- #1
dumb_curiosity
- 14
- 0
Sorry, I wasn't sure of the best way to phrase this. This is a common problem I keep having.
Here's the definition of a norm:
Let E be a vector space V defined over a field F. A norm on V is a function p: [itex]V \rightarrow \mathbb{R}[/itex] such that:
[itex]\forall a \in F[/itex] and [itex]\forall u,b \in V[/itex]:
Select some vector v [itex]\in V[/itex]. Then:
p(v + (-v)) [itex]\leq[/itex] p(v) + p(-v) (by ii)
p(v + (-v)) [itex]\leq[/itex] p(v) + |-1|p(v) = 2p(v) (by i)
p(0) [itex]\leq[/itex] 2p(v)
0 [itex]\leq[/itex] 2p(v) (by iii)
0 [itex]\leq[/itex] p(v)=======
Here's the part I get confused on... How do I know I can actually do all of these steps? For example - I do know that for some arbitrary vector v [itex]\in V[/itex], I can get a -v to use in this little proof. I know I can do this because V is a vector space, and the vectors in a vector space make an abelian group which means that every vector has an inverse. But for example - how do I "know" that I can do division by 2 in that last step? (Going from 0 [itex]\leq[/itex] 2p(v) to 0 [itex]\leq[/itex] p(v)). I think the reason I can do it in this case is because the scalar field F that V is defined over is a division ring (by definition of a vector space)... but I had to go back and look that up before I would allow myself to do this. In fact, is this the reason why I can do it, or is it some other reason? After learning about abstract algebra, I'm just starting to be weary of making any move because I feel like I'm assuming too much. For example - even just doing the step p(v) + |-1|p(v) = 2p(v), I feel uneasy after learning about abstract algebra, because I feel I'm thinking of 1 as I would in the real numbers... or the 0/2 = 0 step... I feel like every move I take, I have to question if I'm doing something because I'm thinking only in terms of number systems I'm familiar with (like the real numbers), and I worry maybe I'm assuming too much.
Here's the definition of a norm:
Let E be a vector space V defined over a field F. A norm on V is a function p: [itex]V \rightarrow \mathbb{R}[/itex] such that:
[itex]\forall a \in F[/itex] and [itex]\forall u,b \in V[/itex]:
(i) p(av) = |a|p(v)
(ii) p(u + v) [itex]\leq[/itex] p(u) + p(v)
(iii) p(v) = 0 [itex] \iff [/itex] v = 0
Now, an obvious property from these axioms is:(ii) p(u + v) [itex]\leq[/itex] p(u) + p(v)
(iii) p(v) = 0 [itex] \iff [/itex] v = 0
(iv) p(v) [itex] \geq [/itex] 0 ([itex]\forall v \in V[/itex])
We can see this as follows:
Select some vector v [itex]\in V[/itex]. Then:
p(v + (-v)) [itex]\leq[/itex] p(v) + p(-v) (by ii)
p(v + (-v)) [itex]\leq[/itex] p(v) + |-1|p(v) = 2p(v) (by i)
p(0) [itex]\leq[/itex] 2p(v)
0 [itex]\leq[/itex] 2p(v) (by iii)
0 [itex]\leq[/itex] p(v)=======
Here's the part I get confused on... How do I know I can actually do all of these steps? For example - I do know that for some arbitrary vector v [itex]\in V[/itex], I can get a -v to use in this little proof. I know I can do this because V is a vector space, and the vectors in a vector space make an abelian group which means that every vector has an inverse. But for example - how do I "know" that I can do division by 2 in that last step? (Going from 0 [itex]\leq[/itex] 2p(v) to 0 [itex]\leq[/itex] p(v)). I think the reason I can do it in this case is because the scalar field F that V is defined over is a division ring (by definition of a vector space)... but I had to go back and look that up before I would allow myself to do this. In fact, is this the reason why I can do it, or is it some other reason? After learning about abstract algebra, I'm just starting to be weary of making any move because I feel like I'm assuming too much. For example - even just doing the step p(v) + |-1|p(v) = 2p(v), I feel uneasy after learning about abstract algebra, because I feel I'm thinking of 1 as I would in the real numbers... or the 0/2 = 0 step... I feel like every move I take, I have to question if I'm doing something because I'm thinking only in terms of number systems I'm familiar with (like the real numbers), and I worry maybe I'm assuming too much.
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