Question

proof for if s belongs to A(v) and if vss*=0 then vs=0​

Answer 1

Answer:

I'm kicking myself over this one, but I just can't seem to make the argument rigorous. From Axler's Linear Algebra Done Right:

for a vector space V with an underlying field F:

Take an element a from F and v⃗ from V. av⃗ =0⃗ ⟹a=0 or v⃗ =0⃗

After only being able to come up with half of a direct proof, I tried doing this by proving the contrapositive a≠0∧v⃗ ≠0⃗ ⟹av⃗ ≠0⃗

Say av⃗ =u⃗ . Since a is non-zero, we can divide both sides by a.

v⃗ =1au⃗

If u⃗ were 0⃗ then by

1a0⃗ =1a(0⃗ +0⃗ )⟹1a0⃗ =0⃗

v would be 0 as well. Since it isn't by assumption, 1au⃗ cannot be zero and so u⃗ cannot be as well.

Is this fully rigorous? It seems like a very simple question, but I'm not sure about it. Namely, the last step of 1au⃗ ≠0⟹u⃗ ≠0 doesn't seem obvious. I think I need to use the 1v⃗ =v⃗ axiom, but I'm not sure how.

Is there a more direct proof? This whole contrapositive business seems a bit clunky for something so simple.