Apply the Gram-Schmidt process to the vectors [1; −2; 0], [1; 0; −1] and [0; 1; 1] .
Let v1 = (1,-2,0),v2 = (1,0,-1) and v3= (0,1,1).
Also, let u1 = v1 = (1,-2,0), u2 = v2 -proju1(v2) = v2 –[(v2.u1)/(u1.u1)]u1 = v2 –[(1+0+0)/(1+4+0)]u1= (1,0,-1)-(1/5) (1,-2,0) = (4/5,2/5,-1) and
u2 = v3 -proju1(v3)- proju2(v3) = v3 -[(v3.u1)/(u1.u1)]u1- [(v3.u2)/(u2.u2)]u2 = v3 -[(0-2+0)/(1+4+0)]u1 –[(0+2/5-1)/(16/25+4/25+1)]u2 = (0,1,1)+(2/5)(1,-2,0)+(1/3) (4/5,2/5,-1)= (2/3,1/3,2/3).
Then {u1,u2,u3} = {1,-2,0), (4/5,2/5,-1), (2/3,1/3,2/3)} is an orthogonal set.
Now, let w1 = u1/||u1|| = (1/√5,-2/√5,0), w2 = u2/||u2|| =(4/5,2/5,-1)= (4/3√5, 2/3√5,-√5/3) and w3 = u3=(2/3,1/3,2/3).
Then { w1, w2, w3}= {(1/√5,-2/√5,0), (4/3√5, 2/3√5,-√5/3), (2/3,1/3,2/3)} is an orthonormal set.
Get Answers For Free
Most questions answered within 1 hours.