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 $\dim_K V = \dim_K W$.

Thus by the definition of dimension:


 * $\mathbb V \sim \mathbb W$

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

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


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

For $\mathbf v \in \mathbb V$ let $l_\mathbf v \in V^\star$ be the unique linear transformation defined on the basis $\mathbb V$ by:
 * $\forall \mathbf v' \in \mathbb V : l_\mathbf v \left({\mathbf v'}\right) = \delta_{\mathbf v, \mathbf v'}$

where $\delta : V \times V \to K$ is the Kronecker delta and $V^*$ is the dual of $V$.

Now:


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

By the definition of $l_{\mathbf v}$ and by Vector Scaled by Zero is Zero Vector, all the terms but the last vanish, and so:


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

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

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

because $\mathbb V$ is linearly independent.

By Vector Scaled by Zero is Zero Vector, $\mathbf x = \mathbf 0$.

That is:


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

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

Recall that $\phi$ is a bijection.

From Inverse of Bijection is Bijection, $\phi$ is invertible.

Suppose $\mathbf y \in W$.

Then:

where this last vector 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 transformation.

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