Show that the vectors \[ \alpha_1 = (1, 0, -1), \quad \alpha_2 = (1, 2, 1), \quad \alpha_3 = (0, -3, 2) \] form a basis for \(\mathbb{R}^3\). Express each of the standard basis vectors \(e_1, e_2, e_3\) as linear combinations of \(\alpha_1, \alpha_2, \alpha_3\).
Solution
Goal: Prove linear independence by computing the non-zero determinant of the matrix formed by \(\alpha_1, \alpha_2, \alpha_3\), and invert the matrix to express standard basis vectors.
1. Basis Verification via Determinant:Proof:
1.1. Form the \(3 \times 3\) matrix \(A = [\alpha_1 \mid \alpha_2 \mid \alpha_3]\) whose columns are the vectors:\[A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 2 & -3 \\ -1 & 1 & 2 \end{bmatrix}.\]
1.2. Compute the determinant of \(A\) along the first row:\[\det(A) = 1 \cdot \det\begin{bmatrix} 2 & -3 \\ 1 & 2 \end{bmatrix} - 1 \cdot \det\begin{bmatrix} 0 & -3 \\ -1 & 2 \end{bmatrix} + 0\]\[\det(A) = 1(4 - (-3)) - 1(0 - 3) = 1(7) - 1(-3) = 7 + 3 = 10.\]
1.3. Since \(\det(A) = 10 \ne 0\), the vectors \(\{\alpha_1, \alpha_2, \alpha_3\}\) are linearly independent.
1.4. Since \(\dim(\mathbb{R}^3) = 3\), three linearly independent vectors form a basis for \(\mathbb{R}^3\).
2. Matrix Inversion to Express \(e_1, e_2, e_3\):Proof:
2.1. To express \(e_j = c_{1j}\alpha_1 + c_{2j}\alpha_2 + c_{3j}\alpha_3\), we find the inverse matrix \(A^{-1}\):\[[\alpha_1 \mid \alpha_2 \mid \alpha_3] A^{-1} = I_3 \implies e_j = A (A^{-1} e_j).\]
2.2. We compute the matrix of cofactors of \(A\):- \(C_{11} = +(4 - (-3)) = 7\)- \(C_{12} = -(0 - 3) = 3\)- \(C_{13} = +(0 - (-2)) = 2\)- \(C_{21} = -(2 - 0) = -2\)- \(C_{22} = +(2 - 0) = 2\)- \(C_{23} = -(1 - (-1)) = -2\)- \(C_{31} = +(-3 - 0) = -3\)- \(C_{32} = -(-3 - 0) = 3\)- \(C_{33} = +(2 - 0) = 2\)
2.3. The adjugate matrix \(\operatorname{adj}(A) = C^T\):\[\operatorname{adj}(A) = \begin{bmatrix} 7 & -2 & -3 \\ 3 & 2 & 3 \\ 2 & -2 & 2 \end{bmatrix}.\]
2.4. The inverse matrix is \(A^{-1} = \frac{1}{\det(A)} \operatorname{adj}(A)\):\[A^{-1} = \frac{1}{10} \begin{bmatrix} 7 & -2 & -3 \\ 3 & 2 & 3 \\ 2 & -2 & 2 \end{bmatrix}.\]
3. Linear Combinations for Standard Basis Vectors:Proof:
3.1. For \(e_1 = (1, 0, 0)\) (first column of \(A^{-1}\)):\[e_1 = \frac{7}{10} \alpha_1 + \frac{3}{10} \alpha_2 + \frac{2}{10} \alpha_3 = \frac{7}{10} \alpha_1 + \frac{3}{10} \alpha_2 + \frac{1}{5} \alpha_3.\]Check: \(\frac{7}{10}(1, 0, -1) + \frac{3}{10}(1, 2, 1) + \frac{2}{10}(0, -3, 2) = (\frac{10}{10}, \frac{6-6}{10}, \frac{-7+3+4}{10}) = (1, 0, 0)\).
3.2. For \(e_2 = (0, 1, 0)\) (second column of \(A^{-1}\)):\[e_2 = -\frac{2}{10} \alpha_1 + \frac{2}{10} \alpha_2 - \frac{2}{10} \alpha_3 = -\frac{1}{5} \alpha_1 + \frac{1}{5} \alpha_2 - \frac{1}{5} \alpha_3.\]Check: \(-\frac{2}{10}(1, 0, -1) + \frac{2}{10}(1, 2, 1) - \frac{2}{10}(0, -3, 2) = (0, \frac{4+6}{10}, \frac{2+2-4}{10}) = (0, 1, 0)\).
3.3. For \(e_3 = (0, 0, 1)\) (third column of \(A^{-1}\)):\[e_3 = -\frac{3}{10} \alpha_1 + \frac{3}{10} \alpha_2 + \frac{2}{10} \alpha_3 = -\frac{3}{10} \alpha_1 + \frac{3}{10} \alpha_2 + \frac{1}{5} \alpha_3.\]Check: \(-\frac{3}{10}(1, 0, -1) + \frac{3}{10}(1, 2, 1) + \frac{2}{10}(0, -3, 2) = (0, \frac{6-6}{10}, \frac{3+3+4}{10}) = (0, 0, 1)\).
4. Conclusion:\(\det[\alpha_1 \mid \alpha_2 \mid \alpha_3] = 10 \ne 0\) confirms the basis; the standard vectors are expressed via the columns of \(A^{-1} = \frac{1}{10}\begin{bmatrix} 7 & -2 & -3 \\ 3 & 2 & 3 \\ 2 & -2 & 2 \end{bmatrix}\). Q.E.D.