Surjective 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:Injective Linear Transformations and note the parallels and the contrasts. In the next section, Section IVLT:Invertible Linear Transformations, we will combine the two properties.
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.
It is perhaps most instructive to examine a linear transformation that is not surjective first.
Example NSAQ: Not surjective, Archetype Q.
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 SAR: Surjective, Archetype R.
Let's now examine a surjective linear transformation between abstract vector spaces.
Example SAV: Surjective, Archetype V.
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's 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 RAO: Range, 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's compute another range, now that we know in advance that it will be a subspace.
Example FRAN: Full 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, $\rng{T}=V$.
Example NSAQR: Not surjective, Archetype Q, revisited.
Example NSAO: Not surjective, Archetype O.
Example SAN: Surjective, Archetype N.
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 $S=\set{\vectorlist{u}{t}}$ 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's an example.
Example BRLT: A basis for the range of a linear transformation.
Elements of the range are precisely those elements of the codomain with non-empty 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*}
Theorem SLTB (Surjective Linear Transformations and Bases) Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and $B=\set{\vectorlist{u}{m}}$ is a basis of $U$. Then $T$ is surjective if and only if $C=\set{\lt{T}{\vect{u}_1},\,\lt{T}{\vect{u}_2},\,\lt{T}{\vect{u}_3},\,\ldots,\,\lt{T}{\vect{u}_m}}$ is a spanning set for $V$.
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 NSDAT: Not 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.
In Subsection LT.NLTFO:Linear Transformations: New Linear Transformations From Old 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.