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Martin Sleziak
Martin Sleziak
7:37 PM
in Mathematics, 13 mins ago, by Ilan Aizelman WS
Help:http://math.stackexchange.com/questions/816217/kernel-of-linear-transformat‌​ion-in-bbb-r3/816300#816300 According to Don's answer he told me that I need to find $dimKerT = 1$ , which is pretty easy because $dimImT = 2$ and $dimR^3 = 3$ so $dimKerT = 1$,which means there's one vector inside $KerT$, but how can I show that this vector equals to $Sp(1,-1,1)$?can I just assume that $(1,-1,1)$ is a basis in KerT because it's dimension equals to one, thus $Sp(1,-1,1)$ inside $KerT$ for sure?
You know the images of the vectors (1,1,0), (1,0,1) and (0,1,1).
Can you express (1,-1,1) as their linear combinations?
Since the vectors are very simple, it is even possible to guess the coefficients.
$$(1,-1,1)=-\frac12 (0,1,1)+\frac32(1,0,1)-\frac12(1,1,0)$$
So by linearity you have
$$T(1,-1,1)=-\frac12 T(0,1,1)+\frac32T(1,0,1)-\frac12T(1,1,0)$$
I get $T(1,-1,1)\ne0$. So that vector is not in the kernel. (Unless I have made a mistake somewhere.)
But now I see that in Anurag's answer you have already been told a vector, which belongs to the kernel.
BTW finding the kernel is nothing more than solving a system of linear equations. You can find several similar questions on MSE.
 

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