Does the set 1 x 3 x2 1 4x x2 span P2 If not then write

Does the set (1 + x, 3 - x^2, 1 + 4x + x^2) span P_2? If not, then write one vector as a linear combin other two vectors, then add one of the standard basis vectors of P_2 to the set to form a basis for P_2. Consider the transformation T: R^3 rightarrow R, T(x, y, z) = (x +2y + 3z, y + z, x 3y + 4z, x + z) (a) Find the image of v (2, 3, -5). (b) Find the preimage of w = (0, -1, -1, 2). (c) basis for the range of T (d) Find a basis for the kernel of T.

Solution

13. We have T(x,y,z) = (x+2y+3z, y+z,x+3y+4z,x+z).

(a). The image of (2,3,-5) under T is ( 2+6-15, 3-5,1+9-20,2-5), i.e. ( -7,-2,-10,-3).

(b).Let the pre-image of w = (0,-1,-1,2) be X =(x,y,z). Then, we have x+2y+3z = 0, y+z = -1,x+3y+4z =-1, and x+z=2. Let A be the augmented matrix of this linear system. Then A=

1

2

3

0

0

1

1

-1

1

3

4

-1

The RREF of A is

1

0

1

2

0

1

1

-1

0

0

0

0

Then the above linear system is equivalent to x+z = 2, or, x = 2-z, andy +z = -1 or y = -1-z. Then X =                  ( 2-z,-1-z,z) where z is arbitrary. Thus, w has several pre-images under T, which are ( 2-z,-1-z,z) where z is arbitrary.

(c ). The standard matrix A of T has T(e₁), T(e₂ ),andT(e₃ ) as columns, where {e₁,e₂,e₃} is the standard basis of R³. Here, T(e₁) = (1,0,1,1), T(e₂)= (2,1,3,0) and T(e₃)= ( 3,1,4,1). Then A =

1

2

3

0

1

1

1

3

4

1

0

1

The RREF of A is

1

0

1

0

1

1

0

0

0

0

0

0

Hence the Range of T , which is same as Col(A) is { (1,0,0,0)T,(0,1,0,0)T}.

(d). Ker(T) is same as Null(A). If X = (x,y,z)^T, then Null (A) is the set of solutions to the equation AX = 0 or, (x+z) = 0, i.e. x = -z, and y +z = 0, i.e. y = -z. Then X = (-z,-z,z)^T = z(-1,-1,1)^T. Hence, a basis for Ker (T) is             {(-1,-1,1)^T}.

1	2	3	0
0	1	1	-1
1	3	4	-1