Let and .
As you noted, and so:
Q: Prove that if and only if and have the same direction.
A: Well, if the same direction of , then must be a multiple of . Thus, where (I am not sure if I lost any generality by choosing ). From here I expanded and ended up with some messy algebra and not the result I was looking for.
In the other direction I started with and worked my way down to and figured I have to introduce the Cauchy-Schawrz Ineguality here, but I am not sure how.
I am having a hard time formalizing this proof. All I have is a bunch of dead und scratch work. It seems that this proof should flow with ease, because, geometrically, it is pretty obviouse.
I would appreciate some guidence.
Thank you.
Hello,
Are you sure you can take wlog this norm ? Though all norms are equivalent in finite dimensions, we're talking about a norm in general, here.
What you didn't say, and that is helpful for the solution, is that k has to be positive.
Otherwise,
They wouldn't have the same direction as they would go in opposit directions.
It's okay, there is no loss of generality. You choose u because it's the assumption of the implication.A: Well, if the same direction of , then must be a multiple of . Thus, where (I am not sure if I lost any generality by choosing ).
Actually, you just have to use the axioms/definition of a norm.From here I expanded and ended up with some messy algebra and not the result I was looking for.
There is the first axiom (positive homogeneity) : , where a is a real number.
So in this case (I'll forget the \vec...too long ^^),
A scalar product gives a norm.In the other direction I started with and worked my way down to and figured I have to introduce the Cauchy-Schawrz Ineguality here, but I am not sure how.
I am having a hard time formalizing this proof. All I have is a bunch of dead und scratch work. It seems that this proof should flow with ease, because, geometrically, it is pretty obviouse.
I would appreciate some guidence.
Thank you.
So let's define <,> as the scalar product associated with this norm.
Now, we assume that and we want to prove that there exists such that
by definition.
Now, we know that
So
And here, apply Cauchy-Schwarz inequality, stating that there is equality if and only if u and v are linearly dependent.
That is to say u=kv, but where k is not necessarily positive.
However, the equivalence only stands for k positive.