[proofplan] The forward direction is immediate: an isomorphism has a two-sided inverse, hence is bijective. The content is the reverse direction: we must show that the set-theoretic inverse $\alpha^{-1}$ of a bijective linear map is automatically linear. This follows by applying $\alpha^{-1}$ to a linear combination in $V$, using linearity of $\alpha$ and injectivity to transfer the computation back to $U$. [/proofplan] [step:Deduce bijectivity from the existence of a linear inverse ($\Rightarrow$)] Suppose $\alpha$ is an isomorphism, so there exists a linear map $\beta: V \to U$ with $\alpha \circ \beta = \operatorname{id}_V$ and $\beta \circ \alpha = \operatorname{id}_U$. Since $\alpha$ has a two-sided inverse as a function, $\alpha$ is bijective. [/step] [step:Construct the set-theoretic inverse of a bijective linear map] Suppose $\alpha: U \to V$ is a bijective linear map. Since $\alpha$ is bijective, the set-theoretic inverse $\alpha^{-1}: V \to U$ exists, satisfying $\alpha \circ \alpha^{-1} = \operatorname{id}_V$ and $\alpha^{-1} \circ \alpha = \operatorname{id}_U$. [/step] [step:Show $\alpha^{-1}$ is linear] Let $v_1, v_2 \in V$ and $\lambda, \mu \in \mathbb{F}$. Set $u_1 = \alpha^{-1}(v_1)$ and $u_2 = \alpha^{-1}(v_2)$, so $\alpha(u_1) = v_1$ and $\alpha(u_2) = v_2$. By linearity of $\alpha$: \begin{align*} \alpha(\lambdau_1 + \muu_2) = \lambda\alpha(u_1) + \mu\alpha(u_2) = \lambdav_1 + \muv_2. \end{align*} Applying $\alpha^{-1}$ to both sides: \begin{align*} \lambdau_1 + \muu_2 = \alpha^{-1}(\lambdav_1 + \muv_2). \end{align*} Substituting back $u_i = \alpha^{-1}(v_i)$: \begin{align*} \alpha^{-1}(\lambdav_1 + \muv_2) = \lambda\alpha^{-1}(v_1) + \mu\alpha^{-1}(v_2). \end{align*} Hence $\alpha^{-1}$ is linear, and $\alpha$ is an isomorphism with inverse $\beta = \alpha^{-1}$. [guided] The key point is that we never need to "prove" the inverse is linear by checking some abstract condition --- the linearity of $\alpha$ itself does the work. Here is the logic in detail. We want to show $\alpha^{-1}(\lambdav_1 + \muv_2) = \lambda\alpha^{-1}(v_1) + \mu\alpha^{-1}(v_2)$. The right-hand side is a specific element of $U$; call it $w = \lambdau_1 + \muu_2$ where $u_i = \alpha^{-1}(v_i)$. We compute $\alpha(w)$ using linearity: \begin{align*} \alpha(w) = \alpha(\lambdau_1 + \muu_2) = \lambda\alpha(u_1) + \mu\alpha(u_2) = \lambdav_1 + \muv_2. \end{align*} Since $\alpha$ is injective (it is bijective by hypothesis), the equation $\alpha(w) = \lambdav_1 + \muv_2$ has a unique solution, which is $w = \alpha^{-1}(\lambdav_1 + \muv_2)$. But we also know $w = \lambdau_1 + \muu_2 = \lambda\alpha^{-1}(v_1) + \mu\alpha^{-1}(v_2)$. Equating the two expressions for $w$ gives the desired linearity. Why does this fail for non-linear bijections in general? It does not --- the argument above uses the linearity of $\alpha$ in the forward direction. A bijective non-linear map may have a non-linear inverse. [/guided] [/step]