Find an orthonormal basis for the kernel of the following ma

Find an orthonormal basis for the kernel of the following matrix. [2 3 1 2 0 -1 -1 1].

Solution

Let the given matrix be denoted by A. Then Ker(A) is the set of solutions to the equation AX = 0. To solve this equation, we will reduce A to its RREF which is

1

0

1

-1

0

1

-2

1

If X = (x,y,z,w)^T, then the equation AX = 0 is equivalent to x+z-w = 0 or, x = -z+w and y-2z+w = 0 or, y = 2z-w. Hence X=(-z+w, 2z-w,z,w)^T=z(-1,2,1,0)^T+w(1,-1,0,1)^T. Then a basis for Ker(A) is {(-1,2,1,0)^T,(1,-1,0,1)^T }= {v₁,v₂} (say).

Now, let u₁ = v₁ =(-1,2,1,0)^T and u₂=v₂–proj_u1(v₂) = v₂ –[(v₂.u₁)/(u₁.u₁)]u₁ = v₂ –[(-1-2+0+0)/(1+4+1+0)]u₁ = (1,-1,0,1)^T +1/2(-1,2,1,0)^T = ( ½,0, ½,1)^T. Then {u₁,u₂} is an orthogonal basis for R³. We will now convert u₁,u₂ into unit vectors.

Let e₁ = u₁/||u₁||= (-1, 2,1,0)^T/(6) = (-1/6.2/6,1/6,0)^T and e₂ = u₂/||u₂||= (1/2,0,1/2, 1)/(3/2) = (1/6,0, 1/6,2/3) Then { e₁,e₂} is the required orthonormal basis for Ker(A).

1	0	1	-1
0	1	-2	1