Three isomorphism theorems in linear algebra

When I first learned about these they were difficult to understand or justify. So I thought I’d write things down, now that I’ve figured them out. These are particular cases of the more general isomorphism theorems in abstract algebra.

Let , be vector spaces over a field . Then we have the following isomorphisms ( ).

First isomorphism theorem

Recall that we know linear maps are not in general bijective (i.e. not isomorphisms). However, for any linear map , the first theorem gives us an isomorphism between its domain modulo its kernel and its image.

Quick notation refresher: let be a vector space over a field and be a subspace of . For , let be a binary relation that is true when . It’s easy to show is an equivalence relation. or denotes the equivalence class with representative ,

The quotient set (read mod ) is

In fact, the quotient set forms a vector space—the quotient space—under the operations and . See Wikipedia for further reading.

Anyways, the first theorem.

Theorem statement: Let be a linear map. Then the map

is a well defined isomorphism.

What we’re really saying here is just that when we mod out the kernel of any linear map, we can define an injective map to its image. Since it’s trivially surjective over its image, it is an isomorphism. This is intuitively very natural when we consider our intuition about the kernel is.

Proof. The proof proceeds by first establishing that such a linear map exists and is well defined, then proving it’s an isomorphism.

Let and be vector spaces over a field . Let be a linear map. Let (for simplicity of notation’s sake). Then, by the universal property of , there exists a unique, well defined linear map , given by . That is, the following diagram commutes (where is the canonical map from to sending each element to its equivalence class).

This establishes the existence of the map we desire, now we need to show it’s both surjective and injective.

We can easily see is surjective, let , then with .

To show injectivity, we use the fact that an injective linear map has the trivial subspace as its kernel. Suppose . Then , so by definition. Then , so . Note that this is because

Hence so it is injective. Therefore, it’s an isomorphism between and . ⁠ 

Second isomorphism theorem

Theorem statement: Let be an vector space. Let , be subspaces of . We have the following isomorphism.

Proof. This proof is a little different. First we’re going to show the following linear map exists.

To see that it exists and is linear, simply note it is the composition

(Where is the inclusion map and is the canonical map sending each element to its equivalence class.)

We know and to be linear, so the composition is also a linear map.

Now that we’ve established the existence of such a linear map , the idea is to first show that is surjective (that is, its image is its codomain ). Then we show that , and apply the first isomorphism theorem to obtain our desired isomorphism.

Suppose . Then , for some and . But , because . So , and is surjective.

Now let .

Therefore . Since we showed is surjective, . Then, by the first isomorphism theorem, there is an isomorphism

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Third isomorphism theorem

This one gives us a sort of “cancellation of fractions” for our quotients.

Theorem statement: Let be an -vector space. Let , be subspaces of . Suppose . Then

Just imagine you’re cancelling

Proof. Again we’re going to try to define a linear map that lets us apply the first isomorphism theorem. No tricks this time, so we’ll attempt to directly construct a linear map and show it is well defined.

Let be a linear map

We need to show is well defined, that is, it doesn’t depend on choice of representatives. Suppose . Then , so . So . Similarly . So is well defined.

Next we show linearity. This is just a simple manipulation. Let , . Then

Now we need surjectivity, which is trivial. Suppose , then obviously .

Finally we need to show .

But actually that last set is precisely .

With all the pieces in place, we now apply the first isomorphism theorem, and we are done

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