- #1
TrapMuzik
- 6
- 0
Hey all,
So I'm just starting a course in linear algebra, but I don't have much experience with proofs. This problem has been giving me some difficulty.
So we have a scalar "a" and vector x. V is a linear space, and x is contained in V. I have to show that if ax=0, where 0 is the zero element of V, then either a=0 (scalar) or x=0 (vector).
However, I must do the proof using the following axioms. I will briefly summarize them.
1. for every v1 and v2 there is a unique element in V equal to the sum of v1 and v2
2. same deal as above but with multiplication of a scalar and a vector
3. v1+v2=v2+v1
4. (v1+v2)+v3= v1 + (v2+v3)
5. existence of the element 0 (v+0=v)
6. v + (-1)v = 0
7. for scalars a and b, a(bv)=(ab)v
8. a(v1+v2)=av1 + av2
9. same deal as 8 but with one vector being multiplied by (a+b)
10. 1v=v
I was messing around with dividing things, but I don't think I can relate it to the axioms
So I'm just starting a course in linear algebra, but I don't have much experience with proofs. This problem has been giving me some difficulty.
So we have a scalar "a" and vector x. V is a linear space, and x is contained in V. I have to show that if ax=0, where 0 is the zero element of V, then either a=0 (scalar) or x=0 (vector).
However, I must do the proof using the following axioms. I will briefly summarize them.
1. for every v1 and v2 there is a unique element in V equal to the sum of v1 and v2
2. same deal as above but with multiplication of a scalar and a vector
3. v1+v2=v2+v1
4. (v1+v2)+v3= v1 + (v2+v3)
5. existence of the element 0 (v+0=v)
6. v + (-1)v = 0
7. for scalars a and b, a(bv)=(ab)v
8. a(v1+v2)=av1 + av2
9. same deal as 8 but with one vector being multiplied by (a+b)
10. 1v=v
I was messing around with dividing things, but I don't think I can relate it to the axioms