Example of a theorem in linear algebra which relies on the isomorphism theorems?

Question

I'm thinking about the isomorphism theorems in linear algebra, and I can't seem to think of a result which relies on them. What are some applications of these results? I don't quite have an understanding of why they're important.

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Theorem (The first isomorphism theorem) Let $\tau: V \rightarrow W$ be a linear transformation. Then the linear transformation $\tau^{\prime}: V / \operatorname{ker}(\tau) \rightarrow W$ defined by $$ \tau^{\prime}(v+\operatorname{ker}(\tau))=\tau(v) $$ is injective and $$ \frac{V}{\operatorname{ker}(\tau)} \approx \operatorname{im}(\tau) $$

Theorem (The second isomorphism theorem) Let $V$ be a vector space and let $S$ and $T$ be subspaces of $V$. Then $$ \frac{S+T}{T} \approx \frac{S}{S \cap T} $$

Answer 1

The first isomorphism theorem implies the rank-nullity theorem which gives a super nice corollary about linear transformations between finite-dimensional spaces of equal dimension:

Theorem.

Let $V,W$ be vector spaces of equal finite-dimension over a field $\Bbb{F}$. For any linear transformation $T:V\to W$, the following statement are equivalent:

• $T$ is injective
• $T$ is surjective
• $T$ is bijective (and hence a linear isomorphism).

This is a fundamental fact about linear transformations which is not true if you deal with other types of functions. This can be phrased in terms of matrices as well (and can be phrased in a variety of equivalent ways).

One of the nice consequences is that to check invertibility, you only need a one-sided inverse; the other follows immediately. So, for example, if $A,B$ are $n\times n$ matrices, then $AB=I$ automatically implies $BA=I$ and hence $A=B^{-1}$ and $B=A^{-1}$. Note very carefully that in general, for two functions, $f\circ g=\text{id}$ does not imply $g\circ f=\text{id}$.

This is of course a simple consequence, but it is used repeatedly.

Also, the rank-nullity theorem itself is super important in linear algebra: it precisely and concisely formalizes the intuitive notion that if you have $m$ equations in $n$ unknowns and only $r$ of those equations are ‘independent’, then you can eliminate $n-r$ many “free” independent variables. Or more geometrically, the solution set of $m$ many linear equations of which $r$ are independent gives an $(n-r)$-dimensional solution space. As a super concrete example, a scalar equation \begin{align} a_1x_1+\dots +a_nx_n&=0, \end{align} where not all the $a_i$’s are $0$, has an $(n-1)$-hyperplane as its solution set.

Of course, there are tons of other examples of the utility of the first isomorphism theorem, but this I think is the ‘most immediately important’ consequence.

Answer 2

A simple but instructive application is just that the isomorphism theorems can save you some tedious work (this is true for the isomorphism theorems in all contexts, not just in the vector space setting). Suppose you are given a vector space $V$ with a subspace $W$, and you wish to describe $V / W$ in more familiar terms.

Often you can make an educated guess about a vector space $Z$ such that $V / W \cong Z$. But proving that the isomorphism holds may be tedious as you have to set up an explicit isomorphism $V / W \to Z$ and verify that it is well defined, i.e. not dependent on choice of coset representative.

Instead, it's typically easier to guess a surjective linear map $\phi \colon V \to Z$ such that $\ker \phi = W$, then the first isomorphism theorem does the work for you.

To illustrate, let $V = \mathbb{R}^2$, $W = \{(x,0) \mid x \in \mathbb{R}\}$. We might guess $V / W \cong \mathbb{R}$. To verify that, define $\phi \colon V \to \mathbb{R}$ by $\phi(x,y) = y$. It's clear that $\phi$ is linear and surjective, and $\ker \phi = W$. So without further ado, the first isomorphism theorem assures us that $V/W \cong \mathbb{R}$ without any tedious checking of well-definedness.

That's a very simple example, but you can play around with others to convince yourself that this kind of use of the isomorphism theorems can save you a lot of effort.

Answer 3

First one I can think of is this :

Let $A$ and $B$ be two linear subspaces of $E$, such that $A \oplus B = E$. Then : $$ E/A \simeq B$$