Let be a vector space. Recall from Lecture 1 (together with Assignment 1, Problem 2) that to say is -dimensional means that contains a linearly independent set of size but does not contain a linearly independent set of size Also recall from Lecture 1 that a basis of is a subset of which is linearly independent and spans The main goal of this lecture is to prove the following theorem.

**Theorem 1:** A vector space is -dimensional if and only if it contains a basis of size

Before going on to the proof of this theorem, let us pause to consider its ramifications. Importantly, Theorem 1 gives a method to calculate the dimension of a given vector space : all we have to do is find a basis of and then count the number of elements in that basis. As an example, let us compute the dimension of This vector space is typically used to model the world in which we live, which consists of four physical dimensions: three for space and one for time. Let us verify that our mathematical definition of vector space dimension matches our physical understanding of dimension.

Let be a vector in Then, we have

where

This shows that the set spans Let us check that is linearly independent. This amounts to performing the above manipulation in reverse. If are numbers such that

then we have

which means that Thus is a linearly independent set which spans i.e. it is a basis of Since has size we conclude from Theorem 1 that

Now let us prove Theorem 1. First, observe that we have already proved that if is -dimensional then it contains a basis of size — this is Theorem 1 from Lecture 1. It remains to prove the converse, which we now state as a standalone result for emphasis and ease of reference.

**Theorem 2:** If contains a basis of size then is -dimensional.

The proof of Theorem 2 is quite subtle. In order to make it easier to understand, it is helpful to first prove the following lemma, in which the main difficulty is concentrated.

**Lemma 1:** If is a linearly independent set in a vector space and is a basis of then

*Proof:* Suppose this were false, i.e. that there exists in the vector space a linearly independent set and a basis such that We will see that this leads to a contradiction. The strategy is the following: we wish to demonstrate that the assumption implies that we can replace of the vectors in with the vectors in in such a way that the resulting set

is linearly independent. If we can show this, we will have obtained the desired contradiction: cannot possibly be independent, because is a basis. In particular, each of the remaining vectors is a linear combination of the vectors in

We will pursue the following strategy to show that the existence of the set as above follows from the hypothesis For each consider the following proposition: there exists a linearly independent set in of the form

where is a subset of of size and Let us call this proposition Now, if we can prove that is true, and that

then we can conclude that is true for all Indeed, we would then have

This is one version of a proof technique known as mathematical induction. But the statement true is exactly what we want, since it gives us a linearly independent set consisting of some number of vectors from together with all vectors in which results in the contradiction explained above.

Let us now implement the above strategy. The first step is to prove that is true. To see that it is, consider the set where and This set is of the required form, and since is linearly independent so is

It remains to prove that if and is true, then is true. Given that is true, there exists a linearly independent set in of the form with

an -element subset of and

Consider the set

If is linearly independent, then so is any subset of it, so in particular the subset

is linearly independent and is true. Now suppose that is linearly dependent. Then, because is linearly independent, must be a linear combination of the vectors in i.e

for some Moreover, there exists a number such that else would be a linear combination of which is impossible because is a linearly independent set. We now claim that the set

is linearly independent. Indeed, if were linearly dependent, then we would have

where not all the numbers are zero. This means that the number since otherwise the above would say that a subset of is linearly dependent, which is false because is linearly independent. Now, if we substitute the representation of as a linear combination of elements given above, this becomes a vanishing linear combination of the vectors in in which the coefficient of is which contradicts the linear independence of So, must be linearly independent. — Q.E.D.

Let us note the following corollary of Lemma 1.

**Corollary 1: **If and are two bases of a vector space then

*Proof: *Since is linearly independent and is a basis, we have by Lemma 1. On the other hand, since is linearly independent and is a basis, we also have by Lemma 1. Thus — Q.E.D.

We now have everything we need to prove Theorem 2.

*Proof of Theorem 2:* Let be a finite-dimensional vector space which contains a basis of size We will prove that is -dimensional using the definition of vector space dimension (Definition 4 in Lecture 1) and Lemma 1. First, since is a linearly independent set of size we can be sure that contains a linearly independent set of size It remains to show that does not contain a linearly independent set of size This follows from Lemma 1: since is a basis of size every linearly independent set in must have size less than or equal to — Q.E.D.

Another very important consequence of Theorem 2 is the following: it reveals that any two vector spaces of the same dimension can more or less be considered the same. Note that this is fundamentally new territory for us; so far, we have only considered one vector space at a time, but now we are going to compare two vector spaces.

To make the above precise, we need to consider functions between vector spaces. In fact, it is sufficient to limit ourselves to the consideration of functions which are compatible with the operations of vector addition and scalar multiplication. This leads to the definition of a linear transformation from a vector space to another vector space If vector spaces are the foundation of linear algebra, then linear transformations are the structures we want to build on these foundations.

**Defintion 1: **A function is said to be a **linear transformation** if it has the following properties:

- where is the zero vector in and is the zero vector in
- For any vectors and any scalars we have

So, a linear transformation is a special kind of function from one vector space to another. The name “linear transformation” comes from the fact that these special functions are generalizations of lines in which pass through the origin More precisely, every such line is the graph of a linear transformation Indeed, a line through the origin in of slope is the graph of the linear transformation defined by

The one exception to this is the vertical line in passing through which has slope . This line is not the graph of any function from to which should be clear if you remember the vertical line test.

To reiterate, a linear transformation from a vector space to is a special kind of function For linear transformations, it is common to write instead of and this shorthand is often used to implicitly indicate that is a linear transformation. In the second half of the course, we will discuss linear transformations in great detail. At present, however, we are only concerned with a special type of linear transformation called an “isomorphism.”

**Defintion 2:** A linear transformation is said to be an **isomorphism** if there exists a linear transformation such that

and

The word “isomorphism” comes from the Greek “iso,” which means “same,” and “morph” which means “shape.” It is not always the case that there exists an isomorphism from to ; when an isomorphism does exist, one says that and are **isomorphic**. Isomorphic vector spaces and have the “same shape” in the sense that there is both a linear transformation which transforms every vector into a vector and an inverse transformation which “undoes” by transforming back into To understand this, it may be helpful to think of isomorphic vector spaces and as two different languages. Any two human languages, no matter how different they may seem, are “isomorphic” in the sense that they describe exactly the same thing, namely the totality of human experience. The isomorphism translates every word in the language to the corresponding word in which means the same thing. The inverse isomorphism translates back from language to language On the other hand, if is a human language and is the language of an alien civilization, then and are not isomorphic, since the experience of membership in human society is fundamentally different from the experience of membership in a non-human society, and this difference is not merely a matter of language.

**Theorem 2**: Any two -dimensional vector spaces and are isomorphic.

*Proof: *Since is -dimensional, it contains a basis by Theorem 1. Likewise, since is -dimensional, it contains a basis Now, since is a basis in for every there exist unique scalars such that

Likewise, since since is a basis in for every there exist unique scalars such that

We may thus define functions

by

and

We claim that is a linear transformation from to To verify this, we must demonstrate that has the two properties stipulated by Defintion 1. First, we have

which verifies the first property. Next, let be any two vectors, and let be any two scalars. Let

be the unique representations of and as linear combinations of the vectors in the basis We then have

which verifies the second property. In the same way, one checks that is a linear transformation from to

To prove that the linear transformation is an isomorphism, it remains only to prove that the linear transformation undoes To see this, let

be an arbitrary vector in expressed as a linear combination of the vectors in the basis We then have

This completes the proof that and are isomorphic vector spaces. — Q.E.D.

To continue with our linguistic analogy, Theorem 2 says that any two -dimensional vector spaces are just different languages expressing the same set of concepts. From this perspective, it is desirable to choose a “standard language” into which everything should be translated. In linguistics, such a language is called a lingua franca, a term which reflects the fact that this standard language was once the historical predecessor of modern French (these days the lingua franca is English, but this too may eventually change).

The lingua franca for -dimensional vector spaces is and it is unlikely that this will ever change. Let be an -dimensional vector space, and let be a basis in . Consider the basis of in which

To be perfectly clear, for each the vector has the number in position and a zero in every other position. The basis is called the **standard basis** of and you should immediately stop reading and check for yourself that it really is a basis. Assuming you have done so, we proceed to define an isomorphism

as follows. Given a vector let

be the unique representation of as a linear combination of vectors in Now set

where the symbol “” means “equal by definition.” The fact that this really is an isomorphism follows from the proof of Theorem 2 above. The isomorphism is called the **coordinate isomorphism** relative to the basis and the geometric vector is called the **coordinate vector** of relative to the basis Because of the special form of the standard basis of we may write the coordinate vector of more concisely as

When working with a given -dimensional vector space, it is often convenient to choose a basis in and use the corresponding coordinate isomorphism to work in the standard -dimensional vector space rather than working in the original vector space . For example, someone could come to you with a strange -dimensional vector space and claim that this vector space is the best model for spacetime. If you find that you disagree, you can choose a convenient basis and use the corresponding coordinate isomorphism to transform their model into the standard model of spacetime.

## 4 Comments