Last time, we began our discussion of linear transformations, and in particular observed that the set of linear transformations from a vector space to a vector space is itself a vector space in a natural way, because there are natural ways to add and scale linear transformations which are compliant with the vector space axioms. In the case that linear transformations are usually referred to as “linear operators,” and the vector space of linear operators on is typically denoted This notation stems from the fact that a fancy name for a linear operator is endomorphism. This term derives from the Greek *endon*, meaning within, and is broadly used in mathematics to emphasize that one is considering a function whose range is contained within its domain. Linear operators are special in that, in addition to being able to scale and add them to one another, we can also multiply them in a natural way. Indeed, given two linear operators we may define their **product** to be their composition, i.e. AB:=A \circ B. Spelled out, this means that is the linear operator defined by

So is a special type of vector space whose vectors (which are operators) can be scaled, added, and multiplied. Such vector spaces warrant their own name.

**Definition 1:** An **algebra** is a vector space together with a multiplication rule

which is bilinear and associative.

Previously, the only algebra we had encountered was and now we find that there are in fact many more algebras, namely all vector spaces for an arbitrary vector space. So, linear operators are in some sense a generalization of numbers.

However, there are some notable differences between numerical multiplication and the multiplication of operators. One of the main differences is that multiplication of linear operators is noncommutative: it is not necessarily the case that

**Exercise 1:** Find an example of linear operators such that

Another key difference between the arithmetic of numbers and the arithmetic of operators is that division is only sometimes possible: it is not the case that all non-zero operators have a multiplicative inverse, which is defined as follows.

**Definition 2:** An operator is said to be **invertible** if there exists an operator such that where is the **identity operator** defined by for all

You should take a moment to compare this definition of invertible linear operator with the definition of a vector space isomorphism from Lecture 2. You will then that being invertible is equivalent to being an isomorphism of with itself. An isomorphism from a vector space to itself is called an **automorphism** where the prefix “auto” is from the Greek work for “self.” The set of all invertible linear operators in is therefore often denoted

**Proposition 1:** If then there is precisely one operator such that

*Proof:* Suppose that are such that

Then we have

— Q.E.D.

Thus, if is an invertible operator, then it has a unique inverse, so it is reasonable to call this “the inverse” of and denote it You should check for yourself that is invertible, and that its inverse is i.e. that

**Exercise 2:** Find an example of a nonzero linear operator which is not invertible.

**Proposition 2: **The set of invertible operators is closed under multiplication: if then

*Proof:* We have

which shows that is invertible, and that .

— Q.E.D.

Proposition 2 shows that the set is an example of a type of algebraic structure called a group, which roughly means a set together with a notion of multiplication in which every element has an inverse. We won’t give the precise definition of a group, since the above is the only example of a group we will see in this course. The subject of group theory is its own branch of algebra, and it has many connections to linear algebra.

All of the above may seem quite abstract, and perhaps it is. However, in the case of finite-dimensional vector spaces, linear transformations can be described very concretely as tables of numbers, i.e. as matrices. Consider the vector space of linear transformations from an -dimensional vector space to an -dimensional vector space Let be a linear transformation, let be a basis of and let be a basis of The transformation is then uniquely determined by the finitely many vectors

Indeed, any vector may be uniquely represented as a linear combination of vectors in

and we then have

Now, we may represent each of the vectors as a linear combination of the vectors in

and we then have

Thus, if

is the unique representation of the vector relative to the basis of So, our computation shows that we have the matrix equation

Schematically, this matrix equation can be expressed as follows: for any we have that

where, denotes the matrix whose entries are the coordinates of the vector relative to the basis of denotes the matrix whose entries are the coordinates of the vector relative to the basis of and is the matrix

whose th column is the matrix What this means at the conceptual level is the following: choosing a basis in results in a vector space isomorphism defined by

choosing a basis in results in a vector space isomorphism defined by

and these two choices together result in a vector space isomorphism defined by

Let us consider how the above works in the special case that and We are then dealing with linear operators and the matrix representing such an operator is the square matrix

For every we have the matrix equation

In this case, there is an extra consideration. Suppose we have two linear operators Then, we also have their product and a natural issue is to determine the relationship between the matrices and Let us now work this out.

Start with a vector and let

be its representation relative to the basis of Let

be the representations of the vectors relative to the basis

We then have

This shows that the matrix of the product transformation relative to the basis is given by the product of the matrices representing and in this basis, i.e.

So, in the case of linear operators, the isomorphism given by

is not just a vector space isomorphism, but a vector space isomorphism compatible with multiplication — an algebra isomorphism.

“AB:=A \circ B” was near the top of the lecture. Was \circ supposed to be compiled using LaTeX?