Vector space isomorphism is an equivalence relation

Dependencies:

Vector space isomorphism is an equivalence relation.

Also, inverse of an isomorphism is an isomorphism and composition of isomorphisms is an isomorphism.

Reflexive: Let $T : U \mapsto U$ be the bijection $T (u) = u$ . Then $T (u_{1} + u_{2}) = u_{1} + u_{2} = T (u_{1}) + T (u_{2})$ and $T (c u) = c u = c T (u)$ . Therefore, $T$ is an isomorphism. Therefore, $U$ is isomorphic to $U$ .
Symmetric:

Let $T : U \mapsto V$ be an isomorphism. Therefore, $T^{- 1}$ exists and is a bijection.

Let $v_{1}, v_{2} \in V$ . $\begin{aligned} (u_{1} = T^{- 1} (v_{1}) \land u_{2} = T^{- 1} (v_{2})) \\ \Rightarrow (T (u_{1}) = v_{1} \land T (u_{2}) = v_{2}) \\ \Rightarrow T (u_{1} + u_{2}) = v_{1} + v_{2} \\ \Rightarrow T^{- 1} (v_{1} + v_{2}) = u_{1} + u_{2} = T^{- 1} (v_{1}) + T^{- 1} (v_{2}) \end{aligned}$

Let $v \in V$ . $u = T^{- 1} (v) \Rightarrow T (u) = v \Rightarrow c v = c T (u) = T (c u) \Rightarrow T^{- 1} (c v) = c u = c T^{- 1} (v)$

Therefore, $T^{- 1}$ is a linear transformation. Therefore, $V$ is isomorphic to $U$ .
Transitive:

Let $S : U \mapsto V$ and $T : V \mapsto W$ be isomorphisms. Therefore, $S T$ is a bijection and a linear transformation. Therefore, $U$ is isomorphic to $W$ .