Same Dimensional Vector Spaces are Isomorphic

Theorem
Let $K$ be a division ring.

Let $V$, $W$ be finite dimensional $K$-vector spaces.

Let $\dim_{K} V = \dim_{K} W$.

Then


 * $V \approx W$

Proof
Let $\mathbb{V}$, $\mathbb{W}$ be bases for $V$, $W$ respectively.

Let $\varphi:\mathbb{V} \rightarrow \mathbb{W}$ be a bijection (which we can do since $\left| \mathbb{V} \right| = \dim_{K} V = \dim_{K} W = \left| \mathbb{W} \right| $).

Define:


 * $\lambda:V \rightarrow W,\sum\limits_{\vec{v} \in \mathbb{V}} a_{\vec{v}}\vec{v} \mapsto \sum\limits_{\vec{v} \in \mathbb{V}} a_{\vec{v}}\varphi(\vec{v})$

Let $l_{\vec{v}} \in V^{\star}:l_{\vec{v}}(\vec{v}') = \delta_{\vec{v}, \vec{v}'} := \begin{cases} 1 & \vec{v} = \vec{v}'\\ 0 & \vec{v} \neq \vec{v}' \end{cases}$

Notice that:


 * $l_{\vec{v}}\left(\sum\limits_{\vec{u} \in \mathbb{V}} a_{\vec{u}}\vec{u}\right)=a_{\vec{v}}\forall \vec{v} \in \mathbb{V}$

For all $\vec{v},\vec{v}' \in V, k \in K$:


 * $\lambda(c\vec{v}+\vec{v}')= \lambda\left(c\sum\limits_{\vec{u} \in \mathbb{V}} l_{\vec{u}}(\vec{v})\vec{u}+\sum\limits_{\vec{u} \in \mathbb{V}} l_{\vec{u}}(\vec{v}')\vec{u}\right)

= \lambda\left(\sum\limits_{\vec{u} \in \mathbb{V}} (cl_{\vec{u}}(v)+l_{\vec{u}}(\vec{v}'))\vec{u}\right) = \sum\limits_{\vec{u} \in \mathbb{V}} (cl_{\vec{u}}(\vec{v})+l_{\vec{u}}(\vec{v}')) \varphi(\vec{u}) = c\sum\limits_{\vec{u} \in \mathbb{V}} l_{\vec{u}}(\vec{v})\varphi(\vec{u})+\sum\limits_{\vec{u} \in \mathbb{V}} l_{\vec{u}}(\vec{v})\varphi(\vec{u}) = c \lambda(\vec{v})+\lambda(\vec{v}')$

Thus $\lambda \in \hom_{K}(V,W)$

Suppose $\vec{x} \in \ker \lambda$.

Then:


 * $\vec{0}=\lambda(\vec{x})=\sum\limits_{\vec{v} \in \mathbb{V}} l_{\vec{v}}(\vec{x})\vec{v} \implies l_{\vec{v}}(\vec{x})=0\forall \vec{v} \in \mathbb{V} \implies \vec{x}=\vec{0}$

So $\ker \lambda = \left\lbrace \vec{0} \right\rbrace$

By Linear Transformation is Injective iff Kernel Contains Only Zero, $\varphi$ is injective.

Suppose $\vec{y} \in W$.

Then:


 * $\vec{y} = \sum\limits_{\vec{w} \in \mathbb{W}} l_{\vec{w}}(\vec{y})\vec{w} = \sum\limits_{\vec{v} \in \mathbb{V}} l_{\varphi^{-1}(\vec{v}}(\vec{y}) \varphi^{-1}(\vec{v}) \in \lambda(V)$

By definition, $\lambda$ is surjective.

By defintion, since $\lambda$ is injective, surjective, and linear; $\lambda$ is an isomorphism.