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Mathematical Expression Editor
Linear Transformations of Abstract Vector Spaces
Recall that a transformation is called a linear transformation if the following are
true for all vectors and in , and scalars .
We generalize this definition as follows.
Let and be vector spaces. A transformation is called a linear transformation if the
following are true for all vectors and in , and scalars .
Recall that is the set of all matrices. In Example ex:setofmatricesvectorspace of Abstract Vector Spaces we
demonstrated that together with operations of matrix addition and scalar
multiplication is a vector space.
Let be a transformation defined by , where is fixed matrix. Show that is a linear
transformation.
Recall that is the set of polynomials of degree or less than . In Example ex:pnisavectorspace of
Abstract Vector Spaces we showed that together with operations of polynomial
addition and scalar multiplication is a vector space.
Suppose is a linear transformation such that
Find the image of under .
Let be a transformation such that . Show that is not linear.
To show that is not
linear it suffices to find two matrices and such that . Observe that if we pick and
so that each has rank we would have while . Clearly . This argument
is sufficient, but if we want to have a specific example, we can find one.
Let
Then
Thus, .
Linear Transformations and Bases
Suppose we want to define a linear transformation by
Is this information sufficient to define ? To answer this question we will try to
determine what does to an arbitrary vector of .
If is a vector in , then can be uniquely expressed as a linear combination of and
By linearity of we have
This shows that the image of every vector of under is completely determined by the
action of on the standard unit vectors and .
Vectors and form a standard basis of . What if we want to use a different
basis?
Let be our basis of choice for . (How would you verify that is a basis of ?) And
suppose we want to define a linear transformation by
Is this enough information to define ?
Because form a basis of , every element of can be written as a unique linear
combination
We can find as follows:
Again, we see how a linear transformation is completely determined by its action on a
basis.
Theorem th:uniquerep assures us that given a basis, every vector has a unique representation as
a linear combination of the basis vectors. Imagine what would happen if
this were not the case. In the first part of this exploration, for instance, we
might have been able to represent as and ( or ). This would have resulted
in mapping to two different elements: and , implying that is not even a
function.
Let be a basis of a vector space . To define a linear transformation , it is sufficient to
state the image of each basis vector under . Once the images of the basis
vectors are established, we can determine the images of all vectors of as
follows:
Given any vector of , write as a linear combination of the elements of
Then
Coordinate Vectors
Transformations that map vectors to their coordinate vectors will prove to be of great
importance. In this section we will prove that such transformations are linear and
give several examples.
If is a vector space, and is an ordered basis for then any vector of can be uniquely
expressed as for some scalars . Vector in given by
is said to be the coordinate vector for with respect to the ordered basis . (See
Definition def:coordvector.)
It turns out that the transformation defined by is linear. Before we prove linearity
of , consider the following examples.
Consider . Let be an ordered basis for . (You should do a quick mental check that
is a legitimate basis.) Define by . Find .
We need to find the coordinate vector for
with respect to .
This gives us:
Let be an ordered basis for . (It is easy to verify that is a basis.) If is
given by , find .
(b)
Let be an ordered basis for . (In Practice Problem prob:linindabstractvsp1, you demonstrated
that is a basis.) If is given by , find .
item:lintranspolycoordvect1 We express as a linear combination of elements of .
Therefore
Note that it is important to keep the basis elements in the same order in which they
are listed, as the order of components of the coordinate vector depends on the order
of the basis elements. We conclude that
item:lintranspolycoordvect2 Our goal is to express as a linear combination of the elements of . Thus, we need to
find coefficients , and such that
This gives us a system of linear equations:
Solving the system yields , and . Thus
Let be an -dimensional vector space, and let be an ordered basis for . Then given
by is a linear transformation.
Proof
First observe that Theorem th:uniquerep of Bases and Dimension of Abstract
Vector Spaces guarantees that there is only one way to represent each element of
as a linear combination of elements of . Thus each element of maps to exactly
one element of , as long as the order in which elements of appear is taken into
account. This proves that is a function, or a transformation. We will now prove
that is linear.
Let be an element of . We will first show that . Suppose , then can be written
as a unique linear combination:
We have:
In our final example, we will consider in the context of a basis of the codomain, as
well as a basis of the domain. This will later help us tackle the question
of the matrix of associated with bases other than the standard basis of
.
Let
Let
Because each of and is linearly independent, let
be ordered bases of and , respectively.
Define a linear transformation by
(a)
Verify that is in and find the coordinate vector .
(b)
Find and the coordinate vector .
item:subtosub1a We need to express as a linear combination of and . This can be done by
observation or by solving the equation
We find that and , so . Thus is in . The coordinate vector for with respect to the
ordered basis is
Let and be vector spaces, and let and be ordered bases of and , respectively.
Suppose is a linear transformation such that:
If , express as a linear combination of vectors of .
Find and