We've actually informally introduced and discussed
about vector spaces when we were proving the
Wedderburn's little theorem, stating that
finite division rings are vector spaces over their centers.
We are using vector space as a tool for deriving
numberical relationship between the elements in the
division ring and and the elements in the center.
Vector spaces and linear transformations are concepts
from undergraduate linear algebra, but in this text,
we will extend them to ring theory discuss them in
depth: first we will give vector spaces a formal and
rigid definition and formalize them as algebraic
objects; then we will discuss their homomorphisms,
which are called the the linear transformations;
finally we will reveal and prove that all linear
transformations on finite dimensional vector spaces
are representable by matrices, from a more abstract
point of view and with more details.
Vector spaces
Vector spaces in ring theory is an extension
of vector spaces of real numbers (like
of -dimensional real vectors)
to arbitrary field, that a vector space
over field is an abelian group of
vectors with vector
addition and additive identity ,
equipped with scalar multiplication
(where is called the scalar),
so that :
- .
- .
- .
- .
By using somebasic trick in ring theory, it
won't be hard to show that for any
, we have
, and
.
Subspace and linear independence
A subspace is defined as a
subgroup of closed under scalar
multiplication, so that
.
A way of constructing subspace of is
to take some vectors
out of , and make the set
of linear combinations of them. Then there's
by:
Such a subspace is called the
spanning set of
,
or the subspace spanned by
.
While there's no constraint on what vectors must
be chosen to span the subspace , without losing
generality, assume non-zero vector
can be represented by other vectors through:
Clearly not all are
zero, otherwise would be zero then.
And For any vector with non-zero
scalar , there's:
So is also a vector representable
by other vectors. And for any vector that can
be located by
,
obviously it can also be located by
.
This is what will happen when there were
"redundant" vectors chosen to span the subspace
. And such phenomenon inspire us to define
the linear (in)dependence for vectors,
that among the vectors
, if
some vectors are representable by other vectors,
then there must be:
And these vectors are said to be linearly
dependent. Contrary to this, if all vectors
are not representable by others remaining, in
order for
,
the only possibility is
. So
we define these vectors to be linearly
independent when there's:
So when linearly dependent vectors are chosen
to span , we can safely remove some
"redundant" vectors from them without affecting
the remaining vectors' capability to span .
And each time when we are able to remove
"redundant" vector, by the rationale above
there're always multiple choices (two or more)
to remove, so there're many possibilities when
we reach the point that remaining vectors are
linearly independent and removing any more of
them will fail to span . But for any two
possibilities among them that
and
,
which are both linearly independent and will
span , there must be .
To prove, first let's show that if is
spanned by linearly independent vectors
,
then vectors
must be linearly dependent if .
Compatible with matrix multiplication there's:
Where .
Compatible with Gauss's elimination
on the first -rows, where must
be invertible. If after the elimination,
is not upper-triangular.
then the last row of must
be all zeroes, and the last row of
must also be .
Let the last row of be
,
which must not be all-zeros, then there's:
So the first vectors of are linearly
dependent, and we are done.
Otherwise If is
upper-triangular, we can furtherly eliminate the
last rows with Gauss's elimination
, so that will be
cleared to zeros after elimination, alongside
with , that:
Noted that there's
,
and since there's
,
we have
.
So among the vectors
,
the last vectors are representable by first
vectors, and they are thus linear dependent.
Then, if subspace is spanned by both
and
,
where they are both linear indepdent but
, without losing generality, let's
assume it's the case that , then
must be linear dependent, which is a contradiction.
Finite dimensional vector spaces and its subspaces
The set of linearly
independent vectors that spans subspace is called
the basis of . Obviously there're always
multiple choices of : given a choice
of basis , linearly combine them so
that they are now linearly dependent, and this
time keep the combined vector and choose another
vector in for removal. But for any
two choices of basis, the number of vectors inside
must always be the same, and is truly an invariance
of subspace spanned by basis. We call it the
dimension of , and denoted it as .
Conversely, given a basis , we can
use it to span a subspace, and such a subspace is
denoted as . For
convenience, if the basis is finite,
then is said to be a
finite dimensional subspace, or subspace of
dimension . And we can treat the
vector space itself as a trivial subspace, calling
it finite dimensional vector space or vector
space of dimension correspondingly.
Specially, there's
.
Any subspace of finite dimensional vector space
is also finite dimensional. This can be shown by
performing the algorithm we've described
here,
by starting at ,
iteratively selecting random vector
, and adding
it to . The randomly selected vector
is guaranteed to be linearly independent
with vectors in already selected basis
,
otherwise there's
,
and
,
which is a contradiction. The algorithm guarantees
that , and will select
no more than vectors, otherwise they will
be linearly dependent, so algorithm is
guaranteed to halt. However the basis of are
not always the basis of : consider the
-dimensional Cartesian coordinate system
with basis
corresponding
to -axis, the vectors on the line
is a subspace of
spanned by , but not
possibly by any one from
.
Just like other algebraic objects, we would like
to define their "homomorphisms". And for vector
spaces, their homomorphisms are conventionally
called the linear transformations. That is,
for two vector spaces over , a linear
transformation is a map fulfilling:
So is a group homomorphism between abelian
groups in the first place, so its
kernel is closed under addition. And
in order for to be the kernel of a
linear transformation, it must fulfil
,
which means it must also be closed under scalar
multiplication. Given that is closed
under addition and scalar multiplication, it
must be subspace of .
Conversely, Consider the canonical homomorphism
of any subspace . When viewing
as quotient group, there's
;
while it's clear that
,
so the quotient group is compatible
with homomorphic scalar multiplication. By now,
we know the canonical homomorphism of any
subspace is a linear transformation.
Rank-nullity theorem
The kernel is conventionally called
the null space of , while the image
is called the range space of .
For finite dimensional vector space , there's
rank-nullity thereom stating that
, where the
dimension of null space is called the
nullity and the dimension of the range
space is called the rank.
To prove, since is a subspace of ,
it is also finite dimensional, let one of its
basis be
,
since vectors in are already
linearly independent and is subspace
of , we can continue on selecting more
vectors from so that
is one of the basis of , and maps
all vectors in by
.
To show the set that set of vectors
are linearly independent, assume they are not
linearly independent instead, so we have
,
for some s that are not all zeroes,
which in turns implies for some s
that are not all zeroes there're
and
.
In this way, the vectors in are not
linearly independent, which is a contradiction.
So we have ,
and by putting all information above together,
we have
,
and we are done.
For linear transformation between finite
dimensional vector spaces that ,
we would like to show that is injective iff
it's surjective: when is surjective,
and , so
it's injective; and when is injective, let
the the basis of be , for any
vector , it's linearly
independent with vectors in , and
now we have linearly
independent vectors in , which is a contradiction.
So must be surjective as long as it's injective.
Quotient space and the first isomorphism theorem
On the other hand, consider the vectors
,
which are extended from the basis
of
to span the whole , after being
attached to as cosets
,
can span all cosets in the quotient space
: for every vector
,
by performing the canonical homomorphism of ,
we will see the portion of
absorbed into , leaving
identifying the coset in . Therefore,
by using
as basis for , we can see they span all
cosets in . Each base vector
for corresponds
to the base vector for ,
fulfilling the first isomorphism
theorem unexceptionally.
Isomorphisms of vector spaces of the same dimension
Let be set of the column vector of
elements from in the form of
.
Compatible with vector addition and scalar
multiplication, is indeed a vector space
of dimension , with vectors in basis
.
Any vector space of dimension is
isomorphic to , since given its basis
,
consider the linear transformation
defined by
,
it's obviously injective otherwise vectors
in will be linearly dependent,
and is thus surjective. So linear transformation
is an isomorphism between and
of dimension .
Sum and intersection of subspaces
For subspaces of vector space ,
when viewing as abelian groups, it's obvious
that and are both normal
subgroups of . And by
,
,
both and are subspaces
of . When is trivial,
is inner direct product of and and
is called the direct sum, denoted as
conventionally. When viewing as
normal subgroups, they obeys the second
isomorphism theorem that
. Let
be finite dimensional vector space, consider
the canonical homomorphism of and
, we have
,
and thus
.
We are almost familiar with linear algebra,
and must have heard (I think those who are not
major in mathematics are generally on the level
of "heard") the statement that all linear
transformations of finite dimensional vector
space are matrices, but we may not know how
they are actually associated and proved they are
truly the same. And let's focus on the connection
between matrices and linear transformations now.
Let and be vector spaces of dimension
and and with basis
and
.
Consider a linear transformationation ,
any vector
will be transformed into
.
For every vectors , if
we know can be represented by
vectors in that
,
then there's
.
And compatible with matrix multiplication there's:
By now, we can clearly see the role matrices
are playing and how they are associated with
linear transformations: for a linear
transformation between finite
dimensional vector spaces, once we are able to
determine basis of and
of , we will know how to
transform their scalars or coordinate, which
can be represented as matrix multiplication.
And the matrix for transforming scalars or
coordinate is called the matrix of
relative to basis
of and basis
of .
Let there be two linear transformation
and with basis
for , basis
for and basis
for respectively, matrix be
matrix of relative to and
, and matrix be
matrix of relative to and
. Given that:
The matrix of is multiplication of
's own matrices when fixing the source,
intermediate and target vector spaces' basis.
This leads us to that all matrix of bijective
linear transformations are invertible: for a
matrix of a bijective linear
transformation relative to basis
and , consider the
matrix of
relative to basis and .
Given that the composition
is the identity linear transformation on ,
whose matrix relative to must
only be the identity matrix ,
there must only be
, so
must be invertible with inverse
, which can be evaluated by
inspecting how each vector in
can be represented as linear combination of
vectors in .
Similarily we define the null space of
matrix as
.
where is matrix of relative
to basis
of and basis
of . The linear transformation
defined by
is an isomorphism, and we will show
:
- ,
,
.
-
.
- .
To soothe our intuition, we could define some kind
of "inner product" between vectors
as
, and two
vectors are said to be orthogonal or perpendicular
when there's
,
then clearly are all vectors
orthogonal to every row vectors of .
Define the row space
of matrix as the subspace of
spanned by row vectors
,
then by
,
any vector from the row space of is
orthogonal to any vector from the null space
of .
On the other hand, define the column space
of matrix
as the subspace of spanned by
its column vectors
,
and the column rank
of matrix
as the dimension of the column space.
Naturally the map defined by
is a linear transformation with its image being
. If is the
matrix of linear transformation
relative to basis
of and basis
of , we've prove that
,
and from this relationship we can decompose
into three linear transformations: first
we extract the coordinate with respect to basis
with linear transformation
defined by
,
it's clear that is bijective; then we
apply the linear transformation naturally
defined by ; finally we reassemble the
coordinate with respect to basis
with linear transformation
defined by
,
it's also bijective for domain and range
. So there's ,
while are both bijective, it's
only possible that
.
In this way we build up the relationship
between the column rank of and
the rank of linear transform .
Row rank equals column rank in a matrix
Given that there's column rank, there's also
row rank
that is defined to be the dimension of the
row space. And for matrix ,
there's always
:
- First let's do the Gauss's elimination
on the rows of ,
so that is in
upper-triangle or staircase shape. When
viewing form the row side, the row vectors
are transformed into
where .
And for each non-zero row vector
whose -th column
is not zero (and there's by it's
in upper-triangle or staircase case), it
is not representable by
since they have non-zero component on the
first columns; it is also not
representable by
since their -th column is zero. So the
number of non-zero row vectors in
is directly
.
Given that
,
defines a bijective linear
transformation between
and ,
and we have
.
- When viewing from the column side, the
column space of
are spanned by vectors
.
Given that
and
,
is
bijectively transformed from
by
and we have
.
- Let's furtherly perform the Gauss's elimination
on the columns of
, so that it becomes
.
Given that
,
defines a bijective linear
transformation between
and
.
By joining all these above together, we have
,
and we are done.
And given that there's
for any matrix , we define the
rank of matrix to be the
column rank or the row rank of ,
denoting it as . In this way,
the rank of the matrix is in accordance with
the rank of the linear transformation
it represents.
Conclusion
In this text, we introduce the concept of
vector spaces and linear transformation from a
more abstract point of view, by studying them
as algebraic objects and their homomorphisms
between the objects.
For vector spaces, the most special kind of
them is the finite dimensional vector space,
in which we can take finite linear independent
vectors to span the whole space. There're
always multiple choices of basis but the
dimension of the space is an invariance,
constraining the exact number of linear
independent vectors required to span the space.
Every subspace of a finite dimensional vector
space can also be shown to be finite dimensional.
For linear transformations, which is the
homomorphism between vector spaces, it can be
shown any subspace of the domain vector space
can be used as kernel as a linear transformation.
For finite dimensional vector spaces, the
rank-nullity theorem must hold, and all
-dimensional vector spaces are isomorphic to
where is the field of scalars.
Finally, we've revealed the interconnection
between matrices and linear transformations,
that when fixing sets of basis of the domain
and range vector spaces, linear transformations
can always be represented by matrices. We've
also inspected the crucial vector spaces
defined by the matrix and related to the
linear transformations: the row space of the
matrix is the vector space spanned by row
vectors of the matrix; the null space of the
matrix is formed by the vectors perpendicular
to the row space, it has the same dimension
as the null space of the linear transformation;
the column space of the matrix is the vector
space spanned by column vectors of the matrix,
it has the same dimension as the rank of the
linear transformation. The dimension of the
row space, which is called the row rank, can
be shown to be equal to the dimension of the
column space, which is called the column rank.
Given that there's such accordance, we call
the row rank and the column rank of the matrix
the rank of the matrix uniformly, and it equals
to the rank of the linear transformation.
Although what's addressed in this text is far
from what one should know about linear algebra,
but I believe it has introduced some fundamental
concepts and can make one's mind clear about
linear algebra: it's a branch of mathematics
studying vector spaces and linear transformations,
rather than merely telling undergraduates how to
manipulate matrices, evaluating determinants,
evaluating eigenvalues and eigen vectors, and
so on. It's also enough for us to "toolizing"
vector spaces and linear transformations in
ring theory for now.