- #1
andrey21
- 476
- 0
Identify the Hermite form of the following linear transformations and the basis for its kernel
(x,y,z) = (x-y+2z,2x+y-z,-3x-6y+9z)
So when finding basis for kernel we have to set equal to 0, giving:
x-y+2z=0 (1)
2x+y-z=0 (2)
-3x-6y+9z=0 (3)
So from (1) we can see:
x=y-2z
Therefore (2) becomes:
2y-4z+y-z = 0
3y=5z
So (3) becomes:
-5z+6z -10z +9z=0
15z = 15z
z=z
This is as far as I can get, not sure of the next step, any help be fantastic:
(x,y,z) = (x-y+2z,2x+y-z,-3x-6y+9z)
So when finding basis for kernel we have to set equal to 0, giving:
x-y+2z=0 (1)
2x+y-z=0 (2)
-3x-6y+9z=0 (3)
So from (1) we can see:
x=y-2z
Therefore (2) becomes:
2y-4z+y-z = 0
3y=5z
So (3) becomes:
-5z+6z -10z +9z=0
15z = 15z
z=z
This is as far as I can get, not sure of the next step, any help be fantastic: