91Ó°ÊÓ

In order to show that the vector spaces V and \(\hat{V}\) are isomorphic, we need to present a function which will map elements from V to their corresponding elements in \(\hat{V}\). Let's define an isomorphism function $$\psi : V \to \hat{V}$$ such that $$\psi(v) = \psi(u+w) = (u, w)$$. Now, let us verify if the function \(\psi\) is linear and bijective.

To verify the linearity of the mapping \(\psi\), we need to confirm whether $$\psi \left( \alpha v_1 + \beta v_2 \right) = \alpha \psi(v_1) + \beta \psi(v_2)$$ for any scalar values \(\alpha, \beta\) and vectors \(v_1, v_2\) in V. For arbitrary vectors \(v_1 = u_1 + w_1\), and \(v_2 = u_2 + w_2\) in V. Let's consider $$\psi \left( \alpha v_1 + \beta v_2 \right)$$: $$\psi \left( \alpha (u_1 + w_1) + \beta (u_2 + w_2) \right) = \psi \left( (\alpha u_1 + \beta u_2) + (\alpha w_1 + \beta w_2) \right)$$. As per the definition of \(\psi\), $$\psi \left( (\alpha u_1 + \beta u_2) + (\alpha w_1 + \beta w_2) \right) = (\alpha u_1 + \beta u_2, \alpha w_1 + \beta w_2)$$. Now, let's consider $$\alpha \psi(v_1) + \beta \psi(v_2)$$: $$\alpha (u_1, w_1) + \beta (u_2, w_2) = (\alpha u_1 + \beta u_2, \alpha w_1 + \beta w_2)$$. Since $$\psi \left( \alpha v_1 + \beta v_2 \right) = \alpha \psi(v_1) + \beta \psi(v_2)$$, the mapping \(\psi\) is linear.

To prove that \(\psi\) is injective, we need to show that for every pair of distinct vectors \(v_1, v_2\) in V, we have \(\psi(v_1) \neq \psi(v_2)\). Let's assume that for two distinct vectors \(v_1 = u_1 + w_1\) and \( v_2 = u_2 + w_2\) in V, the mapping is as follows: $$\psi(v_1) = \psi(u_1+w_1) = (u_1, w_1)$$ and $$\psi(v_2) = \psi(u_2+w_2) = (u_2, w_2)$$. Since \(v_1 \neq v_2\), this implies that either \(u_1 \neq u_2\) or \(w_1 \neq w_2\). Therefore, the mappings \((u_1, w_1) \neq (u_2, w_2)\), which means \(\psi\) is injective.

To prove that \(\psi\) is surjective, we need to show that for every element \((u, w)\) in the external direct sum \(\hat{V}\), there is an element \(v=u+w \in V\) such that \(\psi(v)=(u,w)\). Given any \((u, w)\) in \(\hat{V}\), since \(U\) and \(W\) are subspaces of \(V\), their sum is also in \(V\). So, there exists an element \(v=u+w \in V\) such that, when mapped with the isomorphism function \(\psi\), we get the following: $$\psi(v) = \psi(u+w)=(u, w)$$. Hence, \(\psi\) is surjective.

Since the linear map \(\psi\) is linear, injective, and surjective, it is bijective. Therefore, the correspondence \(v = u+w \leftrightarrow (u,w)\) exhibits an isomorphism between \(V\) and \(\hat{V}\).