91Ó°ÊÓ

To prove reflexivity, we need to show that every vector space \(V\) over a field \(F\) is isomorphic to itself. By definition, an isomorphism between linear spaces is a bijective linear transformation. Let's consider the identity map: $$ I: V \longrightarrow V $$ where \(I(\mathbf{v})=\mathbf{v}\) for all \(\mathbf{v} \in V\). We need to show that \(I\) is both linear and bijective. - Linearity: For any scalars \(a, b \in F\) and vectors \(\mathbf{u}, \mathbf{v} \in V\), we have: $$ I(a\mathbf{u} + b\mathbf{v}) = a\mathbf{u} + b\mathbf{v} = aI(\mathbf{u}) + bI(\mathbf{v}). $$ - Bijectivity: By definition, \(I\) is an identity map, meaning that it is both injective (one-to-one) and surjective (onto). Hence, the identity map \(I\) is a linear isomorphism between \(V\) and itself. Therefore, the relation \(\sim\) is reflexive.

To prove symmetry, we need to show that if a vector space \(U\) over a field \(F\) is isomorphic to a vector space \(V\), then \(V\) is isomorphic to \(U\). Let \(\phi: U \longrightarrow V\) be a linear isomorphism. We need to find a linear isomorphism \(\psi: V \longrightarrow U\). As \(\phi\) is a bijective function, it has an inverse, denoted by \(\phi^{-1}\), which is also bijective. Now, we claim that \(\phi^{-1}: V \longrightarrow U\) is a linear isomorphism. We need to show that \(\phi^{-1}\) is linear. - Linearity: For any scalars \(a, b \in F\) and vectors \(\mathbf{v_1}, \mathbf{v_2} \in V\), we have: \[ \phi^{-1}(a\mathbf{v_1} + b\mathbf{v_2}) = \phi^{-1}(a(\phi(\phi^{-1}(\mathbf{v_1}))) + b(\phi(\phi^{-1}(\mathbf{v_2})))) \] By linearity of \(\phi\), \[ \phi^{-1}(a\mathbf{v_1} + b\mathbf{v_2}) = \phi^{-1}( \phi(a\phi^{-1}(\mathbf{v_1}) + b\phi^{-1}(\mathbf{v_2}))) \] Then, as \(\phi^{-1} \circ \phi = I\), we have: \[ \phi^{-1}(a\mathbf{v_1} + b\mathbf{v_2}) = a\phi^{-1}(\mathbf{v_1}) + b\phi^{-1}(\mathbf{v_2}) \] Therefore, \(\phi^{-1}\) is a linear isomorphism from \(V\) to \(U\), and the relation \(\sim\) is symmetric.

To prove transitivity, we need to show that if a vector space \(U\) is isomorphic to a vector space \(V\) and \(V\) is isomorphic to another vector space \(W\), then \(U\) is isomorphic to \(W\). Let \(\phi: U \longrightarrow V\) and \(\psi: V \longrightarrow W\) be linear isomorphisms. We need to find a linear isomorphism \(\theta: U \longrightarrow W\). Consider the function composition \(\theta = \psi \circ \phi: U \longrightarrow W\). We need to show that \(\theta\) is a linear isomorphism. - Linearity: For any scalars \(a, b \in F\) and vectors \(\mathbf{u_1}, \mathbf{u_2} \in U\), we have: \[ \theta(a\mathbf{u_1} + b\mathbf{u_2}) = (\psi \circ \phi)(a\mathbf{u_1} + b\mathbf{u_2}) \] By linearity of \(\phi\), \[ \theta(a\mathbf{u_1} + b\mathbf{u_2}) = \psi (a(\phi\mathbf{u_1}) + b(\phi\mathbf{u_2})) \] Then, by linearity of \(\psi\), \[ \theta(a\mathbf{u_1} + b\mathbf{u_2}) = a(\psi(\phi(\mathbf{u_1}))) + b(\psi(\phi(\mathbf{u_2}))) = a\theta(\mathbf{u_1}) + b\theta(\mathbf{u_2}) \] - Bijectivity: Both \(\phi\) and \(\psi\) are bijective functions, so their composition \(\theta\) is also a bijective function. Therefore, \(\theta\) is a linear isomorphism from \(U\) to \(W\) and the relation \(\sim\) is transitive. Since the relation \(\sim\) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation on the class of vector spaces over \(F\).