\c}}= \colvector{a-2b-c\\3a-b+2c\\a+b+2c } \end{gather*} \begin{align*} \preimage{S}{\colvector{-2\\5\\3}}& & \preimage{S}{\colvector{-5\\5\\7}}& \end{align*}


If \(\ltdefn{T}{\complex{2}}{\complex{2}}\) satisfies \(\lteval{T}{\colvector{2\\1}} = \colvector{3\\4}\) and \(\lteval{T}{\colvector{1\\1}} = \colvector{-1\\2}\text{,}\) find \(\lteval{T}{\colvector{4\\3}}\text{.}\)


If \(\ltdefn{T}{\complex{2}}{\complex{3}}\) satisfies \(\lteval{T}{\colvector{2\\3}} = \colvector{2\\2\\1}\) and \(\lteval{T}{\colvector{3\\4}} = \colvector{-1\\0\\2}\text{,}\) find the matrix representation of \(T\text{.}\)


Define \(\ltdefn{T}{M_{22}}{\complex{1}}\) by \(\lteval{T}{\begin{bmatrix} a & b \\ c & d \end{bmatrix}} = a + b + c - d\text{.}\) Find the pre-image \(\preimage{T}{3}\text{.}\)


Define \(\ltdefn{T}{P_3}{P_2}\) by \(\lteval{T}{a + bx + cx^2 + dx^3} = b + 2cx + 3dx^2\text{.}\) Find the pre-image of \(\mathbf{0}\text{.}\) Does this linear transformation seem familiar?


Define two linear transformations, \(\ltdefn{T}{\complex{4}}{\complex{3}}\) and \(\ltdefn{S}{\complex{3}}{\complex{2}}\) by \begin{align*} \lteval{S}{\colvector{x_1\\x_2\\x_3}} &= \colvector{ x_1-2x_2+3x_3\\ 5x_1+4x_2+2x_3 } & \lteval{T}{\colvector{x_1\\x_2\\x_3\\x_4}} &= \colvector{ -x_1+3x_2+x_3+9x_4\\ 2x_1+x_3+7x_4\\ 4x_1+2x_2+x_3+2x_4 }\text{.} \end{align*} Using the proof of Theorem MLTCV compute the matrix representations of the three linear transformations \(T\text{,}\) \(S\) and \(\compose{S}{T}\text{.}\) Discover and comment on the relationship between these three matrices.


Suppose \(U\) and \(V\) are vector spaces and define a function \(\ltdefn{Z}{U}{V}\) by \(\lteval{Z}{\vect{u}}=\zerovector_{V}\) for every \(\vect{u}\in U\text{.}\) Prove that \(Z\) is a (stupid) linear transformation. (See Exercise ILT.M60, Exercise SLT.M60, Exercise IVLT.M60.)


Use the conclusion of Theorem LTLC to motivate a new definition of a linear transformation. Then prove that your new definition is equivalent to Definition LT. (Proof Technique D and Proof Technique E might be helpful if you are not sure what you are being asked to prove here.)

Theorem SER established three properties of matrix similarity that are collectively known as the defining properties of an “equivalence relation”. Exercises T30 and T31 extend this idea to linear transformations.


Suppose that \(\ltdefn{T}{U}{V}\) is a linear transformation. Say that two vectors from \(U\text{,}\) \(\vect{x}\) and \(\vect{y}\text{,}\) are related exactly when \(\lteval{T}{\vect{x}}=\lteval{T}{\vect{y}}\) in \(V\text{.}\) Prove the three properties of an equivalence relation on \(U\text{:}\) (a) for any \(\vect{x}\in U\text{,}\) \(\vect{x}\) is related to \(\vect{x}\text{,}\) (b) if \(\vect{x}\) is related to \(\vect{y}\text{,}\) then \(\vect{y}\) is related to \(\vect{x}\text{,}\) and (c) if \(\vect{x}\) is related to \(\vect{y}\) and \(\vect{y}\) is related to \(\vect{z}\text{,}\) then \(\vect{x}\) is related to \(\vect{z}\text{.}\)


Equivalence relations always create a partition of the set they are defined on, via a construction called equivalence classes. For the relation in the previous problem, the equivalence classes are the pre-images. Prove directly that the collection of pre-images partition \(U\) by showing that (a) every \(\vect{x}\in U\) is contained in some pre-image, and that (b) any two different pre-images do not have any elements in common.