Consider \(\RR^4\) with the standard inner product. Let \(W\) be the subspace of \(\RR^4\) consisting of all vectors which are orthogonal to both \(\alpha = (1, 0, -1, 1)\) and \(\beta = (2, 3, -1, 2)\). Find a basis for \(W\).
Solution
Let \(x=(x_1,x_2,x_3,x_4)\). Orthogonality to \[ \alpha=(1,0,-1,1), \qquad \beta=(2,3,-1,2) \] gives \[ x_1-x_3+x_4=0 \] and \[ 2x_1+3x_2-x_3+2x_4=0. \] Subtracting twice the first equation from the second yields \[ 3x_2+x_3=0, \] so \(x_3=-3x_2\). Substituting into the first equation gives \[ x_1=-3x_2-x_4. \] Thus, with \(s=x_2\) and \(t=x_4\), \[ x=s(-3,1,-3,0)+t(-1,0,0,1). \] The two displayed vectors are linearly independent, hence \[ \boxed{ W=\operatorname{span}\{(-3,1,-3,0),\,(-1,0,0,1)\}.} \] Therefore they form a basis of \(W\) [@HK71].