We met linear transformations for the first time back in Lecture 2, and have encountered them a handful of times in subsequent lectures. In this lecture, we begin a systematic study of linear transformation that will take up the rest of the course. As with the study of any new species, there will be a lot of classifying: we will divide up linear transformations into various classes defined by certain common features which to a large extent determine their behavior. Before classifying linear transformations into various special species and families, let us consider the ecosystem of all linear transformations as a collective whole.
Let and
be vector spaces, and let
be the set of all linear transformations
This seemingly strange notation stems from the fact that is a fancy alternative name for linear transformations is homomorphisms, which roughly means
“same shape” in Greek.
Theorem 1: is a vector space.
Proof: In order to promote from simply a set to a vector space, we need to give a rule for adding and scaling linear transformations. These rules are simply and natural. If
linear transformations and
are scalars we define a new linear transformation
by
One must check first that the function from to
so defined satisfies the linear transformation axioms, and then that the set
equipped with these notions of addition and scalar multiplication satisfies the vector spaces axioms. This is left as an exercise to the reader (it is a long but easy exercise which you should do at least once).
— Q.E.D.
Now let us start asking questions about the vectors in i.e. considering what properties a linear transformation
from
to
may or may not have. Given
one of the first things we would like to know is whether it is injective and/or surjective. These questions reduce to questions about certain subspaces of
and
associated to
Definition 1: The kernel of is the subset of
defined by
and the image of is the subset of
defined by
You can think of the kernel and image of as the solution sets to two different equations involving
The kernel in terms of certain vector equations associated to
First, the kernel of
is the set of solutions to the equation
where is an unknown vector in
Second, the image of
is the set of all
such that the equation
has a solution.
Proposition 1: The kernel of is a subspace of
, and the image of
is a subspace of
Proof: Since is a linear transformation, it is by definition the case that
so that
Now let
be any two vectors in the kernel of
and let
be any two scalars. We then have
which shows that is closed under taking linear combinations. Hence,
is a subspace of
Now consider the image of Since
is a linear transformation, it is by definition the case that
so that
Now let
be any two vectors in the image of
Then, by definition of
there exist vectors
such that
So, for any scalars
we have that
which shows that is closed under taking linear combinations.
— Q.E.D.
Proposition 2: The linear transformation is injective if and only if
and surjective if and only if
Proof: First we consider the relationship between the injectivity of and its kernel. Suppose that
Let
be vectors such that
This is equivalent to
which says that Since the only vector in
is
this forces
which means that
and we conclude that
is injective. Conversely, suppose we know that
is injective. Let
be any two vectors in the kernel of
Then
Since
is injective, this forces
which says that any two vectors in
are equal to one another. But this means that any vector in
is equal to
and we conclude that
Now let us consider the relationship between the surjectivity of and its image. Suppose that
This is exactly what it means for the function
to be surjective: every element in the codomain of
is in fact in the range of
Conversely, suppose we know that
is surjective. By definition of surjectivity, this means that
— Q.E.D.
We can use the above propositions to gain some insight into linear equations, meaning equations of the form
where is a given linear transformation from
to
is a given vector in
and
is an unknown vector in
Theorem 1: The solution set of the above linear equation is i.e. the set of of all vectors in
of the form
where
is any particular vector such that
and
Proof: We begin by first consider the homogenous equation associated to the linear equation we want to solve, which is
By definition, the solution set of this homogeneous equation is Now, if
is any vector such that
then we also have
which shows that all vectors in are solutions of the equation
Conversely, suppose that is a vector in
such that
Then, we have
which shows that is a vector in
That is,
for some
or equivalently
This shows that all solutions of
belong to the set
— Q.E.D.
We can now make the following statements about solutions of the linear equation :
- The equation has a solution if and only if
.
- If the equation has a solution, this solution is unique if and only if
- If the equation has a solution but the solution is not unique, then the equation has infinitely many solutions.
Exercise 1: Arrange the above information into a tree diagram which shows the relationship between the various cases.
In Lecture 12, we will think in more detail about the structure of the vector space in the case that
and
are finite-dimensional. In the finite-dimensional case, we shall see that everything we might want to know can be expressed in the language of matrices.
Hi I think there are typos in the two displayed equation in proposition 1.
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Corrected, thanks.
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