$\displaystyle A$ = $\displaystyle \left(\begin{array}{ccc}1&{-a}&{a}\\{-1}&a&{a+2}\\1&{2a+3}&{-3a-4}\end{array}\right)$

$\displaystyle a \in R$

$\displaystyle A$ describes the linear map $\displaystyle T: R^3 \rightarrow R^3$ for the basis B = {(1,0,-1), (1,-1,0), (1,1,1)}

dimKerT = 2

Find $\displaystyle a$ and T(x,y,z) for every (x,y,z) in $\displaystyle R^3$

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 $\displaystyle a$ 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!!!