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We find the matrix of a linear transformation with respect to arbitrary bases, and
find the matrix of an inverse linear transformation.
LTR-0080: Matrix of a Linear Transformation with Respect to Arbitrary
We know that every linear transformation from into is a matrix transformation
(Theorem th:matlin of LTR-0020). What about linear transformations between vector spaces
other than ? In this module we will learn to represent linear transformations between
arbitrary finite-dimensional vector spaces using matrices. To do so, we will use the
fact that every -dimensional vector space is isomorphic to (Corollary cor:ndimisotorn of LTR-0060).
What we do here will serve as yet another example of how isomorphisms can be used
to translate problems in one vector space to another, more convenient, vector
In LTR-0060 we introduced the transformation
(Expression eq:justforfuniso2 of LTR-0060).
You should verify that is linear. (See Practice Problem prob:taulinear.)
We will examine in an effort to find a way to represent it with a matrix. (In the
process, we will also end up proving that is an isomorphism, which is what you were
challanged to do in LTR-0060.)
We will start by selecting a basis for each of and . We can choose any basis for either
space, but we will choose bases that will make computations easier.
be our bases of choice for and , respectively.
Recall that a coordinate vector isomorphism maps a vector to its coordinate vector
with respect to the given basis (Example ex:coordmapiso of LTR-0060). In the diagram
below, let and be coordinate vector isomorphisms with respect to and
Observe that . Because is a composition of linear transformations, itself is linear
(Theorem th:complinear of LTR-0030). Thus, we should be able to find the standard matrix for .
To do this, find the images of the standard unit vectors and use them to create the
standard matrix for .
We say that is the matrix of with respect to and .
As a side-note, observe that . Observe also that is invertible because is
invertible. So, is an isomorphism. As a composition of isomorphisms, is an
isomorphism (Theorem th:isocompisiso of LTR-0060). While we could have proved this result
directly, as you were challenged to do in LTR-0060, this approach is much less
In Example ex:subtosub1 of LTR-M-0025 we defined and as follows:
Geometrically, and are planes in . We chose
as bases of and , respectively.
We also defined a linear transformation by
Our goal now is to find a matrix for with respect to and .
The information given in this problem is slightly different from the information in
Exploration init:taumatrix. Instead of being given an expression for the image of a generic vector of
, we are only given the images of the two basis vectors of . But this information is
sufficient to determine the linear transformation.
As before, we will map vectors of and to their coordinate vectors. Where are the
coordinate vectors located?
and are elements of , ,
and are elements of , ,
Now we find the coordinate vectors.
Here is a diagram that summarizes this information. (Press the arrow on the right to
Define by . is a linear transformation that maps and to and , respectively. Thus,
the standard matrix for is:
We say that is a matrix for with respect to and .
Let’s take a look at what this matrix can do for us. Recall that in Example ex:subtosub1 of
LTR-M-0025 we found that the image of is
This result can be obtained by finding the product of with the coordinate vector of
Given any vector of , we can find as follows:
This gives us
Let and be vector spaces with bases and , respectively. Define a linear
Find the matrix of with respect to and , and use it to find . Verify your answer by
We start with a diagram:
Looking at the images of the standard unit vectors in , we can construct the standard
matrix of with respect to and .
Applying this matrix to the coordinate vector of we get
This means that . We can verify this by direct computation as follows:
The Matrix of a Linear Transformation
In this section we will formalize the process for finding the matrix of a linear
transformation with respect to arbitrary bases that we established through earlier
Let and be vector spaces with bases and , respectively. Suppose is a linear
transformation. Our goal is to find a matrix for with respect to and .
Observe that and . Let and be coordinate vector isomorphisms defined
We know that is an isomorphism and exists. Consider the transformation
As a composition of linear transformation, is linear and thus has a standard matrix.
To find it, we need to determine the images of standard unit vectors under . We have
Vectors will become the columns of the standard matrix. We summarize this
discussion as a theorem.
Let and be finite-dimensional vector spaces with bases and , respectively. Suppose
is a linear transformation.
Then for all vectors in .
In conclusion, observe how isomorphisms helped us solve the matrix of a linear
transformation problem. The coordinate mappings and are isomorphisms. This
means that and are isomorphic and have the same structural properties. The same
is true for and . In this abstract discussion, we do not know anything about the
elements of and , but isomorphisms allow us to take a problem that we do not
know much about and transform it to a familiar problem involving familiar
The Inverse of a Linear Transformation and its Matrix
Let and be vector spaces. Suppose is an invertible linear transformation. This, of
course, means that is an isomorphism, which means that
Let and , respectively. We can find the matrix of with respect to and by finding
the standard matrix of the linear transformation .
Observe that is the inverse of . So, if is the standard matrix of , then
is the standard matrix of . Thus, is the matrix of with respect to and
In Exploration Problem init:subtosub of LTR-M-0030, we introduced the following set
Define a linear transformation
In Example ex:subtosubinvert of LTR-M-0035, we proved that is invertible. Find the matrix of with
respect to bases and
Consider the diagram:
This gives us the matrix of with respect to and :
We now find .
is the matrix of with respect to and .