User:Linus44/Sandbox

Theorem
Let $K$ be a division ring.

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

Then:
 * $\dim_{K} V = \dim_{K} W$ if and only if $V \simeq W$

where:
 * $V \simeq W$ indicates that $V$ and $W$ are isomorphic, and
 * $\dim_{K} V$ and $\dim_{K} W$ are the dimensions of $V$ and $W$ respectively.

Sufficient condition
Suppose that $V$ and $W$ have bases of the same cardinality.

Let $\mathbf e = \left({ e_1,\ldots, e_n}\right)$ and $\mathbf f = \left({ f_1,\ldots,f_n }\right)$ be bases for $V$ and $W$ respectively.

Define:


 * $\lambda:V \rightarrow W$
 * $:\sum\limits_{i = 1}^n v_i e_i \mapsto \sum\limits_{i = 1}^n v_if_i$

Then by construction $\lambda$ maps the basis $\mathbf e$ to the basis $\mathbf f$.

Therefore by Linear Transformation of Vector Space Equivalent Statements it suffices to show that $\lambda$ is a linear transformation.

Let:
 * $\displaystyle v = \sum_{i = 1}^n v_i e_i,\ v'= \sum_{i = 1}^n v_i' e_i \in V,\ k \in K$:

We have:

Thus $\lambda$ is linear.

Necessary condition
Suppose that there exists an isomorphism $\phi : V \to W$.

Then by Linear Transformation of Vector Space Equivalent Statements we have that $\phi$ maps every basis of $V$ to a basis of $W$.

Since $\phi$ is a bijection this means that bases of $V$ and $W$ have the same cardinality.

Thus $V$ and $W$ have the same dimension.