SectionSLTSurjective Linear Transformations

The companion to an injection is a surjection. Surjective linear transformations are closely related to spanning sets and ranges. So as you read this section reflect back on Section ILT and note the parallels and the contrasts. In the next section, Section IVLT, we will combine the two properties.

SubsectionSLTSurjective Linear Transformations

As usual, we lead with a definition.

Definition SLT Surjective Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is surjective if for every $\vect{v}\in V$ there exists a $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$.

Given an arbitrary function, it is possible for there to be an element of the codomain that is not an output of the function (think about the function $y=f(x)=x^2$ and the codomain element $y=-3$). For a surjective function, this never happens. If we choose any element of the codomain ($\vect{v}\in V$) then there must be an input from the domain ($\vect{u}\in U$) which will create the output when used to evaluate the linear transformation ($\lt{T}{\vect{u}}=\vect{v}$). Some authors prefer the term onto where we use surjective, and we will sometimes refer to a surjective linear transformation as a surjection.

SubsectionESLTExamples of Surjective Linear Transformations

It is perhaps most instructive to examine a linear transformation that is not surjective first.

Example NSAQNot surjective, Archetype Q

Here is a cartoon of a non-surjective linear transformation. Notice that the central feature of this cartoon is that the vector $\vect{v}\in V$ does not have an arrow pointing to it, implying there is no $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$. Even though this happens again with a second unnamed vector in $V$, it only takes one occurrence to destroy the possibility of surjectivity.

DiagramNSLTNon-Surjective Linear Transformation

To show that a linear transformation is not surjective, it is enough to find a single element of the codomain that is never created by any input, as in Example NSAQ. However, to show that a linear transformation is surjective we must establish that every element of the codomain occurs as an output of the linear transformation for some appropriate input.

Example SARSurjective, Archetype R

Here is the cartoon for a surjective linear transformation. It is meant to suggest that for every output in $V$ there is at least one input in $U$ that is sent to the output. (Even though we have depicted several inputs sent to each output.) The key feature of this cartoon is that there are no vectors in $V$ without an arrow pointing to them.

DiagramSLTSurjective Linear Transformation

Let us now examine a surjective linear transformation between abstract vector spaces.

SubsectionRLTRange of a Linear Transformation

For a linear transformation $\ltdefn{T}{U}{V}$, the range is a subset of the codomain $V$. Informally, it is the set of all outputs that the transformation creates when fed every possible input from the domain. It will have some natural connections with the column space of a matrix, so we will keep the same notation, and if you think about your objects, then there should be little confusion. Here is the careful definition.

Definition RLT Range of a Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then the range of $T$ is the set \begin{equation*} \rng{T}=\setparts{\lt{T}{\vect{u}}}{\vect{u}\in U} \end{equation*}

Example RAORange, Archetype O

We know that the span of a set of vectors is always a subspace (Theorem SSS), so the range computed in Example RAO is also a subspace. This is no accident, the range of a linear transformation is always a subspace.

Theorem RLTS Range of a Linear Transformation is a Subspace

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then the range of $T$, $\rng{T}$, is a subspace of $V$.

Let us compute another range, now that we know in advance that it will be a subspace.

Example FRANFull range, Archetype N

In contrast to injective linear transformations having small (trivial) kernels (Theorem KILT), surjective linear transformations have large ranges, as indicated in the next theorem.

Theorem RSLT Range of a Surjective Linear Transformation

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is surjective if and only if the range of $T$ equals the codomain, $\rng{T}=V$.

SubsectionSSSLTSpanning Sets and Surjective Linear Transformations

Just as injective linear transformations are allied with linear independence (Theorem ILTLI, Theorem ILTB), surjective linear transformations are allied with spanning sets.

Theorem SSRLT Spanning Set for Range of a Linear Transformation

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and \begin{align*} S&=\set{\vectorlist{u}{t}} \end{align*} spans $U$. Then \begin{align*} R&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_t}} \end{align*} spans $\rng{T}$.

Theorem SSRLT provides an easy way to begin the construction of a basis for the range of a linear transformation, since the construction of a spanning set requires simply evaluating the linear transformation on a spanning set of the domain. In practice the best choice for a spanning set of the domain would be as small as possible, in other words, a basis. The resulting spanning set for the codomain may not be linearly independent, so to find a basis for the range might require tossing out redundant vectors from the spanning set. Here is an example.

Example BRLTA basis for the range of a linear transformation

Elements of the range are precisely those elements of the codomain with nonempty preimages.

Theorem RPI Range and Pre-Image

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then \begin{equation*} \vect{v}\in\rng{T}\text{ if and only if }\preimage{T}{\vect{v}}\neq\emptyset \end{equation*}

Now would be a good time to return to Diagram KPI which depicted the pre-images of a non-surjective linear transformation. The vectors $\vect{x},\,\vect{y}\in V$ were elements of the codomain whose pre-images were empty, as we expect for a non-surjective linear transformation from the characterization in Theorem RPI.

Theorem SLTB Surjective Linear Transformations and Bases

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and \begin{align*} B&=\set{\vectorlist{u}{m}} \end{align*} is a basis of $U$. Then $T$ is surjective if and only if \begin{align*} C&=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}} \end{align*} is a spanning set for $V$.

SubsectionSLTDSurjective Linear Transformations and Dimension

Theorem SLTD Surjective Linear Transformations and Dimension

Suppose that $\ltdefn{T}{U}{V}$ is a surjective linear transformation. Then $\dimension{U}\geq\dimension{V}$.

Example NSDATNot surjective by dimension, Archetype T

Notice that the previous example made no use of the actual formula defining the function. Merely a comparison of the dimensions of the domain and codomain are enough to conclude that the linear transformation is not surjective. Archetype O and Archetype P are two more examples of linear transformations that have “small” domains and “big” codomains, resulting in an inability to create all possible outputs and thus they are non-surjective linear transformations.

SubsectionCSLTComposition of Surjective Linear Transformations

In Subsection LT.NLTFO we saw how to combine linear transformations to build new linear transformations, specifically, how to build the composition of two linear transformations (Definition LTC). It will be useful later to know that the composition of surjective linear transformations is again surjective, so we prove that here.

Theorem CSLTS Composition of Surjective Linear Transformations is Surjective

Suppose that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are surjective linear transformations. Then $\ltdefn{(\compose{S}{T})}{U}{W}$ is a surjective linear transformation.