1. **Statement of the problem:** Given a system of linear homogeneous equations $$A^1x_1 + A^2x_2 + \cdots + A^m x_m = 0 \in M_{n \times 1}(K)$$ prove that the solution set forms a vector subspace of $\mathbb{R}^m$. 2. **Formula and rules:** - The solution set of a homogeneous linear system is the null space (kernel) of the matrix formed by the coefficients. - The null space is always a vector subspace. 3. **Proof:** - Let $X, Y$ be solutions, i.e., $A^1 x_1 + \cdots + A^m x_m = 0$ and $A^1 y_1 + \cdots + A^m y_m = 0$. - For any scalars $\alpha, \beta \in K$, consider $Z = \alpha X + \beta Y$. - Then $$A^1 z_1 + \cdots + A^m z_m = A^1 (\alpha x_1 + \beta y_1) + \cdots + A^m (\alpha x_m + \beta y_m) = \alpha (A^1 x_1 + \cdots + A^m x_m) + \beta (A^1 y_1 + \cdots + A^m y_m) = 0$$ - Hence, $Z$ is also a solution. - The zero vector is trivially a solution. - Therefore, the solution set is closed under addition and scalar multiplication, so it is a vector subspace. 4. **Second part:** Given $y_p$ a particular solution of the nonhomogeneous system $$A^1 x_1 + A^2 x_2 + \cdots + A^m x_m = B \in M_{n \times 1}(K)$$ and $Y_H$ the general solution of the corresponding homogeneous system $$A^1 x_1 + A^2 x_2 + \cdots + A^m x_m = 0,$$ prove that $$Y_H + y_p$$ is the general solution of the nonhomogeneous system. 5. **Proof:** - Let $Y$ be any solution of the nonhomogeneous system. - Then $$A^1 y_1 + \cdots + A^m y_m = B$$ - Consider $Y - y_p$: $$A^1 (y_1 - y_{p1}) + \cdots + A^m (y_m - y_{pm}) = B - B = 0$$ - So $Y - y_p$ is a solution of the homogeneous system. - Hence, any solution $Y$ can be written as $$Y = y_p + Y_H$$ - This shows that the general solution of the nonhomogeneous system is the sum of a particular solution and the general solution of the homogeneous system. 6. **Example: Solve the system** $$\begin{cases} 4x + y = 0 \\ x - 2y = 0 \end{cases}$$ - From the second equation: $x = 2y$. - Substitute into the first: $4(2y) + y = 8y + y = 9y = 0 \Rightarrow y = 0$. - Then $x = 0$. - The only solution is the trivial solution $(0,0)$. - The solution space is the zero vector, dimension $0$. 7. **Example: Solve the system** $$\begin{cases} x - 2y = 0 \\ 2x + y = 0 \end{cases}$$ - From the first: $x = 2y$. - Substitute into the second: $2(2y) + y = 4y + y = 5y = 0 \Rightarrow y = 0$. - Then $x = 0$. - The solution space is trivial, dimension $0$. 8. **Example: Solve the system** $$\begin{cases} 4x - 3y + z = 0 \\ x - 2y + 2z = 0 \\ 3x + y - z = 0 \end{cases}$$ - Use elimination or matrix methods to find the solution space. - The system matrix is $$A = \begin{bmatrix}4 & -3 & 1 \\ 1 & -2 & 2 \\ 3 & 1 & -1 \end{bmatrix}$$ - Row reduce to find rank and nullity. - The solution space dimension is $3 - \text{rank}(A)$. 9. **Summary:** - The solution set of homogeneous linear systems is a vector subspace. - The general solution of a nonhomogeneous system is a particular solution plus the homogeneous solution. - The dimension of the solution space equals the number of variables minus the rank of the coefficient matrix. **Final answer:** - The solution set of the homogeneous system forms a vector subspace. - The general solution of the nonhomogeneous system is $Y_H + y_p$. - For the example systems, the solution spaces are as computed above.