4. Let \( V \) be a vector space, and \( f: V \rightarrow R, g: V \rightarrow R \) two linear mappings. Let \( F: V \rightarrow R^{2} \) be the mapping defined by \( F(v)=(f(v), g(v)) \). Show that \( F \) is linear.

To show that the mapping \( F: V \rightarrow \mathbb{R}^2 \) defined by \( F(v) = (f(v), g(v)) \) is linear, we need to verify two properties of linearity: 1. **Additivity**: For all \( u, v \in V \), we need to show that \( F(u + v) = F(u) + F(v) \). 2. **Homogeneity**: For all \( v \in V \) and all scalars \( c \in \mathbb{R} \), we need to show that \( F(cv) = cF(v) \). ### Step 1: Additivity Let \( u, v \in V \). We compute \( F(u + v) \): \[ F(u + v) = (f(u + v), g(u + v)) \] Since \( f \) and \( g \) are linear mappings, we can use the property of linearity: \[ f(u + v) = f(u) + f(v) \] \[ g(u + v) = g(u) + g(v) \] Thus, we have: \[ F(u + v) = (f(u) + f(v), g(u) + g(v)) = (f(u), g(u)) + (f(v), g(v)) = F(u) + F(v) \] ### Step 2: Homogeneity Let \( v \in V \) and \( c \in \mathbb{R} \). We compute \( F(cv) \): \[ F(cv) = (f(cv), g(cv)) \] Again, using the linearity of \( f \) and \( g \): \[ f(cv) = c f(v) \] \[ g(cv) = c g(v) \] Thus, we have: \[ F(cv) = (c f(v), c g(v)) = c (f(v), g(v)) = c F(v) \] ### Conclusion Since both properties of linearity (additivity and homogeneity) are satisfied, we conclude that the mapping \( F: V \rightarrow \mathbb{R}^2 \) is linear. Therefore, we have shown that: \[ F \text{ is linear.} \]