Archive for the ‘linear algebra’ 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 1 Let
and
be vector spaces over the field
. A linear transformation
is 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 2 Let
and
be vector spaces over a field
with
finite-dimensional and let
be a linear transformation. Then,
- The range of
, denoted range
, is the set
. The rank of
is dimension of range
.
- The null space of
, denoted null
, is the set
. The nullity of
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 3 Let
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.