91Ó°ÊÓ

A function F: V → U is called a linear mapping if it satisfies the following two conditions: 1. \(F(\alpha_1 v_1 + \alpha_2 v_2) = \alpha_1 F(v_1) + \alpha_2 F(v_2)\) for all vectors \(v_1, v_2 \in V\) and all scalars \(\alpha_1, \alpha_2\). 2. If F is one-to-one, then for every vector u in U, there exists a unique vector v in V such that F(v) = u.

Given that F is one-to-one and onto, this means that we can define an inverse mapping, denoted as \(F^{-1}: U \rightarrow V\), that satisfies the following condition: \(F^{-1}(F(v)) = v\) for all vectors \(v \in V\), and \(F(F^{-1}(u)) = u\) for all vectors \(u \in U\).

To show that the inverse mapping \(F^{-1}\) is linear, we need to prove that it satisfies the two conditions mentioned in Step 1. Let \(u_1, u_2 \in U\), and \(\beta_1, \beta_2\) be scalars. We need to show that: \(F^{-1}(\beta_1 u_1 + \beta_2 u_2) = \beta_1 F^{-1}(u_1) + \beta_2 F^{-1}(u_2)\) Since F is onto and one-to-one, then for each \(u_i\) in U, there exists a unique \(v_i\) in V such that: \(F(v_i) = u_i\) Now we will apply F to both sides of the expression we want to prove: \(F(F^{-1}(\beta_1 u_1 + \beta_2 u_2)) = F(\beta_1 F^{-1}(u_1) + \beta_2 F^{-1}(u_2))\) Using the properties of the inverse mapping: \(\beta_1 u_1 + \beta_2 u_2 = F(\beta_1 F^{-1}(u_1) + \beta_2 F^{-1}(u_2))\) We know that F is linear, so we can apply the properties of a linear mapping: \(\beta_1 u_1 + \beta_2 u_2 = \beta_1 F(F^{-1}(u_1)) + \beta_2 F(F^{-1}(u_2))\) Using the properties of the inverse mapping again: \(\beta_1 u_1 + \beta_2 u_2 = \beta_1 u_1 + \beta_2 u_2\) Since the above expression is true for every choice of \(u_1, u_2 \in U\) and \(\beta_1, \beta_2\), this demonstrates that \(F^{-1}\) is a linear mapping from U to V.