Question

1. Let B { 1, х, хг} et S {x2 +x, 2-1, x+1 } be two basis of P2. Let T : P2 P2 be a linear transformation such that 3,S 2 2 -2 Find a basis of Ker(T), a basis of Im(T) and T^b 2. Let Let : P1 → P1 be a linear transformation such that 4 -3 where B-[1, x,} et S - {2c - 1,x - 1} be two basis of P1. Find A2 and T2. Can you explain the apparent difference in vour findings?

Answer 1

1. Here, in view of [T]_B,S, we have T(1) = 1(x²+x)-1(2x-1)+2(x+1)=x²+x -2x+1+2x+2=3+x+x² ; T(x)=2(x²+x) +3(2x-1)+0(x+1) = 2x²+2x+6x-3 = -3+8x+2x² and T(x²) = 0(x²+x)+5(2x-1)-2(x+1) = 10x-5-2x-2 = -7+8x. Therefore, standard basis of P₂ is [T]_B = A =

3	-3	-7
1	8	8
1	2	0

Since [T]_B = A is the standard matrix of T, therefore, Ker(T) = Ker(A) and Im(T) = Col(A). The RREF of A is

1	0	0
0	1	0
0	0	1

Hence a basis for Im(T) = Col(A) is {(3,1,1)^T,(-3,8,2)^T ,(-7,8,0)^T}. Further, Ker(A) is the set of solutions to the equation AX = 0 . If X = (x,y,z)^T, then this equation is equivalent to x = 0 y = 0 and z = 0.Then X = (0,0,0)^T Therefore, AX = 0 has only the trivial solution so that a basis for Ker(T) = Ker(A) has no vector as the zero vector cannnot be part of any basis.

Also, as above, [T]_B = A =

3	-3	-7
1	8	8
1	2	0

2 Please post the 2^nd question again separately.