From A First Course in Linear Algebra
Version 2.30
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/
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.
As usual, we lead with a definition.
Definition SLT
Surjective Linear Transformation
Suppose is a linear
transformation. Then is
surjective if for every
there exists a
so that .
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 and the codomain element ). For a surjective function, this never happens. If we choose any element of the codomain () then there must be an input from the domain () which will create the output when used to evaluate the linear transformation (). 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
Archetype Q is the linear transformation
We will demonstrate that
is an unobtainable element of the codomain. Suppose to the contrary that is an element of the domain such that . Then
Now we recognize the appropriate input vector as a solution to a linear system of equations. Form the augmented matrix of the system, and row-reduce to
With a leading 1 in the last column, Theorem RCLS tells us the system is inconsistent. From the absence of any solutions we conclude that no such vector exists, and by Definition SLT, is not surjective.
Again, do not concern yourself with how was selected, as this will be explained shortly. However, do understand why this vector provides enough evidence to conclude that is not surjective.
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
Archetype R is the linear transformation
To establish that is surjective we must begin with a totally arbitrary element of the codomain, and somehow find an input vector such that . We desire,
We recognize this equation as a system of equations in the variables , but our vector of constants contains symbols. In general, we would have to row-reduce the augmented matrix by hand, due to the symbolic final column. However, in this particular example, the coefficient matrix is nonsingular and so has an inverse (Theorem NI, Definition MI).
so we find that
This establishes that if we are given any output vector , we can use its components in this final expression to formulate a vector such that . So by Definition SLT we now know that is surjective. You might try to verify this condition in its full generality (i.e. evaluate with this final expression and see if you get as the result), or test it more specifically for some numerical vector (see Exercise SLT.C20).
Let’s now examine a surjective linear transformation between abstract vector spaces.
Example SAV
Surjective, Archetype V
Archetype V is defined by
To establish that the linear transformation is surjective, begin by choosing an arbitrary output. In this example, we need to choose an arbitrary matrix, say
and we would like to find an input polynomial
so that . So we have,
Matrix equality leads us to the system of four equations in the four unknowns, ,
which can be rewritten as a matrix equation,
The coefficient matrix is nonsingular, hence it has an inverse,
so we have
So the input polynomial will yield the output matrix , no matter what form takes. This means by Definition SLT that is surjective. All the same, let’s do a concrete demonstration and evaluate with ,
For a linear transformation , the range is a subset of the codomain . 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
is a linear transformation. Then the range of
is the
set
(This definition contains Notation RLT.)
Example RAO
Range, Archetype O
Archetype O is the linear transformation
To determine the elements of in , find those vectors such that for some ,
This says that every output of () can be written as a linear combination of the three vectors
using the scalars . Furthermore, since can be any element of , every such linear combination is an output. This means that
The three vectors in this spanning set for form a linearly dependent set (check this!). So we can find a more economical presentation by any of the various methods from Section CRS and Section FS. We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to Theorem BRS, so we can describe the range of with a basis,
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
is a linear transformation. Then the range of
,
, is a
subspace of .
Proof We can apply the three-part test of Theorem TSS. First, and by Theorem LTTZZ, so and we know that the range is non-empty.
Suppose we assume that . Is ? If then we know there are vectors such that and . Because is a vector space, additive closure (Property AC) implies that . Then
So we have found an input, , which when fed into creates as an output. This qualifies for membership in . So we have additive closure.
Suppose we assume that and . Is ? If , then there is a vector such that . Because is a vector space, scalar closure implies that . Then
So we have found an input () which when fed into creates as an output. This qualifies for membership in . So we have scalar closure and Theorem TSS tells us that is a subspace of .
Let’s compute another range, now that we know in advance that it will be a subspace.
Example FRAN
Full range, Archetype N
Archetype N is the linear transformation
To determine the elements of in , find those vectors such that for some ,
This says that every output of () can be written as a linear combination of the five vectors
using the scalars . Furthermore, since can be any element of , every such linear combination is an output. This means that
The five vectors in this spanning set for form a linearly dependent set (Theorem MVSLD). So we can find a more economical presentation by any of the various methods from Section CRS and Section FS. We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to Theorem BRS, so we can describe the range of with a (nice) basis,
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 is a linear
transformation. Then is surjective
if and only if the range of
equals the codomain, .
Proof () By Definition RLT, we know that . To establish the reverse inclusion, assume . Then since is surjective (Definition SLT), there exists a vector so that . However, the existence of gains membership in , so . Thus, .
() To establish that is surjective, choose . Since we are assuming that , . This says there is a vector so that , i.e. is surjective.
Example NSAQR
Not surjective, Archetype Q, revisited
We are now in a position to revisit our first example in this section,
Example NSAQ. In that example, we showed that Archetype Q is not
surjective by constructing a vector in the codomain where no element of the
domain could be used to evaluate the linear transformation to create the
output, thus violating Definition SLT. Just where did this vector come
from?
The short answer is that the vector
was constructed to lie outside of the range of . How was this accomplished? First, the range of is given by
Suppose an element of the range has its first 4 components equal to , in that order. Then to be an element of , we would have
So the only vector in the range with these first four components specified, must have in the fifth component. To set the fifth component to any other value (say, 4) will result in a vector ( in Example NSAQ) outside of the range. Any attempt to find an input for that will produce as an output will be doomed to failure.
Whenever the range of a linear transformation is not the whole codomain, we can employ this device and conclude that the linear transformation is not surjective. This is another way of viewing Theorem RSLT. For a surjective linear transformation, the range is all of the codomain and there is no choice for a vector that lies in , yet not in the range. For every one of the archetypes that is not surjective, there is an example presented of exactly this form.
Example NSAO
Not surjective, Archetype O
In Example RAO the range of Archetype O was determined to be
a subspace of dimension 2 in . Since , Theorem RSLT says is not surjective.
Example SAN
Surjective, Archetype N
The range of Archetype N was computed in Example FRAN to be
Since the basis for this subspace is the set of standard unit vectors for (Theorem SUVB), we have and by Theorem RSLT, is surjective.
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 is a linear
transformation and
spans .
Then
spans .
Proof We need to establish that , a set equality. First we establish that . To this end, choose . Then there exists a vector , such that (Definition RLT). Because spans there are scalars, , such that
Then
which establishes that (Definition SS). So .
To establish the opposite inclusion, choose an element of the span of , say . Then there are scalars so that
This demonstrates that is an output of the linear transformation , so . Therefore , so we have the set equality (Definition SE). In other words, spans (Definition TSVS).
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
Define the linear transformation
by
A convenient spanning set for is the basis
So by Theorem SSRLT, a spanning set for is
The set is not linearly independent, so if we desire a basis for , we need to eliminate some redundant vectors. Two particular relations of linear dependence on are
These, individually, allow us to remove and from with out destroying the property that spans . The two remaining vectors are linearly independent (check this!), so we can write
and see that .
Elements of the range are precisely those elements of the codomain with non-empty preimages.
Theorem RPI
Range and Pre-Image
Suppose that
is a linear transformation. Then
Proof () If , then there is a vector such that . This qualifies for membership in , and thus the preimage of is not empty.
() Suppose the preimage of is not empty, so we can choose a vector such that . Then .
Theorem SLTB
Surjective Linear Transformations and Bases
Suppose that is a linear
transformation and
is a basis of . Then
is surjective if
and only if is a
spanning set for .
Proof () Assume is surjective. Since is a basis, we know is a spanning set of (Definition B). Then Theorem SSRLT says that spans . But the hypothesis that is surjective means (Theorem RSLT), so spans .
() Assume that spans . To establish that is surjective, we will show that every element of is an output of for some input (Definition SLT). Suppose that . As an element of , we can write as a linear combination of the spanning set . So there are are scalars, , such that
Now define the vector by
Then
So, given any choice of a vector , we can design an input to produce as an output of . Thus, by Definition SLT, is surjective.
Theorem SLTD
Surjective Linear Transformations and Dimension
Suppose that is a surjective
linear transformation. Then .
Proof Suppose to the contrary that . Let be a basis of , which will then contain vectors. Apply to each element of to form a set that is a subset of . By Theorem SLTB, is spanning set of with or fewer vectors. So we have a set of or fewer vectors that span , a vector space of dimension , with . However, this contradicts Theorem G, so our assumption is false and .
Example NSDAT
Not surjective by dimension, Archetype T
The linear transformation in Archetype T is
Since , cannot be surjective for then it would violate Theorem SLTD.
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 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
and
are surjective linear transformations. Then
is a surjective linear
transformation.
Proof That the composition is a linear transformation was established in Theorem CLTLT, so we need only establish that the composition is surjective. Applying Definition SLT, choose .
Because is surjective, there must be a vector , such that . With the existence of established, that is surjective guarantees a vector such that . Now,
This establishes that any element of the codomain () can be created by evaluating with the right input (). Thus, by Definition SLT, is surjective.
C10 Each archetype below is a linear transformation. Compute the range for
each.
Archetype M
Archetype N
Archetype O
Archetype P
Archetype Q
Archetype R
Archetype S
Archetype T
Archetype U
Archetype V
Archetype W
Archetype X
Contributed by Robert Beezer
C20 Example SAR concludes with an expression for a vector
that we believe will
create the vector when
used to evaluate . That
is, . Verify this assertion
by actually evaluating
with . If you
don’t have the patience to push around all these symbols, try choosing a numerical instance
of , compute
, and then
compute , which
should result in .
Contributed by Robert Beezer
C22 The linear transformation is not surjective. Find an output that has an empty pre-image (that is .)
Contributed by Robert Beezer Solution [1547]
C23 Determine whether or not the following linear transformation is surjective:
Contributed by Chris Black Solution [1549]
C24 Determine whether or not the linear transformation below is surjective:
Contributed by Chris Black Solution [1551]
C25 Define the linear transformation
Find a basis for the range of ,
. Is
surjective?
Contributed by Robert Beezer Solution [1551]
C26 Let be
given by . Find
a basis of .
Is
surjective?
Contributed by Chris Black Solution [1552]
C27 Let be
given by . Find
a basis of .
Is
surjective?
Contributed by Chris Black Solution [1553]
C28 Let be
given by . Find
a basis of .
Is
surjective?
Contributed by Chris Black Solution [1554]
C29 Let be
given by . Find
a basis of .
Is
surjective?
Contributed by Chris Black Solution [1555]
C30 Let be
given by , where
is the derivative.
Find a basis of .
Is
surjective?
Contributed by Chris Black Solution [1555]
C40 Show that the linear transformation is not surjective by finding an element of the codomain, , such that there is no vector with .
Contributed by Robert Beezer Solution [1556]
M60 Suppose and
are vector spaces.
Define the function
by for every
. Then by
Exercise LT.M60,
is a linear transformation. Formulate a condition on
that is
equivalent to
being an surjective linear transformation. In other words, fill in the
blank to complete the following statement (and then give a proof):
is surjective
if and only if
is __________________. (See Exercise ILT.M60, Exercise IVLT.M60.)
Contributed by Robert Beezer
T15 Suppose that that and are linear transformations. Prove the following relationship between ranges.
Contributed by Robert Beezer Solution [1557]
T20 Suppose that is an matrix. Define the linear transformation by
Prove that the range of
equals the column space of ,
.
Contributed by Andy Zimmer Solution [1558]
C22 Contributed by Robert Beezer Statement [1541]
To find an element of
with an empty pre-image, we will compute the range of the linear transformation
and
then find an element outside of this set.
By Theorem SSRLT we can evaluate with the elements of a spanning set of the domain and create a spanning set for the range.
So
This spanning set is obviously linearly dependent, so we can reduce it to a basis for using Theorem BRS, where the elements of the spanning set are placed as the rows of a matrix. The result is that
Therefore, the unique vector in with a first slot equal to 6 and a second slot equal to 15 will be the linear combination
So, any vector with first two components equal to 6 and 15, but with a third component different from 9, such as
will not be an element of the range of and will therefore have an empty pre-image. Another strategy on this problem is to guess. Almost any vector will lie outside the range of , you have to be unlucky to randomly choose an element of the range. This is because the codomain has dimension 3, while the range is “much smaller” at a dimension of 2. You still need to check that your guess lies outside of the range, which generally will involve solving a system of equations that turns out to be inconsistent.
C23 Contributed by Chris Black Statement [1542]
The linear transformation
is surjective if for any ,
there is a vector
in so
that .
We need to be able to solve the system
This system has an infinite number of solutions, one of which is , , , and , so that
Thus, is surjective, since for every vector , there exists a vector so that .
C24 Contributed by Chris Black Statement [1542]
According to Theorem SLTD, if a linear transformation
is surjective, then
. In this example,
has dimension 4, and
has dimension 5, so
cannot be surjective.
(There is no way can
“expand” the domain
to fill the codomain .)
C25 Contributed by Robert Beezer Statement [1543]
To find the range of ,
apply to the elements
of a spanning set for
as suggested in Theorem SSRLT. We will use the standard basis vectors
(Theorem SUVB).
Each of these vectors is a scalar multiple of the others, so we can toss two of them in reducing the spanning set to a linearly independent set (or be more careful and apply Theorem BCS on a matrix with these three vectors as columns). The result is the basis of the range,
With , , so Theorem RSLT says is not surjective.
C26 Contributed by Chris Black Statement [1543]
The range of
is
Since the vectors and are linearly independent (why?), a basis of is . Since the dimension of the range is 2 and the dimension of the codomain is 3, is not surjective.
C27 Contributed by Chris Black Statement [1544]
The range of
is
By row reduction (not shown), we can see that the set
are linearly independent, so is a basis of . Since the dimension of the range is 3 and the dimension of the codomain is 4, is not surjective. (We should have anticipated that was not surjective since the dimension of the domain is smaller than the dimension of the codomain.)
C28 Contributed by Chris Black Statement [1544]
The range of
is
Can you explain the last equality above?
These three matrices are linearly independent, so a basis of is . Thus, is not surjective, since the range has dimension 3 which is shy of . (Notice that the range is actually the subspace of symmetric matrices in .)
C29 Contributed by Chris Black Statement [1544]
If we transform the basis of ,
then Theorem SSRLT guarantees we will have a spanning set of
. A basis
of is
.
If we transform the elements of this set, we get the set
which is a
spanning set for .
These three vectors are linearly independent, so
is a
basis of .
C30 Contributed by Chris Black Statement [1544]
If we transform the basis of ,
then Theorem SSRLT guarantees we will have a spanning set of
. A basis
of is
.
If we transform the elements of this set, we get the set
which is a
spanning set for .
Reducing this to a linearly independent set, we find that
is a basis
of . Since
and
both have
dimension 4,
is surjective.
C40 Contributed by Robert Beezer Statement [1544]
We wish to find an output vector
that has no associated input. This is the same as requiring that there is no
solution to the equality
In other words, we would like to find an element of not in the set
If we make these vectors the rows of a matrix, and row-reduce, Theorem BRS provides an alternate description of ,
If we add these vectors together, and then change the third component of the result, we will create a vector that lies outside of , say .
T15 Contributed by Robert Beezer Statement [1545]
This question asks us to establish that one set
() is a subset
of another ().
Choose an element in the “smaller” set, say
. Then we know that
there is a vector
such that
Now define , so that then
This statement is sufficient to show that , so is an element of the “larger” set, and .
T20 Contributed by Andy Zimmer Statement [1546]
This is an equality of sets, so we want to establish two subset conditions
(Definition SE).
First, show . Choose . Then by Definition CSM and Definition MVP there is a vector such that . Then
This statement qualifies as a member of (Definition RLT), so .
Now, show . Choose . Then by Definition RLT, there is a vector in such that . Then
So by Definition CSM and Definition MVP, qualifies for membership in and so .