Let A1 3 3 0 0 10 5 15 1 1 1 6 5 5 10 15 Find orthogonal bas

Let A=[-1 -3 -3 0 0 10 5 -15 1 -1 1 6 -5 -5 -10 -15] Find orthogonal bases of the kernel and image of A. Basis of the kernel: [], []. Basis of the image: [], [].

Solution

To answer the given questions, we will reduce A to its RREF as under:

Multiply the 1st row by -1

Add 3 times the 1st row to the 2nd row

Add 3 times the 1st row to the 3rd row

Multiply the 2nd row by 1/10

Add -5 times the 2nd row to the 3rd row              

Add 15 times the 2nd row to the 4th row

Then the RREF of A is

1

0

-1

5

0

1

-2/5

1

0

0

0

0

0

0

0

0

Now, if X = (x,y,z,w)^T, then the equation AX= 0 is equivalent to x –z+5w= 0 or, x = z-5w and y-2z/5+w = 0 or, y=2z/5-w so that X = (z-5w,2z/5-w, z,w)^T = z/5(5,2,5,0)^T+w(-5,-1,0,1)^T. Hence a basis for Ker(A) is {(5,2,5,0)^T,(-5,-1,0,1)^T }= {v₁,v₂} (say). Further, let u₁ = v₁= (5,2,5,0)^T and u₂ = v₂-proj_u1(v₂) = v₂-[(v₂.u₁)/(u₁.u₁)]u₁ = v₂-[(-25-2+0+0)/(25+4+25+0)]u₁= v₂+(27/54)u₁ =(-5,-1,0,1)^T+(1/ 2)( 5,2,5,0)^T=       (-5/2, 0,5/2,1)^T. Then {(5,2,5,0)^T , (-5/2, 0,5/2,1)^T} is an orthogonal basis for ker(A).

Im(A) is same as Col(A). It is apparent from the RREF of A that the 3^rd and the 4^th columns of A are linear combination s of its first 2 columns so that a basis for Col(A) is {(1,0,0,0)^T,(0,1,0,0)^T} or, { (-1,-3,-3,0)^T, (0, 10,5,-15)^T} = {w₁,w₂} (say). The first basis for Col(A) is already orthogonal. We will orthogonalize the 2^nd basis for Col(A) as under. Let z₁ = w₁ = (-1,-3,-3,0)^T and z₂ = w₂- proj _z1(w₂) = w₂-[(w₂.z₁)/(z₁.z₁)]z₁ = w₂-[(0-30-15+0)/ (1+9+9+0)]z₁= w₂+(45/19)z₁ = (0,10,5,-15)^T +(45/19)(-1,-3,-3,0)^T= (-45/19, 55/19,-40/19,-15)^T. Then {(-1,-3,-3,0)^T ,(-45/19, 55/19,-40/19,-15)^T} is an orthogonal basis for Im(A) or, Col(A).

1	0	-1	5
0	1	-2/5	1
0	0	0	0
0	0	0	0