Apply the Gram-Schmidt process to the vectors \(\beta_1 = (1, 0, 1)\), \(\beta_2 = (1, 0, -1)\), \(\beta_3 = (0, 3, 4)\), to obtain an orthonormal basis for \(\RR^3\) with the standard inner product.
Solution
Set \[ \beta_1=(1,0,1),\qquad \beta_2=(1,0,-1),\qquad \beta_3=(0,3,4). \] Since \[ \beta_1\cdot\beta_2=0, \qquad \|\beta_1\|=\|\beta_2\|=\sqrt2, \] the first two normalized vectors are \[ u_1=\frac1{\sqrt2}(1,0,1), \qquad u_2=\frac1{\sqrt2}(1,0,-1). \]
For the third vector, \[ (\beta_3\mid u_1)=2\sqrt2, \qquad (\beta_3\mid u_2)=-2\sqrt2. \] Therefore the Gram–Schmidt residual is \[ \begin{aligned} v_3 &=\beta_3-(\beta_3\mid u_1)u_1-(\beta_3\mid u_2)u_2\\ &=(0,3,4)-(2,0,2)-(-2,0,2)\\ &=(0,3,0). \end{aligned} \] Hence \(u_3=(0,1,0)\). Thus an orthonormal basis obtained from the prescribed order is \[ \boxed{ \left\{ \frac1{\sqrt2}(1,0,1), \frac1{\sqrt2}(1,0,-1), (0,1,0) \right\}.} \] Each pair has zero dot product and each vector has norm one, so the result is indeed orthonormal [@HK71].