0
$\begingroup$

I'm not sure if I understand what it means that in finite dimensional vector spaces all norms are equivalent. Does this mean, no matter what we do, we can just pick any norm without destroying our result? As long as we pick the same norm on all our results?

So for instance if we do the Gram Schmidt method we could choose the maximums norm and not the Euclidean norm?

Definition: Two norm $ \|\cdot\|_1$ and $ \| \cdot\|_2 $ are called equivalent, if there exist positive numbers $c,C $ such that $c \cdot \| x\|_2 \leq \| x\|_1 \leq C \cdot \| x\|_2 \text{ for all } x \in X $

  • 0
    It also implies that when something converges in one norm (space) it also converges in the other (space). – 2017-02-14
  • 0
    @JaneMaths Thanks, but would I lose some properties like for instance orthogonality of two vectors? (When I normalize them) – 2017-02-14
  • 0
    Othogonality is related to an inner product, not to a norm. An inner product defines a norm, but not every norm comes from an inner product. – 2017-02-14
  • 0
    @JuliánAguirre Thank you – 2017-02-14

1 Answers 1

0

It is mostly about the existence of constants $m$, $M$ such that $$ m \lVert . \rVert_1 \le \lVert . \rVert_2 \le M \lVert . \rVert_1 $$ So your Gram Schmidt would end up with different vectors for different norms, but statements about convergence etc, where the above inequalites play a role, should give the same result for different norms.

  • 0
    What do you mean with I would end up with different vectors? Would I get different vectors who are still orthogonal or would I get vectors which would lose their orthogonality? (In that case it would be useless) – 2017-02-14
  • 0
    Gram Schmidt is implicitly about a 2 norm, which comes from an inner product. Different definitions of inner product on the same space (e.g. using different weightings) would indeed give different sets of orthogonal vectors (with respect to that inner product). You would still have equivalence in terms of continuity though. – 2017-02-14
  • 0
    @Paul Thank you. So when I used a given inner product for Gram Schmidt then I also use the same inner product $\sqrt{\langle a ,a \rangle }$ to normalize? – 2017-02-14
  • 0
    @WaldoRozir Yes, that is right. – 2017-02-14