Let \(\alpha_1 = (1, 0, -1, 2)\) and \(\alpha_2 = (2, 3, 1, 1)\), and let \(W\) be the subspace of \(\RR^4\) spanned by \(\alpha_1\) and \(\alpha_2\). Which linear functionals \(f\): \[ f(x_1, x_2, x_3, x_4) = c_1 x_1 + c_2 x_2 + c_3 x_3 + c_4 x_4 \] are in the annihilator of \(W\)?
Solution
Write \[ f(x_1,x_2,x_3,x_4)=c_1x_1+c_2x_2+c_3x_3+c_4x_4. \] The condition \(f\in W^0\) is equivalent to \[ f(\alpha_1)=c_1-c_3+2c_4=0, \] \[ f(\alpha_2)=2c_1+3c_2+c_3+c_4=0. \] Solving gives \[ (c_1,c_2,c_3,c_4) =s(1,-1,1,0)+t(-2,1,0,1). \] Hence \(W^0\) consists exactly of the functionals \[ f_{s,t}(x)=s(x_1-x_2+x_3)+t(-2x_1+x_2+x_4), \qquad s,t\in\RR. \] In particular, a basis of \(W^0\) is given by the two coefficient vectors \[ (1,-1,1,0),\qquad (-2,1,0,1). \]