# Linear Algebra Composition Isomorphism Question

• Nexttime35
In summary, the conversation discusses the possibility of the composition of two linear maps, S and T, being an isomorphism. It is determined that this is not possible if the dimension of the range of T is smaller than the dimension of the domain of S. An example is given to further explain this concept. It is also noted that while an isomorphism does exist between U and W in this scenario, TS is not an isomorphism.
Nexttime35

## Homework Statement

Let S : U →V and T : V →W be linear maps.

Given that dim(U) = 2, dim(V ) = 1, and dim(W) = 2, could T composed of S be an isomorphism?

## Homework Equations

If Dim(v) > dim(W), then T is 1-1
If Dimv < dim(w), then T is not onto.

## The Attempt at a Solution

So this seems like a tricky question, but I am having trouble proving whether or not this is an isomorphism. While I know that T composed of S (u) = T(S(u)), S: U→V and T:V→W, this would be an isomorphism if dim(U) = Dim(W). But does the fact that S maps to V, which is one dimension less than U and W, affect that this is an isomorphism?

I think an example could be helpful in helping me understand this problem.
Thank you.

If Dim(v) > dim(W), then T is 1-1

That should read "If Dim(V) > Dim(W), then T is not 1-1". I'm sure that's what you meant, but just making sure.

Anyway, let's suppose that TS is an isomorphism. Which would mean that given w $\in$ W, there is a u $\in$ U. Such that TS(u) = w. Is this a problem?

That makes sense, kduna, thank you. So Basically this composition bypasses the use of V, and there exists an isomorphism from U to W. Thanks.

Nexttime35 said:
That makes sense, kduna, thank you. So Basically this composition bypasses the use of V, and there exists an isomorphism from U to W. Thanks.

Actually no. w = TS(u) = T(S(u)). So w $\in$ Range(T). But this can be done for any w $\in$ W. This would mean T is surjective, a contradiction since dim(V) < dim(W).

Therefore TS cannot be an isomorphism.

Note that since dim(U) = 2 = dim(W), an isomorphism does exist between U and W. But TS is not an isomorphism.

Ahh, I see. That makes much more sense. Thank you for your help, kduna.

## 1. What is linear algebra composition?

Linear algebra composition is the process of combining two linear transformations to create a new linear transformation. This is done by applying one transformation to the output of the other transformation.

## 2. What is an isomorphism in linear algebra?

An isomorphism in linear algebra is a type of linear transformation that preserves the structure of the vector space it is applied to. This means that the properties of the original vector space, such as dimensionality and linear independence, are maintained in the transformed space.

## 3. How is linear algebra composition related to isomorphism?

Linear algebra composition and isomorphism are related because an isomorphism can be created by composing two linear transformations. This is known as an isomorphism of linear transformations. Additionally, composition is used to prove that two vector spaces are isomorphic.

## 4. Why is isomorphism important in linear algebra?

Isomorphism is important in linear algebra because it allows us to study and understand different vector spaces by relating them to each other. It also allows us to prove that two vector spaces are equivalent, even if they may look different at first glance.

## 5. Can you give an example of linear algebra composition and isomorphism?

Yes, an example of linear algebra composition and isomorphism is the rotation and scaling transformations in two-dimensional space. If we rotate a vector and then scale it, or vice versa, the result will be the same as if we had applied a single transformation that combines both rotation and scaling. This shows that the two transformations are isomorphic and can be composed to create a new transformation.

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