Math 31AH: Lecture 11

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.