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 $\mathbf v, \mathbf v' \in V, c \in K$: {{begin-eqn}} {{eqn | l = \lambda \left({c \mathbf v + \mathbf v'}\right)       | r = \lambda \left({c \sum \limits_{\mathbf u \mathop \in \mathbb V} l_\mathbf u \left({\mathbf v}\right) \mathbf u + \sum \limits_{\mathbf u \mathop \in \mathbb V} l_\mathbf u \left({\mathbf v'}\right) \mathbf u}\right) }} {{eqn | r = \lambda \left({\sum \limits_{\mathbf u \mathop \in \mathbb V} \left({c l_\mathbf u \left({\mathbf v}\right) + l_\mathbf u \left({\mathbf v'}\right)}\right) \mathbf u}\right)  }} {{eqn | r = \sum \limits_{\mathbf u \mathop \in \mathbb V} \left({c l_\mathbf u \left({\mathbf v}\right) + l_\mathbf u \left({\mathbf v'}\right)}\right) \phi \left({\mathbf u}\right)  }} {{eqn| r = c \sum \limits_{\mathbf u \mathop \in \mathbb V} l_\mathbf u \left({\mathbf v}\right) \phi \left({\mathbf u}\right) + \sum \limits_{\mathbf u \mathop \in \mathbb V} l_\mathbf u \left({\mathbf v}\right) \phi \left({\mathbf u}\right) }} {{eqn | r = c \lambda \left({\mathbf v}\right) + \lambda \left({\mathbf v'}\right) }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \mathbf y       | r = \sum \limits_{\mathbf w \mathop \in \mathbb W} l_\mathbf w \left({\mathbf y}\right) \mathbf w }} {{eqn | r = \sum \limits_{\mathbf v \mathop \in \mathbb V} l_{\phi^{-1} \left({\mathbf v}\right)} \left({\mathbf y}\right) \phi^{-1} \left({\mathbf v}\right) }} {{end-eqn}} 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. {{explain|The definition for isomorphism as given by that link is more explicit than using the term "linear" - it boils down to the same thing of course, but this needs to be made explicit. '''I don't see the difference''' --Linus44 (talk) 14:22, 21 May 2013 (UTC)}} {{qed}} Category:Linear Algebra mvq7dnwbbihlhiav0zro3khtti05ym8