Dick said:
Okay, we're moving somewhere.

For convenience I will copy paste the second part of the proof and
continue from there.
T(v1)=c1v1+0+c3v3
Now mapping from {v1, v2} to {v1,v2}
T(v1)=c1v1+c2v2+c3v3
we found that c2v2=0 since c2=0
and because c3v3 is not within the span {v1,v2}
as it is not a linear combination of its vectors
unless c3=0, c3 must be zero.
Now, we found that c2 and c3 are both zero.
So T(v1)=c1v1+0v2+0v3=c1v1
Now we take T(v2)=c1v1+c2v2+c3v3
map from {v2, v3} to {v2, v3}
c2v2+c3v3 is a linear combination in {v2,v3}
Since we have found that T(v1)=c1v1
and c1v1 is not within {v2,v3}, c1=0 for the
same reasons as c2 and c3 are.
Thus T(v2)=0*v1+c2v2+0*v1
T(v2)=c2v2
Since c2=0, T(v2)=0*v2=0
T(v2)=0
Now that we found that c1=0,
T(v1)=c1*v1
T(v1)=0*v1
T(v1)=0
Now mapping from {1,3} to {1,3}
T(v3)=c1v1+c2v2+c3v3
Since we have shown that c1 and c2 are 0,
T(v3)=0*v1+0*v2+c3v3
T(v3)=c3v3
Since we have shown that c3=0
T(v3)=0*v3
T(v3)=0
So T(v1)=c1v1+c2v2+c3v3
T(v1)=T(v1)+T(v2)+T(v3)
Since T(v1),T(v2),T(v3) all =0
T(v1)=T(v1+v2+v3)
T(v1)=0*(v1+v2+v3)
can be shown for the others