Vector space isomorphism is an equivalence relation

Dependencies:

1. Composition of linear transformations
2. /sets-and-relations/equivalence-relation
3. /sets-and-relations/composition-of-bijections-is-a-bijection

Vector space isomorphism is an equivalence relation.

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

Proof

• 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 $T(cu) = cu = cT(u)$. 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{align} & (u_1 = T^{-1}(v_1) \wedge u_2 = T^{-1}(v_2)) \\ &\Rightarrow (T(u_1) = v_1 \wedge 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{align}

  Let $v \in V$. \[ u = T^{-1}(v) \Rightarrow T(u) = v \Rightarrow cv = cT(u) = T(cu) \Rightarrow T^{-1}(cv) = cu = cT^{-1}(v) \]

  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, $ST$ is a bijection and a linear transformation. Therefore, $U$ is isomorphic to $W$.

Dependency for:

1. Canonical decomposition of a linear transformation

Info:

• Depth: 6
• Number of transitive dependencies: 8

Transitive dependencies:

1. /sets-and-relations/composition-of-bijections-is-a-bijection
2. /sets-and-relations/equivalence-relation
3. Group
4. Ring
5. Field
6. Vector Space
7. Linear transformation
8. Composition of linear transformations