In \(\RR^3\), let \(\alpha_1 = (1, 0, 1)\), \(\alpha_2 = (0, 1, -2)\), \(\alpha_3 = (-1, -1, 0)\).
If \(f\) is a linear functional on \(\RR^3\) such that \[ f(\alpha_1) = 1, \quad f(\alpha_2) = -1, \quad f(\alpha_3) = 3, \] and if \(\alpha = (a, b, c)\), find \(f(\alpha)\).
Describe explicitly a linear functional \(f\) on \(\RR^3\) such that \[ f(\alpha_1) = f(\alpha_2) = 0 \quad \text{but} \quad f(\alpha_3) \neq 0. \]
Let \(f\) be any linear functional such that \[ f(\alpha_1) = f(\alpha_2) = 0 \quad \text{and} \quad f(\alpha_3) \neq 0. \] If \(\alpha = (2, 3, -1)\), show that \(f(\alpha) \neq 0\).
Solution
Any linear functional on \(\RR^3\) has the form \[ f(x,y,z)=px+qy+rz. \] The conditions in (a) give \[ p+r=1,\qquad q-2r=-1,\qquad -p-q=3. \] Solving yields \((p,q,r)=(4,-7,-3)\). Hence \[ f(a,b,c)=4a-7b-3c. \]
For (b), to annihilate \(\alpha_1\) and \(\alpha_2\) we require \[ p+r=0,\qquad q-2r=0. \] Taking \(r=1\) gives \[ f(x,y,z)=-x+2y+z. \] Then \(f(\alpha_3)=-1\ne0\).
For (c), every functional annihilating \(\alpha_1,\alpha_2\) has the form \[ f(x,y,z)=r(-x+2y+z) \] for some scalar \(r\). The hypothesis \(f(\alpha_3)\ne0\) forces \(r\ne0\). Therefore \[ f(2,3,-1)=r(-2+6-1)=3r\ne0. \]