Homework SourseGiven the matrix A below find an orthogonal basis for each o
Solution
We will reduce A to its RREF as under:
Add -2 times the 1st row to the 2nd row ; Add -2 times the 1st row to the 3rd row
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
Then, a basis for C(A) is {v1,v3,v5} = { (1,2,2,-1)T, (4,9,9,-4)T, (5,8,9,-5)T }. An othogonal basis for C(A) is { (1,0,0,0)T, (0,1,0,0)T,(0,0,1,0)T } .
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 +4x6 = 0, and x5+5x6=0. Now, let x2 =r, x4 =s, and x6=t. Then, 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.Then a basis for N(A) is {(3,1,0,0,0,0)T,(14,0,-3,1,0,0)T,( 37,0,-4,0,-5,1)T }. Let B be the 6X3 matrix with the basis vectors in N(A) as columns. Then the RREF of B is
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
Thus, { {(1,0,0,0,0,0)T,(0,1,0,0,0,0)T,( 0,0,1,0,0)T }.
Also, the RREF of AT 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
Hence, an orthogonal basis for R(A) is{ (1,0,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0,0)}.
I do not know what is L(A).
| 1 | -3 | 0 | -14 | 0 | -37 |
| 0 | 0 | 1 | 3 | 0 | 4 |
| 0 | 0 | 0 | 0 | 1 | 5 |
| 0 | 0 | 0 | 0 | 0 | 0 |
