Change of basis: T(x,y,z)=(-x-y-z,x+y-5z,-3x-3y+3z)

[jacko]

Hey i was just doping someone wouldnt mind looking over my working to see if im 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!

 

[tkhunny]

*T(x,y,z)=(-x-y-z,x+y-5z,-3x-3y+3z) is a linear transformation.

T(0,0,0)=(-0-0-0,0+0-0,-0-0+0)= (0,0,0)

T(x+a,y+b,z+c)=(-(x+a)-(y+b)-(z+c),(x+a)+(y+b)-5(z+c),-3(x+a),-3(y+b)+3(z+c)) = (-x-y-z,x+y-5z,-3x-3y+3z) + (-a-b-c,a+b-5c,-3a-3b+3c)

T(tx,ty,tz)=(-tx-ty-tz,tx+ty-5tz,-3tx-3ty+3tz) = t(-x-y-z,x+y-5z,-3x-3y+3z)