Question

Chapter 3.6, Problem 20E in Introduction to Linear Algebra (5th Edition)

Find the basis for the null space and the range of the given matrix. Then use Gram-Schmidt to obtain the orthagonal bases.

1	3	10	11	9
-1	2	5	4	1
2	-1	-1	1	4

Answer 1

Let the given matrix be denoted by A> Then the null space of A is the set of solutions to the equation AX = 0. To solve this equation, we will reduce A to its RREF as under:

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

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

Multiply the 2nd row by 1/5

Add 7 times the 2nd row to the 3rd row

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

Then the RREF of A is

1	0	1	2	3
0	1	3	3	2
0	0	0	0	0

Now, if X = (x,y,z,w,u)^T, then the equation AX=0 is equivalent to x+z+2w+3u=0 or, x=-z-2w-3u and y+3z+3w+2u =0 or, y =-3z-3w-2u. Hence X =(-z-2w-3u, -3z-3w-2u,z,w,u)^T = z(-1,-3,1,0,0)^T+w(-2,-3,0,1,0)^T +       u(-3,-2,0,0,1)^T. Thus, a basis for the Null space of A is { v₁,v₂,v₃} = {(-1,-3,1,0,0)^T,(-2,-3,0,1,0)^T,(-3,-2,0,0,1)^T }.

Now, let u₁ = v₁=(-1,-3,1,0,0)^T, u = v₂ –[(v₂.u₁)/(u₁.u₁)]u₁ = v₂ –[(2+9+0+0)/(1+9+1+0+0)]u₁ = v₂ –u₁ =(-2,-3,0,1,0)^T-(-1,-3,1,0,0)^T = (-1,0,-1,1,0)^T and u₃ = v₃ - [(v₃.u₁)/(u₁.u₁)]u_1-[(v₃.u₂)/(u₂.u₂)]u₂ = v₃ - [(3+6+0+0+0)/ (1+9+1+0+0)]u₁ –[( 3+0+0+0+0)/(1+0+1+1+0)]u₂ = v₃ -9u₁/10u₁-u₂ = (-3,-2,0,0,1)^T+(9/10,27/10,-9/10,0,0)+(1,0,1,-1,0)^T = (-11/10, 7/10,1/10,-1,1)^T.

Then {u₁,u₂,u₃} = {(-1,-3,1,0,0)^T , (-1,0,-1,1,0)^T , (-11/10, 7/10,1/10,-1,1)^T} is the required orthogonal basis .