First Isomorphism Theorem/Vector Spaces

Theorem

Let $K$ be a field.

Let $X$ and $Y$ be vector spaces over $K$.

Let $T : X \to Y$ be a linear transformation.

Let $\ker T$ be the kernel of $T$.

Let $X/\ker T$ be the quotient vector space of $X$ modulo $\ker T$.

Then $X/\ker T$ is isomorphic to $\Img T$ as a vector space.

Proof

From Image of Linear Transformation is Submodule, we assure ourselves that $\Img T$ is indeed a vector space over $K$.

Let $\pi : X \to X/\ker T$ be the quotient mapping.

From Condition for Mapping from Quotient Vector Space to be Well-Defined, there exists a linear transformation $\Lambda : X/\ker T \to \Img T$ such that:

$\map \Lambda {\map \pi x} = T x$ for each $x \in X$.

To finish, we need to show that $\Lambda$ is injective and surjective.

From Linear Transformation is Injective iff Kernel Contains Only Zero, to show $\Lambda$ is injective it suffices to show that:

$\ker \Lambda = \set { {\mathbf 0}_{X/\ker T} }$

Suppose that $\map \Lambda {\map \pi x} = {\mathbf 0}_Y$.

Then $T x = {\mathbf 0}_Y$.

So $x \in \ker T$.

That is, by Kernel of Quotient Mapping, we have $\map \pi x = {\mathbf 0}_{X/\ker T}$.

So we have $\ker \Lambda = \set { {\mathbf 0}_{X/\ker T} }$, and so $\ker \Lambda$ is injective by Linear Transformation is Injective iff Kernel Contains Only Zero.

Now let $y \in \Img T$.

Then there exists $x \in X$ such that $y = T x$.

Then, we have $\map \Lambda {\map \pi x} = T x = y$.

So $\Img T \subseteq \Img \Lambda$.

Since we already have $\Img \Lambda \subseteq \Img T$, we obtain $\Img \Lambda = \Img T$.

So $\Lambda : X/\ker T \to \Img T$ is a vector space isomorphism.

$\blacksquare$