Question

For problems 1-4 the linear transformation T/; R^n - R^m  is defined by T(v)=AV with

A=[-1 -2 -1 -1

2 1 0 -4

4 6 1 15]

1.   Find the dimensions of R^n and R^m (3 pts)

2.   Find T(<-2, 1, 4, -1>) (5 pts)

3.   Find the preimage of <-6, 12, 4> (8 pts)

4.   Find the Ker(T) (8 pts)

Answer 1

T: Rⁿ →R^m is defined by T(V) = Av, where A =

-1	-2	-1	-1
2	1	0	-4
4	6	1	15

1. Since A is a 3x4 matrix, hence it can be multiplied to the right only by a 4-vector and the result will be a 3-vector. Hence n= 4 and m = 3.

2. Let v = (-2,1,4,-1)^T. Then T(v) = Av = (-3,1,-13)^T.

3. Let the pre-image of (-6,12,4)^T be v = (a,b,c,d)^T. Then T(v) =Av = (-6,12,4)^T or, (-a-2b-c-d, 2a+b-4d, 4a+6b+c+15d)^T = (-6,12,4)^T . Hence, -a,-2b-c-d = -6 or, a+2b+c+d = 6…(1), 2a+b-4d = 12…(2) and 4a+6b+c+15d = 4…(3).

The augmented matrix of the above linear system is M =

1	2	1	1	6
2	1	0	-4	12
4	6	1	15	4

The RREF of M is

1	0	0	-6	10
0	1	0	8	-8
0	0	1	-9	12

Hence a-6d = 10 or, a = 10+6d, b+8d = -8 or, b = -8-8d and c-9d = 12 or, c =12+9d. On assuming d = 1, we have a = 16, b = -16, c = 21.

Thus, ( 16,-16,21,1)^T is a pre-image of (-6,12,4)^T.

4. Ker(T) is the set of solutions to the equation T(X) = AX = 0. To solve this equation, we have to reduce A to its RRREF which is

1	0	0	-6
0	1	0	8
0	0	1	-9

Now, if X = (x,y,z,w)^T, then the equation T(X) = AX = 0 is equivalent to x-6w = 0 or, x = 6w, y+8w = 0 or, y = -8w and z-9w = 0 or, z = 9w. Therefore, X = (6w,-8w,9w,w)^T = w( 6,-8,9,1)^T. Hence, every solution to T(X0 = 0 is a scalar multiple of the vector ( 6,-8,9,1)^T so that ker(T) = span{( 6,-8,9,1)^T }.