Find all solutions in \(\mathbb{C}^2\) to the homogeneous linear system: \[ \begin{aligned} (1 - i) x_1 - i x_2 &= 0 \\ 2 x_1 + (1 - i) x_2 &= 0. \end{aligned} \]
Solution
Goal: Solve the \(2 \times 2\) homogeneous complex linear system \(A x = 0\) by determining the determinant and row echelon reduction.
1. Matrix Form and Determinant Computation:Proof:
1.1. The system can be written as \(A x = 0\) where \(x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \in \mathbb{C}^2\) and:\[A = \begin{bmatrix} 1 - i & -i \\ 2 & 1 - i \end{bmatrix}.\]
1.2. We compute the determinant of the coefficient matrix \(A\):\[\det(A) = (1 - i)(1 - i) - (-i)(2) = (1 - i)^2 + 2i.\]
1.3. Expanding the complex square:\[(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i.\]
1.4. Substituting this into the determinant:\[\det(A) = -2i + 2i = 0.\]
1.5. Since \(\det(A) = 0\), the matrix \(A\) is singular (\(\operatorname{rank}(A) < 2\)), so the system has non-trivial solutions.
2. Row Reduction and Parametric Solution:Proof:
2.1. Multiplying the first equation by \((1 + i)\):\[(1 + i)(1 - i) x_1 - (1 + i) i x_2 = 0 \implies 2 x_1 - (i - 1) x_2 = 0 \implies 2 x_1 + (1 - i) x_2 = 0.\]
2.2. This is identical to the second equation! Thus the second row is linearly dependent on the first.
2.3. The single independent equation is:\[(1 - i) x_1 - i x_2 = 0 \implies i x_2 = (1 - i) x_1.\]
2.4. Multiplying both sides by \(-i\) (since \(-i \cdot i = 1\)):\[x_2 = -i (1 - i) x_1 = (-i - 1) x_1 = -(1 + i) x_1.\]
2.5. Setting \(x_1 = t \in \mathbb{C}\) as the free parameter:\[x_2 = -(1 + i) t.\]
3. Solution Space:Proof:
3.1. The complete solution set is the 1-dimensional complex subspace:\[\operatorname{Sol} = \left\{ t \begin{bmatrix} 1 \\ -(1 + i) \end{bmatrix} \;\middle|\; t \in \mathbb{C} \right\} = \operatorname{span}_\mathbb{C} \left\{ \begin{bmatrix} 1 \\ -1 - i \end{bmatrix} \right\}.\]
4. Conclusion:The solutions are all scalar multiples \(x = t \begin{bmatrix} 1 \\ -1 - i \end{bmatrix}\) for \(t \in \mathbb{C}\). Q.E.D.