We presume that the triplets mentioned are the columns of the
matrix.
(a). Let A =
The RREF of A is
|
1
|
0
|
0
|
4/5
|
|
0
|
1
|
0
|
8/5
|
|
0
|
0
|
1
|
-4/5
|
Thus, a basis for col(A) is {
v1,v2,v3} = {
(1,1,1)T,(1,0,2)T,(3,1,0)T} .
Let
u1=v1=(1,1,1)T,u2=v2–proju1(v2)=v2–[(v2.u1)/(u1.u1)]u1=v2–[(1+0+2)/(1+1+1)]u1=(1,0,2)T-(1,1,1)T=
(0,-1, 1)Tand
u3=v3–proju1(v3)=v3–[(v3.u1)/(u1.u1)]u1–[(v3.u2)/(u2.u2)]u2=v3–[(3+1+0)/(1+1+1)]u1-[(0-1+0)/
(0+1+1)]u2 = (3,1,0)T –(4/3)(
1,1,1)T -(1/2)( 0,-1, 1)T=
(5/3,-5/6,-5/6)T. Then {
u1,u2,u3}= {(1,1,1)T,
(0,-1, 1)T, (5/3,-5/6,-5/6)T } is an
orthogonal basis for col(A).
Further, let e1 = u1/||u1||
=(1/?3,1?3,1/?3)T e2 =
u2/||u2|| =(0,-1?2, 1/?2)Tand
e3 = u3/||u3|| =
(5/3,-5/6,-5/6)T= (?2/?3, 1/?6,-1/?6)T. Then
{e1,e2,e3} =
{(1/?3,1?3,1/?3)T, (0,-1?2, 1/?2)T, (?2/?3,
1/?6,-1/?6)T} is an orthonormal basis for col(A).
Please post the other question again.
