Let \(W\) be the subspace of \(\RR^5\) which is spanned by the vectors \[ \alpha_1 = \epsilon_1 + 2\epsilon_2 + \epsilon_3, \quad \alpha_2 = \epsilon_2 + 3\epsilon_3 + 3\epsilon_4 + \epsilon_5, \] \[ \alpha_3 = \epsilon_1 + 4\epsilon_2 + 6\epsilon_3 + 4\epsilon_4 + \epsilon_5. \] Find a basis for \(W^0\).
Solution
Identify a linear functional on \(\RR^5\) with its coefficient vector \(c=(c_1,\ldots,c_5)\). The condition \(c\in W^0\) is \[ \begin{bmatrix} 1&2&1&0&0\\ 0&1&3&3&1\\ 1&4&6&4&1 \end{bmatrix}c^{\mathsf T}=0. \] Row reduction gives a two-dimensional nullspace with basis \[ (-4,3,-2,1,0),\qquad (-3,2,-1,0,1). \] Therefore a basis for \(W^0\) is \[ f_1(x)=-4x_1+3x_2-2x_3+x_4, \] \[ f_2(x)=-3x_1+2x_2-x_3+x_5. \] Direct substitution in each \(\alpha_i\) verifies that both functionals annihilate \(W\); their coefficient vectors are independent, and \(\dim W^0=5-\dim W=2\).