Question

1. Let F: R4-R3 be a linear transformation satisfying F(1,1,1,1) (0, 1,2), F(1,1,0, 1)(0, 0,2) F(0,1,0, 0) 1,0,0) F(1,1,0,0) (0,0,0), (a) Calculate F(x, y, z, w) (b) Calculate ker(F) and R(F)

Answer 1

1. (a). Let v₁ = (1,1,1,1), v₂ = (1,1,0,1), v₃ = (1,1,0 0) and v₄ = (0,1,0,0). Also, let A =

1	1	1	0	x
1	1	1	1	y
1	0	0	0	z
1	1	0	0	w

The RREF of A is

1	0	0	0	z
0	1	0	0	w-z
0	0	1	0	x-w
0	0	0	1	y-x

This implies that (x,y,z,w) = zv₁+ (w-z)v₂ +(x-w)v₃+ (y-x)v₄.

Now, since F is a linear transformation, hence F(x,y,z,w) = F(zv₁+ (w-z)v₂ +(x-w)v₃+ (y-x)v₄) = zF(v₁)+ (w-z)F(v₂) +(x-w)F(v₃)+ (y-x)F(v₄) = z(0,1,2)+(w-z)(0,0,2)+(x-w)(0,0,0)+(y-x)(-1,0,0) = (x-y, z, 2w).

(b). We have F(e₁) = F(1,0,0,0) = (1,0,0), F(e₂) = F(0,1,0,0) = (-1,0,0), F(e₃) = F(0,0,1,0) = (0,1,0) and F(e₄) = F(,0,0,0,1) = (0,0,2).

Hence the standard matrix of F is M(say) = [F(e₁), F(e₂), F(e₃), F(e₄)] =

1	-1	0	0
0	0	1	0
0	0	0	2

The RREF of M is

1	-1	0	0
0	0	1	0
0	0	0	1

Now, Ker(F)= Ker(A) is the set of solutions to the equation AX = 0. If X = (p,q,r,s)^T, then this equation is equivalent to p-q = 0 or, p = q , r = 0 and s = 0. Then X = (q,q0,0)^T = q(1,1,0,0)^T. This implies that every solution to the equation AX= 0 is a scalar multiple of the vector (1,1,0,0)^T. Hence, ker(F) = span{(1,1,0,0)^T}.

The 1^st, 3^rd and 4^th columns of A are linearly independeny and the 2^nd column is a scalar multiple of its 1^st column.

If R(F) means the range of F = col(A), then , R(F) = span{(1,0,0), (0,1,0),(0,0,2)}.