a=0 or a=-1
=
describes the linear map for the basis B = {(1,0,-1), (1,-1,0), (1,1,1)}
dimKerT = 2
Find and T(x,y,z) for every (x,y,z) in
Attempt:
First off, I don't understand how the hint dimKerT = 2 helps us. I know it means that dimImT = 1 since dimImt + dimkerT = dimR^3...
* Anyway, I got (and i believe these calculations are correct...)
T(1,0,-1) = (1,2,0)
T(1,-1,0) = (2a+3,a+3,3a+3)
T(1,1,1) = (-a-2,-4a-6, -4a-4)
** since B is a basis of R^3, every vector (x,y,z) can be represented by B:
(x,y,z) = a(1,0,-1) + b(1,-1,0) +c(1,1,1) --- and after calculations I replace a,b,c and get:
(x,y,z,) = (-x-y+2z)(1,0,-1) + (x-z)(1,-1,0) + (x+y-z)(1,1,1)
so T(x,y,z) = (-x-y+2z)T(1,0,-1) + (x-z)T(1,-1,0) + (x+y-z)T(1,1,1)
T(x,y,z) = (-x-y+2z)(1,2,0) + (x-z)(2a+3,a+3,3a+3) + (x+y-z)(-a-2,-4a-6, -4a-4)
So as you can see, I probably have to find before I get up to step **. I am sure the dimkerT = 2 hint is useful, but I don't know why. Can someone please help?
Thanks!!!