1. **Problem statement:** Check if the mapping $F : \mathbb{R}^4 \to M(2 \times 2, \mathbb{R})$ defined by $$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} \mapsto \begin{pmatrix} x_1 & x_1 - x_4 \\ x_2 - x_3 & x_4 \end{pmatrix} $$ is linear and if it is an isomorphism. 2. **Recall:** A function $F$ is linear if for all vectors $u,v$ and scalars $\alpha$, $$ F(u + v) = F(u) + F(v) \quad \text{and} \quad F(\alpha u) = \alpha F(u). $$ An isomorphism is a linear bijection (one-to-one and onto). 3. **Check additivity:** Let $u = (u_1,u_2,u_3,u_4)^T$, $v = (v_1,v_2,v_3,v_4)^T$. Then $$ F(u+v) = \begin{pmatrix} u_1 + v_1 & (u_1 + v_1) - (u_4 + v_4) \\ (u_2 + v_2) - (u_3 + v_3) & u_4 + v_4 \end{pmatrix}. $$ On the other hand, $$ F(u) + F(v) = \begin{pmatrix} u_1 & u_1 - u_4 \\ u_2 - u_3 & u_4 \end{pmatrix} + \begin{pmatrix} v_1 & v_1 - v_4 \\ v_2 - v_3 & v_4 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 & (u_1 - u_4) + (v_1 - v_4) \\ (u_2 - u_3) + (v_2 - v_3) & u_4 + v_4 \end{pmatrix}. $$ Since $$ (u_1 + v_1) - (u_4 + v_4) = (u_1 - u_4) + (v_1 - v_4), $$ and $$ (u_2 + v_2) - (u_3 + v_3) = (u_2 - u_3) + (v_2 - v_3), $$ additivity holds. 4. **Check homogeneity:** For scalar $\alpha$, $$ F(\alpha u) = \begin{pmatrix} \alpha u_1 & \alpha u_1 - \alpha u_4 \\ \alpha u_2 - \alpha u_3 & \alpha u_4 \end{pmatrix} = \alpha \begin{pmatrix} u_1 & u_1 - u_4 \\ u_2 - u_3 & u_4 \end{pmatrix} = \alpha F(u). $$ So homogeneity holds. 5. **Conclusion on linearity:** Since both additivity and homogeneity hold, $F$ is linear. 6. **Check if $F$ is an isomorphism:** - $F$ maps from $\mathbb{R}^4$ to $M(2 \times 2, \mathbb{R})$, which also has dimension 4. - To be an isomorphism, $F$ must be bijective. 7. **Check injectivity:** Suppose $F(x) = 0$, i.e., $$ \begin{pmatrix} x_1 & x_1 - x_4 \\ x_2 - x_3 & x_4 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}. $$ This gives the system: $$ x_1 = 0, \quad x_1 - x_4 = 0, \quad x_2 - x_3 = 0, \quad x_4 = 0. $$ From $x_1 = 0$ and $x_1 - x_4 = 0$, we get $x_4 = 0$. From $x_2 - x_3 = 0$, we get $x_2 = x_3$. So the kernel is all vectors $(0, x_2, x_2, 0)^T$, which is not just the zero vector. 8. **Therefore, $F$ is not injective and hence not an isomorphism.**