The First Isomorphism Theorem

Let’s return to isomorphisms. Let $V,W$ be vector spaces over a field $\mathbb K$ . When are $V$ and $W$ isomorphic? When we can find a vector space isomorphism $T :V \to W$ between them. Well, how do we find such an isomorphism?

In general, this is a rather challenging task. However, the first isomorphism theorem yields us a partial solution. But first, let’s review some ideas in discrete mathematics.

Lemma 1. Let $f : K \to L$ be a surjective function. Define the quotient set $K/{\sim} := \{f^{-1}(\{y\}) : y \in L\}$ and the (surjective) projection map $\pi : K \to K/{\sim}$ by $\pi(x) = f^{-1}( \{ f(x) \})$ . Then, there exists a unique bijection $\phi : K/{\sim} \to L$ such that $f = \phi \circ \pi$ .

The proof of Lemma 1 can furnished by students in discrete mathematics using the notions of equivalence classes and partitions. For our discussion on linear algebra, we will accept this idea and use it to make sense of vector space isomorphisms.

Theorem 1 (First Isomorphism Theorem). Let $T : V \to W$ be a surjective linear transformation. Let $\phi : V/{\sim} \to W$ be the canonical bijection in Lemma 1. Then $V/{\sim} \equiv V/{\ker(T)}$ can be equipped with a vector space structure, called a quotient space, and $\phi$ becomes an isomorphism. Furthermore, for any $\mathbf v \in V$ ,

$[\mathbf v] = \{\mathbf v\} + \ker(T) =: \mathbf v + \ker(T).$

Elements of the form $\mathbf v + \ker(T)$ are known as cosets. Concisely, if $T$ is a surjective linear transformation, then $V/{\ker(T)} \cong W$ .

Proof. Since there exists a bijection from $V/{\sim}$ to the vector space $W$ , we can equip $V/{\sim}$ with the vector space structure defined by

$\begin{aligned} \phi([\mathbf u] + [\mathbf v]) &:= \phi([\mathbf u]) + \phi([\mathbf v]) \\ &= T(\mathbf u) + T(\mathbf v) \\ &= T(\mathbf u + \mathbf v) = \phi([\mathbf u + \mathbf v]), \end{aligned}$

and

$\begin{aligned} \phi(c[\mathbf v] ) &:= c\phi([\mathbf v]) = cT(\mathbf v) = Tc(\mathbf v) = \phi([c\mathbf v]), \end{aligned}$

so that $[\mathbf u]+[\mathbf v] = [\mathbf u + \mathbf v]$ and $c[\mathbf v] = [c\mathbf v]$ .

To describe the elements of $V/{\sim}$ , we observe that for each $\mathbf v \in V$ ,

$\begin{aligned} [\mathbf v] &= T^{-1}(\{ T(\mathbf v) \}) \\ &= \{\mathbf w \in V : T(\mathbf u) = T(\mathbf v)\} \\ &= \{\mathbf w \in V : T(\mathbf w - \mathbf v) = \mathbf 0\} \\ &= \{\mathbf w \in V : \mathbf w - \mathbf v \in \ker(T)\} \\ &= \{\mathbf w \in V : \mathbf w = \mathbf v + \mathbf u, \mathbf u \in \ker(T)\} \\ &= \{\mathbf v + \mathbf u : \mathbf u \in \ker(T)\} = \mathbf v + \ker(T).\end{aligned}$

Therefore, we recover the usual definitions of the quotient space $V/{\sim}$ and coset arithmetic:

$\begin{aligned} (\mathbf u + \ker(T)) + (\mathbf v + \ker(T)) &= [\mathbf u]+[\mathbf v] \\ &= [\mathbf u + \mathbf v] \\ &= (\mathbf u + \mathbf v) + \ker(T)\end{aligned}$

and

$c(\mathbf v + \ker(T)) = c[\mathbf v] = [c\mathbf v] = c\mathbf v + \ker(T).$

The isomorphism theorem, somewhat surprisingly, is the reason behind why we need to put a “ $+C$ ” at the end of our integrals. In fact, it is behind a more fundamental process: solving systems of linear equations like $T(\mathbf x) = \mathbf y$ .

Corollary 1. Let $T : V \to W$ be a linear transformation, and suppose for the vector $\mathbf y \in W$ , there exists a vector $\mathbf u \in V$ such that $T(\mathbf u) = \mathbf y$ . Then the set of solutions to the equation $T(\mathbf x) = \mathbf y$ is $\mathbf u + \ker(T)$ .

Proof. Using ideas in the isomorphism theorem, the set of solutions is

$\mathbf x = T^{-1}(\mathbf y) = \phi^{-1}(T(\mathbf u)) = \pi(\mathbf u) = \mathbf u + \ker(T).$

Corollary 2. Let $\mathcal C^1 := \{f \in \mathcal D : f' \in \mathcal C\}$ denote the subspace of continuously differentiable functions on $\mathbb R$ . Then $\mathcal C^1/{\mathbb R} \cong \mathcal C$ .

Proof. Consider the linear transformation $D : \mathcal C^1 \to \mathcal C$ defined by $D(F) = F'$ . By the fundamental theorem of calculus, $D$ is surjective. By the first isomorphism theorem,

$\mathcal C^1/{\ker(D)} \cong \mathcal C.$

By the mean value theorem, $\ker(D) = \mathbb R$ , so that $\mathcal C^1/{\mathbb R}=\mathcal C^1/{\ker(D)} \cong \mathcal C.$

Furthermore, $D(F) = f'$ implies that the indefinite integral

$\displaystyle \int f = D^{-1}(f) = F + \mathbb R,$

so that we write

$\displaystyle \int f(x)\, \mathrm dx = F(x) + C,$

where $C \in \mathbb R$ is an arbitrary constant.

Finally, we can use the first isomorphism theorem to prove the rank-nullity theorem.

Corollary 3. Suppose $W \subseteq V$ as a subspace and $V$ is finite-dimensional. Then $V/W$ is finite-dimensional and $\dim(V/W) = \dim(V) - \dim(W)$ .

Proof. Since $V$ is finite-dimensional, so is $W$ . Let $\{\mathbf w_1,\dots,\mathbf w_k\}$ be a basis for $W$ , and extend it to a basis $\{\mathbf w_1,\dots,\mathbf w_k, \mathbf u_1,\dots,\mathbf u_m\}$ of $V$ , so that $\dim(W) = k$ and $\dim(V) = \dim(W) + m$ . Define $U := \mathrm{span}\{\mathbf u_1,\dots,\mathbf u_m\}$ and the surjective linear transformation $T : V \to U$ by

$T(\mathbf w_i) = \mathbf 0,\quad T(\mathbf u_j) = \mathbf u_j.$

It is obvious that $\ker(T) = W$ , so that the first isomorphism theorem yields $V/W \cong U$ , hence

$\dim(V/W) = \dim(U) = m = \dim(V) - \dim(W).$

Corollary 4. Let $T : V \to W$ be a linear transformation, and suppose $V$ is finite-dimensional. Define $\text{rank}(T) := \dim(T(V))$ and $\text{nullity}(T) := \dim(\ker(T))$ . Then