91Ó°ÊÓ

A linear functional in \(\mathbf{R}^{3}\) can be represented as \(\phi(x, y, z) = a x + b y + c z\), where a, b and c are coefficients, and (x, y, z) are the coordinates of the vector to be evaluated by the functional. The dual basis is: \[\left\\{\phi_{1}(x, y, z) = a_{1}x + b_{1}y + c_{1}z,\right.\] \[\left. \phi_{2}(x, y, z) = a_{2}x + b_{2}y + c_{2}z,\right.\] \[\left. \phi_{3}(x, y, z) = a_{3}x + b_{3}y + c_{3}z\right\\}\]

Use the condition \(\phi_{i}(v_{j}) = \delta_{ij}\) to find the values of the coefficients. For our given basis, this will give: For i = 1, j = 1: \(\phi_{1}(v_{1}) = a_{1}(1) + b_{1}(-1) + c_{1}(3) = 1 \) For i = 1, j = 2: \(\phi_{1}(v_{2}) = a_{1}(0) + b_{1}(1) + c_{1}(-1) = 0 \) For i = 1, j = 3: \(\phi_{1}(v_{3}) = a_{1}(0) + b_{1}(3) + c_{1}(-2) = 0 \) For i = 2, j = 1: \(\phi_{2}(v_{1}) = a_{2}(1) + b_{2}(-1) + c_{2}(3) = 0 \) For i = 2, j = 2: \(\phi_{2}(v_{2}) = a_{2}(0) + b_{2}(1) + c_{2}(-1) = 1 \) For i = 2, j = 3: \(\phi_{2}(v_{3}) = a_{2}(0) + b_{2}(3) + c_{2}(-2) = 0\) For i = 3, j = 1: \(\phi_{3}(v_{1}) = a_{3}(1) + b_{3}(-1) + c_{3}(3) = 0 \) For i = 3, j = 2: \(\phi_{3}(v_{2}) = a_{3}(0) + b_{3}(1) + c_{3}(-1) = 0 \) For i = 3, j = 3: \(\phi_{3}(v_{3}) = a_{3}(0) + b_{3}(3) + c_{3}(-2) = 1 \)

Now we have 9 linear equations with 9 unknowns (a_{1}, b_{1}, c_{1}, a_{2}, b_{2}, c_{2}, a_{3}, b_{3}, c_{3}). Solve this system of equations to find the values of the coefficients. In this case, we can solve each set of three equations separately: For \(\phi_{1}\): \[a_{1} - b_{1} + 3c_{1} = 1,\quad b_{1} - c_{1} = 0, \quad 3b_{1} - 2c_{1} = 0\] Solving this system, we find: \(a_{1} = -1, \quad b_{1} = 3, \quad c_{1} = 3\). For \(\phi_{2}\): \[a_{2} - b_{2} + 3c_{2} = 0,\quad b_{2} - c_{2} = 1, \quad 3b_{2} - 2c_{2} = 0\] Solving this system, we find: \(a_{2} = 0, \quad b_{2} = 2, \quad c_{2} = 1\). For \(\phi_{3}\): \[a_{3} - b_{3} + 3c_{3} = 0,\quad b_{3} - c_{3} = 0, \quad 3b_{3} - 2c_{3} = 1\] Solving this system, we find: \(a_{3} = 1, \quad b_{3} = -1, \quad c_{3} = -1\).

Now that we have found all the coefficients, we can write down the dual basis: \[\left\\{\phi_{1}(x, y, z) = -x + 3y + 3z, \quad \phi_{2}(x, y, z) = 2y + z, \quad \phi_{3}(x, y, z) = x - y - z\right\\}\]