[proofplan] The forward direction uses the fact that isomorphisms preserve bases (hence the dimension of the image equals the dimension of the domain). The reverse direction explicitly constructs an isomorphism: choose bases for both spaces, define a linear map sending one basis to the other, and verify it is bijective. [/proofplan] [step:Show isomorphic spaces have equal dimension ($\Rightarrow$)] Suppose $\alpha: U \to V$ is an isomorphism. Let $\{e_1, \ldots, e_n\}$ be a basis for $U$. By [Linear Maps Preserve Linear Structure](/theorems/379) part (iii), $\{\alpha(e_1), \ldots, \alpha(e_n)\}$ is a basis for $V$. Hence $\dim V = n = \dim U$. [/step] [step:Construct an isomorphism between spaces of equal dimension ($\Leftarrow$)] Suppose $\dim U = \dim V = n$. Choose a basis $\{e_1, \ldots, e_n\}$ for $U$ and a basis $\{f_1, \ldots, f_n\}$ for $V$. Define $\alpha: U \to V$ as the unique linear map satisfying $\alpha(e_i) = f_i$ for each $i \in \{1, \ldots, n\}$. This map exists and is unique by [Unique Representation by a Basis](/theorems/372): every $u \in U$ has a unique expansion $u = \sum_{i=1}^{n}\lambda_i e_i$, and we set $\alpha(u) = \sum_{i=1}^{n}\lambda_i f_i$. [claim:The map $\alpha$ is bijective] The linear map $\alpha: U \to V$ defined by $\alpha(e_i) = f_i$ is bijective. [/claim] [proof] **Injectivity.** Suppose $\alpha(u) = \mathbf{0}$. Write $u = \sum_{i=1}^{n}\lambda_i e_i$. Then $\alpha(u) = \sum_{i=1}^{n}\lambda_i f_i = \mathbf{0}$. Since $\{f_1, \ldots, f_n\}$ is linearly independent, $\lambda_i = 0$ for all $i$, so $u = \mathbf{0}$. **Surjectivity.** Let $v \in V$. Write $v = \sum_{i=1}^{n}\mu_i f_i$ (using the basis for $V$). Set $u = \sum_{i=1}^{n}\mu_i e_i \in U$. Then $\alpha(u) = \sum_{i=1}^{n}\mu_i f_i = v$. [/proof] By the [Isomorphism iff Linear Bijection](/theorems/378) theorem, $\alpha$ is an isomorphism. Hence $U \cong V$. [guided] The construction is natural: given bases of equal size, the map that sends one basis to the other is the canonical candidate for an isomorphism. Why must such a map be bijective? For injectivity: if $\alpha(u) = \mathbf{0}$, then expanding $u = \sum \lambda_i e_i$ gives $\sum \lambda_i f_i = \mathbf{0}$. The independence of $\{f_1, \ldots, f_n\}$ forces all $\lambda_i = 0$, hence $u = \mathbf{0}$. The kernel is trivial, so $\alpha$ is injective. For surjectivity: every $v \in V$ has a unique representation $v = \sum \mu_i f_i$ (by [Unique Representation by a Basis](/theorems/372)). The preimage $u = \sum \mu_i e_i$ satisfies $\alpha(u) = v$. This result shows that, up to isomorphism, the only invariant of a finite-dimensional vector space is its dimension. Two spaces over the same field with the same dimension are "the same" algebraically --- every such space is isomorphic to $\mathbb{F}^n$. [/guided] [/step]