Finding [T(e2)]B for Linear Transformation

[jacko_20]

Hey i was just doping someone wouldn't mind looking over my working to see if I am on the right track!
*T(x,y,z)=(-x-y-z,x+y-5z,-3x-3y+3z) is a linear transformation.
S is the standard basis, S={e1,e2,e3} and B is another basis, B={v1,v2,v3} where:
e1=(1,0,0) e2=(0,1,0) e3=(0,0,1) v1=(1,1,1,) v2=(1,-1,0) v3=(0,1,-1)
- [T]S->S = [1 0 0
0 1 0
0 0 1]
-P B->S = [1 1 0
1 -1 1
1 0 -1]
-P S->B = [1/3 1/3 1/3
2/3 -1/3 -1/3
1/3 1/3 -2/3]

-[e2]B = P S->B.[e2]S
= (1/3,-1/3,1/3)
-[T(e2)]B =? what does this refer to? Do I have to refer to the equation in any part of these? as in the matrix [-1 -1 -1
1 1 -5
-3 -3 3]
Any help is greatly appreciated!

 

[EnumaElish]

What is the problem to be solved?

 

[jacko_20]

The question is: What is [T(e2)]B?
Thanks.

 

[HallsofIvy]

A good way of finding a matrix form for a linear transformation, in a given basis, is to apply the transformation to each of the basis vectors, in turn, and write the result in terms of the given basis. Each of those will be one column of the matrix.
For example, what do you get if you apply this transformation to v1= (1, 1, 1)? Now write that result as av1+ bv2+ cv3. The numbers a, b, c will be the first column of the matrix.

(This problem has obviously been set up to make it easy to do that. Applying the transformation to v2 is particularly interesting.)