a) You can foil out the LHS to show that they are NOT equal. The function is NOT a homomorphism.
b) Yes, now you need to show that .
c) Correct.
My problem is, that my mathematical understanding is full of gaps and I am stuck at some steps or I dont know if I am finished or not.
Sorry for the text.
a)f:ℝ³-->ℝ² f((x,y,z)) = (xy,x+y)
f(0)ofℝ³ = f(0,0,0) = (0*0, 0+0) = (0,0) = ℝ²
Showing homomorphism
v1=(x,y,z), v2=(x',y',z') ∈ ℝ³
Left side:
f((x,y,z)+(x',y',z')) = f(x+x', y+y', z+z') = ((x+x')*(y+y'), (x+x')+(y+y'))
Right side:
f(x,y,z)+f(x',y',z') = (xy, x+y) + (x'y', x'+y') = ((xy+x'y'), (x+y+x'+y'))
Am I done here or can I transform the term to make both parts equal?
b)f:ℝ³-->ℝ³ f((x,y,z)) = (0, -x+y, 3x-5y)
f(0)ofℝ³ = f(0,0,0) = (0, 0+0, 3*0-3*0) = (0,0,0) = ℝ³
Showing homomorphism
v1=(x,y,z), v2=(x',y',z') ∈ ℝ³
Left side:
f((x,y,z)+(x',y',z')) = f(x+x', y+y', z+z') = (0 , ((-x+(-x'))+(y+y')) , (3*(x+x')-5*(y+y')))
Right side:
f(x,y,z)+f(x',y',z') = (0 , -x+y , 3x-5y) + (0 , -x'+y' , 3x'-5y') = (0 , ((-x+(-x'))+(y+y')) , (3*(x+x')-5*(y+y')))
Its the same as far as I see it. It this correct and can I continue the proof now?
c)f:ℝ³-->ℝ³ f((x,y,z)) = (0,0,1)
f(0)ofℝ³ = f(0,0,0) ≠ (0,0,1)
- Not a linear function between ℝ³ and ℝ³
for b) Is this correct? I think I messed up somewhere in the middle:
f(c*(x,y,z))
= f(cx,cy,cz)
= (0, cx, cy)
= (0 , (c(-x+y)) , (c(3x-5y))
= c*(0 , (-x+y) , (3x-5y)
= cf(x,y,z)