In \(C^3\), let \[ \alpha_1 = (1, 0, -i), \quad \alpha_2 = (1 + i, 1 - i, 1), \quad \alpha_3 = (i, i, i). \] Prove that these vectors form a basis for \(C^3\). What are the coordinates of the vector \((a, b, c)\) in this basis?
Solution
1. Prove that \(\{\alpha_1, \alpha_2, \alpha_3\}\) forms a basis of \(\mathbb{C}^3\):
1.1. Form the matrix \(A\) having \(\alpha_1, \alpha_2, \alpha_3\) as columns:\[ A = \begin{pmatrix} 1 & 1+i & i \\ 0 & 1-i & i \\ -i & 1 & i \end{pmatrix}. \]
matrix representation of column vectors.
1.2. Compute the determinant of \(A\) by cofactor expansion along the first column:\[ \det(A) = 1 \cdot \det\begin{pmatrix} 1-i & i \\ 1 & i \end{pmatrix} - i \cdot \det\begin{pmatrix} 1+i & i \\ 1-i & i \end{pmatrix}. \]Compute the \(2 \times 2\) determinants:\[ \det\begin{pmatrix} 1-i & i \\ 1 & i \end{pmatrix} = (1-i)i - i = i + 1 - i = 1, \]\[ \det\begin{pmatrix} 1+i & i \\ 1-i & i \end{pmatrix} = (1+i)i - i(1-i) = i - 1 - i - 1 = -2. \]Thus:\[ \det(A) = 1(1) - i(-2) = 1 + 2i. \]
cofactor expansion.
1.3. Since \(\det(A) = 1 + 2i \neq 0\), the matrix \(A\) is invertible, so the vectors \(\{\alpha_1, \alpha_2, \alpha_3\}\) are linearly independent.Since \(\dim(\mathbb{C}^3) = 3\), they form a basis for \(\mathbb{C}^3\).
non-zero determinant characterization of bases.
2. Compute the coordinates \((z_1, z_2, z_3)\) of \((a, b, c)\) in the basis:
2.1. The coordinates satisfy \(A \begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}\), so \(\begin{pmatrix} z_1 \\ z_2 \\ z_3 \end{pmatrix} = A^{-1} \begin{pmatrix} a \\ b \\ c \end{pmatrix}\).
definition of coordinate vector.
2.2. Compute the inverse matrix \(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A) = \frac{1-2i}{5} C^T\):The cofactor matrix of \(A\) is:\[ C = \begin{pmatrix} 1 & 1 & 1+i \\ 1 & i-1 & -i \\ -2 & -i & 1-i \end{pmatrix} \implies \operatorname{adj}(A) = C^T = \begin{pmatrix} 1 & 1 & -2 \\ 1 & i-1 & -i \\ 1+i & -i & 1-i \end{pmatrix}. \]
computation of cofactor transpose.
2.3. Multiply \(A^{-1} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \frac{1-2i}{5} \begin{pmatrix} 1 & 1 & -2 \\ 1 & i-1 & -i \\ 1+i & -i & 1-i \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}\):\[ z_1 = \frac{1-2i}{5}(a + b - 2c), \]\[ z_2 = \frac{1-2i}{5}\big(a + (i-1)b - ic\big) = \frac{(1-2i)a + (1+3i)b - (2+i)c}{5}, \]\[ z_3 = \frac{1-2i}{5}\big((1+i)a - ib + (1-i)c\big) = \frac{(3-i)a - (2+i)b - (1+3i)c}{5}. \]
matrix-vector multiplication.
3. Conclusion:\(\{\alpha_1, \alpha_2, \alpha_3\}\) is a basis, and the coordinates are \((z_1, z_2, z_3)\) as given above. Q.E.D.
1 and 2.