## Transcript — Introduction

In the last lecture, we conjectured that for two vector spaces to be isomorphic, they must have the same dimension. Related to this, we conjectured that an isomorphism must map basis vectors to basis vectors. In this lecture, we will prove both of these conjectures, and more.

## Theorem 8.4.2

Theorem 8.4.2: If V and W are finite-dimensional vector spaces, then V and W are isomorphic if and only if they have the same dimension.

Proof: We start by assuming that V and W are isomorphic, and since V is finite-dimensional, we can assume that the dimension of V is n. We want to prove that the dimension of W is also n. So, we think to ourselves, how can we prove the dimension of W is n? By definition of dimension, we just need to find a basis for W with n vectors in it. From our experience in the examples last lecture, we know how to do this. In particular, we believe that an isomorphism maps basis vectors to basis vectors, so this is what we will try.

Since V and W are isomorphic, there exists an isomorphism L from V to W. If B = {v1 to vn} is a basis for V, then we believe that C = {L(v1) up to L(vn)} is a basis for W. Of course, we have to prove this. For linear independence, consider c1L(v1) + up to cnL(vn) = the 0 vector. Since L is an isomorphism, it is linear, so this gives L(c1v1 + up to cnvn) = the 0 vector. Therefore, by definition, c1v1 + up to cnvn is in the kernel of L. But we also know that L is one-to-one, and hence the kernel of L only contains the 0 vector by Lemma 8.4.1. Thus, c1v1 + up to cnvn = the 0 vector, which implies that c1 = up to cn = 0 since B is linearly independent. And thus, C is linearly independent.

As usual, to prove C spans W, we need to show that if we pick any vector in W, then it can be written as a linear combination of the vectors in C. Let w be any vector in W. Because L is onto, there exists a vector v in V such that L(v) = w. Since B is a basis for V, we can write v as a linear combination of the vectors in B—say, v = d1v1 + up to dnvn. Hence, we have w = L(d1v1 + up to dnvn) = d1L(v1) + up to dnL(vn) since L is linear. Thus, we have written w as a linear combination of the vectors in C, so C also spans W, and hence is a basis for W.

Consequently, the dimension of W equals n, which equals the dimension of V as required.

Before we prove the other direction, I want to take a minute for us to think about the first part of this proof. First, notice that we have proven more than just that the dimension of V equals the dimension of W. We have, in fact, proven our other claim that an isomorphism L maps basis vectors to basis vectors. Also observe that the proof shows that there is a connection between onto and spanning, and a connection between one-to-one and linear independence. Take a minute to look over the first part of the proof.

Now let’s prove the other direction. Assume that the dimension of V equals the dimension of W. We want to prove that this implies that V and W are isomorphic. Thus, we need to construct an isomorphism between them. The first part of the proof teaches us how to do this. Let B = {v1 to vn} be a basis for V, and C = {w1 to wn} be a basis for W. We define a mapping L from V to W which maps the basis vectors in V to the basis vectors in W. In particular, we define L(c1v1 + up to cnvn) = c1w1 + up to cnwn. To prove that L is an isomorphism, we follow our steps exactly as we were doing with the computational problems last lecture.

I will leave the proof that this mapping is linear as an easy exercise. So, one-to-one: Assume that x = c1v1 + up to cnvn is in the kernel of L. Then we have the 0 vector = L(x) = L(c1v1 + up to cnvn), which, by definition of the mapping, equals c1w1 + up to cnwn. Notice that this gives us a linear combination of the vectors in C which equals to the 0 vector. Since C is linearly independent, we get that c1 = up to cn = 0. This implies that x is the 0 vector, so the kernel of L only contains the 0 vector, and hence L is one-to-one by Lemma 8.4.1. Onto: For this proof, we use the Rank-Nullity Theorem to make it even easier. Since the kernel of L only contains the 0 vector, by the Rank-Nullity Theorem we get that the rank of L equals the dimension of V minus the nullity of L, which is the dimension of W. Since the range of L is a subspace of W, and the dimension of the range equals the dimension of W, we get that the range of L must equal W, and hence, L is also onto.

Therefore, L is an isomorphism, and so V and W are isomorphic.

Take a minute to look over the entire proof.

## More on Isomorphisms

You may have noticed that we could have shortened the first part of the proof greatly by using the Rank-Nullity Theorem. The reason that I didn’t do this was because then I wouldn’t have proven that an isomorphism maps basis vectors to basis vectors. Our proof of onto in the second part shows us how to prove the following useful theorem.

Theorem 8.4.3: If V and W are both n-dimensional vector spaces, and L is a linear mapping from V to W, then L is one-to-one if and only if L is onto. This theorem shows us that if we already know that two vector spaces are isomorphic, then proving that a linear mapping L between them is an isomorphism is easier. In particular, we would just need to show that L is either one-to-one or onto, and then the theorem tells us that the other must also be true. Again, it is very important to remember that just because two vector spaces V and W are isomorphic, it does not mean that every linear mapping L from V to W is an isomorphism.

We can use isomorphisms to help us prove many useful results. For example, one nice result that is relatively easy to prove using isomorphisms is Theorem 8.4.4. If B is any basis for an n-dimensional vector space V, C is any basis for an m-dimensional vector space W, and L is a linear mapping from V to W, then the rank of L equals the rank of the matrix of L with respect to bases B and C.

I will leave the proof of this theorem as a somewhat challenging exercise. I will, however, give you some hints at how to proceed. By definition, we know that the rank of L equals the dimension of the range of L, and we know that the rank of the matrix of L with respect to bases B and C is the dimension of the columnspace of the matrix. Thus, to prove this theorem, we just need to prove that the range of L is isomorphic to the columnspace of the matrix of L with respect to bases B and C. As usual, the tricky part is in defining the isomorphism. You need to think how you can easily relate vectors in the range to vectors in the columnspace of the matrix. In particular, you need to think carefully about the definitions of these, and how they are related.

This ends this lecture.