thank you Given the matrix A below find an orthogonal basis

thank you

Given the matrix A below find an orthogonal basis for each of C(A), R(A), L(A), N(A). A = (1 2 2 -1 -3 -6 -6 3 4 9 9 -4 -2 -1 -1 2 5 8 9 -5 4 2 7 -4)

Solution

We will first reduce A to its RREF as under:

Add 1 times the 1st row to the 4th row

Add -1 times the 2nd row to the 3rd row

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

Add -5 times the 3rd row to the 1st row

Add -4 times the 2nd row to the 1st row

Then the RREF of A is

1

-3

0

-14

0

-37

0

0

1

3

0

4

0

0

0

0

1

5

0

0

0

0

0

0

Apparently, C(A) = span{ (1,2,2,-1)^T, (4,9,9,-4)^T, (5,8,9,-5)^T} or span { (1,0,0,0)^T,( 0,1,0,0)^T, ( 0,0,1,0)^T }. Further, { (1,0,0,0)^T,( 0,1,0,0)^T, ( 0,0,1,0)^T } is an orthogonal basis for C(A).

Also, N(A) is the set of solutions to the equation AX = 0. If X = (x₁,x₂,x₃,x₄,x₅,x₆)^T , then this equation is equivalent to x₁ -3x₂ -14x₄ -37x₆ = 0, x₃+3x₄ +4x₅ = 0 and x₅+5x₆ = 0. Now, let x₂ =r, x₄ = s and x₆ = t. Then we have x₁ = 3r+14s+37t , x₃ = -3s-4t and x₅ = -5t so that X = (3r+14s+37t, r, -3s-4t, s,-5t,t)^T = r(3,1,0,0,0,0)^T+ s( 14, 0,-3,1,0,0)^T +t ( 37,0,-4,0-5,1)^T}. Hence N(A) = span {v₁,v₂,v₃} where v₁ = (3,1,0,0,0,0)^T, v₂ = ( 14, 0,-3,1,0,0)^T and v₃ = ( 37,0,-4,0-5,1)^T. We will use the Gram-Schmidt process to orthogonalise the basis for N(A). Let u₁ = v₁, u₂ = v₂ - Proj_u1(v₂ ) = v₂ –[(v₂.u₁)/(u₁.u₁)]u₁ = v₂ –( 26/5)u₁ = (14, 0,-3,1,0,0)^T –(26/5) (3,1,0,0,0,0)^T = (-8/5, -26/5, -3,1,0,0)^T. Also, u₃ = v₃ - Proj_u1(v₃ )- Proj_u2(v₃) = v₃ –[(v₃.u₁)/(u₁.u₁)]u₁–[(v₃.u₂)/(u₂.u₂)]u₂= v₃-(111/10)u₁+(118/495)u₂ = ( 37,0,-4,0-5,1)^T – (111/10) (3,1,0,0,0,0)^T +(118/495) (-8/5, -26/5, -3,1,0,0)^T= ( 37,0,-4,0-5,1)^T –(333/10, 111/10,0,0,0,0)^T+ (-944/2475,-3068/2475,-354/495,118/495,0,0)^T=                                                                                    (16427/4950,-61081/4950,-23340/4950,1180/4950,-5,1)^T. Then, {u₁,u₂,u₃} is an orthogonal basis for N(A).

Further, we will reduce A^T to its RREF as under:

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

Add -4 times the 1st row to the 3rd row

Add 2 times the 1st row to the 4th row

Add -5 times the 1st row to the 5th row

Add -4 times the 1st row to the 6th row

Interchange the 2nd row and the 3rd row

Add -3 times the 2nd row to the 4th row

Add 2 times the 2nd row to the 5th row

Add 6 times the 2nd row to the 6th row

Interchange the 3rd row and the 5th row

Add -5 times the 3rd row to the 6th row

Add -1 times the 3rd row to the 2nd row

Add -2 times the 3rd row to the 1st row

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

Then the RREF of A^T is

1

0

0

-1

0

1

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

0

0

It is now apparent that R(A) = span { (1,-3,4,-2,5,4),(2,-6,9,-1,8,2),(2,-6,9,-1,9,7)} or,                                           span{ (1,0,0,0,0,0), (0,1,0,0,0,0), (0,0,1,0,0,0)}. An orthogonal basis for R(A) is the set {(1,0,0,0,0,0), (0,1,0,0,0,0), (0,0,1,0,0,0)}.

I do not know what L(A) means.

1	-3	0	-14	0	-37
0	0	1	3	0	4
0	0	0	0	1	5
0	0	0	0	0	0