Same Dimensional Vector Spaces are Isomorphic

Theorem
Let $K$ be a division ring.

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

Suppose that $\dim_K V = \dim_K W$.

Then:


 * $V \cong W$

That is, $V$ and $W$ are isomorphic.

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

By hypothesis $\mathbb V$ and $\mathbb W$ have the same cardinality.

Therefore we can choose a bijection $\phi: \mathbb V \to \mathbb W$.

Define the mapping $\lambda: V \to W$ as:


 * $\displaystyle \lambda \left({\sum \limits_{\mathbf v \mathop \in \mathbb V} a_{\mathbf v} \mathbf v}\right) = \sum \limits_{\mathbf v \mathop \in \mathbb V} a_\mathbf v \phi \left({\mathbf v}\right)$

Let $l_\mathbf v \in V^\star$ be defined by:
 * $l_\mathbf v \left({\mathbf v'}\right) = \delta_{\mathbf v, \mathbf v'}$

where $\delta$ is the Kronecker delta.

Notice that:


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

For all $\mathbf v, \mathbf v' \in V, k \in K$:

Thus:
 * $\lambda \in \hom_K \left({V, W}\right)$

$\lambda$ has thus been shown to be linear.

Let $\mathbf x \in \ker \lambda$ where $\ker \lambda$ denotes the kernel of $\lambda$.

Then:


 * $\displaystyle \mathbf 0 = \lambda \left({\mathbf x}\right) = \sum \limits_{\mathbf v \mathop \in \mathbb V} l_\mathbf v \left({\mathbf x}\right) \mathbf v$

Therefore:
 * $\forall \mathbf v \in \mathbb V : l_\mathbf v \left({\mathbf x}\right) = 0$

so $\mathbf x = \mathbf 0$.

So:
 * $\ker \lambda = \left\{{\mathbf 0}\right\}$

By Linear Transformation is Injective iff Kernel Contains Only Zero, it follows that $\phi$ is injective.

Suppose $\mathbf y \in W$.

Then:

where this last quantity belongs to $\lambda \left({V}\right)$.

Thus $\lambda$ is surjective.

$\lambda$ has been shown to be injective and surjective, and so is a bijection.

$\lambda$ has also been shown to be linear.

Thus, by definition, $\lambda$ is an isomorphism.