## Archive for the ‘prelim’ Tag

### Linear Transformations, Rank-Nullity, and Matrix Representations (Part 1)

One of the most fundamental theorems in Linear Algebra is the Rank-Nullity, or Dimension, Theorem. This post and the next (in an attempt to study for my preliminary exams) will discuss this theorem, along with linear transformations and their relationship to matrices. In what follows, we follow Hoffman and Kunze by letting greek letters be vectors and roman lowercase letters be scalars.

We begin with a few preliminary remarks.

Definition 1Let and be vector spaces over the field . Alinear transformationis a mapping such that

for all and .

Linear Transformations are wildly important in all of mathematics. In , for example, any function which defines a line through the origin is a linear transformation. Why only lines that pass through the origin, rather than *any* line? Say there were a linear transformation such that . Then, , so by linearity, . This is only true if . Indeed, for any linear transformation , , where and are the zero-elements of and , respectively. A more important example of a linear transformation is the derivative function from the space of differentiable functions to itself. Finally, any matrix is a linear transformation from to . We’ll see later that any linear transformation can be written as such a matrix.

Two important properties of any linear transformation are its rank and null space (or kernel).

Definition 2Let and be vector spaces over a field with finite-dimensional and let be a linear transformation. Then,

- The
rangeof , denoted range, is the set . Therankof is dimension of range.- The
null spaceof , denoted null, is the set . Thenullityof is the dimension of null.

Notice that by naming the their dimensions, we’ve implicitly assumed that the range and nullspace of a linear transformation are themselves vector spaces. This is not too hard to justify, so we’ll do it here.

Theorem 3Let and be vector spaces over the filed and let be a linear transformation. Then, range is a subspace of and null is a subspace of .

*Proof:* To show our result, we use the subspace test. So, we will show that range and null are nonempty sets which are closed under vector addition and scalar multiplication.

First recall from above that . This shows that both the range and nullspace of are never empty.

Now, we want to show that if range, then range as well. So, assume and so that range. Then, by linearity, . This means, however, that there is an such that whenever and are in range – i.e., is in the range of . This gives us that range is a subspace of .

We follow the same process for the nullspace. So, assume null. Then, So, null whenver and are. That is, null is a subspace of .

With all of this in hand, we’re finally ready to state and prove the Rank-Nullity theorem for transformations.

Theorem 4 (Rank-Nullity)Let and be vector spaces over the field , finite-dimensional, and let be a linear transformation. Then,

*Proof:* Our proof will proceed as follows: first, we’ll assume a basis for null. We know this exists, since is finite-dimensional and null is a subspace of . From this and the basis for , we’ll construct a spanning set for range and show that it’s linearly independent.

Assume that dim. Then, assume is a basis for null (certainly, ). We contend that every non-empty linearly indepedent set of vectors in a finite dimensional vector space is part of a basis for that vector space. Why? Assume not. If only one element, say , from this linearly independent set, say , is absent from the basis, then either would be linearly independent from the basis, which would be a contradiction, or there would be an element of the basis, say linearly dependent on . If this were the case, however, could take the place of in the basis. This argument can be extended to more elements of until we have shown that no element of can be in the basis. This is clearly absurd. So, since are a linearly independent subset of , we can construct a basis for as .

Now, consider the set . This certainly spans range. Moreover, since are in the nullspace of , for . We have, then, that is a spanning set of range and want to show that it’s linearly independent. So consider the linear combination

for scalars . We want to show that for all . Since is linear, (3) is equivalent to

Thus, is an element of the nullspace of . Since is a basis for the nullspace, we know that

The form a linearly independent set, however, so for all . Thus, we have that is a linearly indepenent set and so a basis for range. Therefore,

as desired.

In introductory Linear Algebra classes and texts, the Rank-Nullity Theorem is usually written in terms of solvability of a matrix. We in the next post, we discuss the representation of transformations by matrices. We will see that this can always be done. Certainly, it’s not hard to see that there are times when it is easier to manipulate matrices than it is linear transformations, and vise versa. Indeed, in practice, to quote Hoffman and Kunze, “We may do this when it is convenient, and when it is not convenient we shall not.” The following post will discuss how to write transformations in terms of matrices as well as how to write the Rank-Nullity Theorem in terms of matrices.