Matrices of Linear Transformations with Respect to Arbitrary Bases

We know that every linear transformation from into is a matrix transformation (Theorem th:matlin of Standard Matrix of a Linear Transformation from to ). What about linear transformations between vector spaces other than ? In this section 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 Isomorphic Vector Spaces). 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 space.

In Isomorphic Vector Spaces we introduced the transformation defined by

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 challenged to do in Isomorphic Vector Spaces.)

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.

Let be our ordered bases or choice for and , respectively.

Recall that a coordinate vector isomorphism maps a vector to its coordinate vector with respect to the given ordered basis (Theorem ex:coordmapiso of Isomorphic Vector Spaces). In the diagram below, let and be coordinate vector isomorphisms with respect to and .

Define by Observe that . Because is a composition of linear transformations, itself is linear (Theorem th:complinear of Composition and Inverses of Linear Transformations). 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 ordered bases 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 Isomorphic Vector Spaces). While we could have proved this result directly, as you were challenged to do in Isomorphic Vector Spaces, this approach is much less tedious.

Let

In Example ex:subtosub1 we defined and as follows:

Geometrically, and are planes in . We chose as ordered 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?

(a)
and are elements of , ,
(b)
and are elements of , ,

Now we find the coordinate vectors.

(a)
(b)

Here is a diagram that summarizes this information. (Press the arrow on the right to expand.)

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 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

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 examples.

Let and be vector spaces with ordered 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 by

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 the following:

Vectors will become the columns of the standard matrix. We summarize this discussion as a theorem.

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 spaces.

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 be ordered bases of 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 .

Practice Problems

Verify that of Exploration init:taumatrix is linear.
Define by
Show that is linear.
Is an isomorphism? Yes, No
Let be ordered bases for and , respectively.

Find the matrix for with respect to and .

Answer:

Let and be vector spaces with ordered bases and , respectively. Define a linear transformation by Find the matrix of with respect to and , and use it to find . Verify your answer by computing directly.

Let Let and be subspaces of with ordered bases respectively.

Let be a linear transformation such that

Show that lie in by expressing them as linear combinations of and .
Find the matrix of with respect to and .

Find the matrix of with respect to and .

Find .