1 point 2 3 13 13 Let A 2 1 3 11 Find orthogonal bases of th

(1 point) -2 -3 -13 -13 Let A 2 1 3 11 Find orthogonal bases of the kernel and image of A. Basis of the kernel: 1 -5/sort(27) 1/sqrt(27) 1/sqrt(27) -5/sqrt (27) Basis of the image: 1/sqrt(27) 1/sqrt(27)

Solution

The Image of A is same as its column space i.e. Col(A). To determine Ker(A) and Col(A), we will reduce A to its RREF as under:

Multiply the 1st row by -1

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

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

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

Multiply the 2nd row by -1/3

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

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

Then the RREF of A is

1

0

-1

5

0

1

5

1

0

0

0

0

0

0

0

0

Now, apparently, the 3^rd and the 4^th columns of A are linear combinations of its first two columns and only its first two columns are linearly independent. Hence a basis for Col(A) is B₁={ (-1,-2,-1,2)^T,(0,-3,-3,1)^T} or B₂= {( 1,0,0,0)^T, (0,1,0,0)^T} . Also, B₂ is an orthogonal basis for Im(A) i.e. Col(A). To orthogonalise the basis B₁, we will use the Gram-Schmidt process. Let u₁=v₁= (-1,-2,-1,2)^T and v₂ = (0,-3,-3,1)^T. Also, let u₂= v₂-proj_u1 (v₂)= v₂-[(v₂.u₁)/(u₁.u₁)]u₁ = v₂-[(0+6+3+2)/(1+4+1+4)]u₁= (0,-3,-3,1)^T –(11/10) (-1,-2,-1,2)^T= ( 11/10, -8/10,-19/10, -12/10)^T. Then {u₁,u₂} = {(-1,-2,-1,2)^T, ( 11/10, -8/10,-19/10, -12/10)^T } is an orthogonal basis for Im(A) or Col(A).To determine an orthonormal basis, we need to convert these vectors into unit vectors.

Let e₁ = u₁/||u₁||= u₁/(1+4+1+4) =1/ 10(-1,-2,-1,2)^T = (-1/10, -2/10,-1/10,2/10)^T and,

e₂ = u₂/||u₂||= u₂/( 121/100+64/100+361/100+144/100)= u₂/(365/100) = 10u₂/365 = 10/365( 11/10, -8/10,-19/10, -12/10)^T = (11/365,-8/365,-19/365,-12/365)^T. Then {e₁,e₂} is an orthonormal basis for Im(A)or Col(A).

Further, Ker(A) is the set of solutions to the equation AX = 0. If X = (x,y,z,w)^T, then this equation is equivalent to x-z+5w = 0or, x = z-5w, and y+5z+w = 0 or, y = -5z-w. Then X = (z-5w, -5z-w,z,w)^T = z(1,-5,1,0)^T+w( -5,-1,0,1)^T. Thus, a basis for Ker(A) is {(1,-5,1,0)^T,( -5,-1,0,1)^T }. Since (1,-5,1,0)^T.( -5,-1,0,1)^T =-5+5+0+0= 0, therefore, {v₁,v₂}= {(1,-5,1,0)^T,( -5,-1,0,1)^T } is an orthogonal basis for Ker(A). To determine an orthonormal basis, we need to convert these vectors into unit vectors.

Let p₁= v₁/||v₁||= v₁/(1+25+1+0) = v₁/27= (1/33)(1,-5,1,0)^T= (1/33,-5/33,1/33,0)^T and

p₂= v₂/||v₂||= v₂/(25+1+0+1) = v₂/27 = (1/33)(-5,-1,0,1)^T = (-5/33,-1/33,0, 1/33)^T. Then {p₁,p₂} is an orthonormal basis for Ker(A).

From the RREF of A, it is apparent that a basis for Row(A) is{w₁,w₂} = {(1,0,-1,5),(0,1,5,1)}. Since (1,0,-1,5).(0,1,5,1)= 0, this is an orthogonal basis for . To create an orthonormal basis for Row(A), we will convert w₁,w₂ into unit vectors as under:

let q₁= w₁/||w₁||= w₁/(1+0+1+25) = w₁/27= (1/33)(1,0,-1,5)= (1/33,0,-1/33,5/33) and

q₂= w₂/||w₂||= w₂/(0+1+25+1) = w₂/27= (1/33)(0,1,5,1)= (0, 1/33, 5/33,1/33)

Then {q₁,q₂} is an orthonormal basis for Row(A).

1	0	-1	5
0	1	5	1
0	0	0	0
0	0	0	0