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June 18th, 2018, 11:10 AM   #1
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Norms and the Cauchy Schwartz Inequality

I'm having a bit of an issue with the conceptual idea of Norms. I understand that it is used to define distances in $\displaystyle R^n$. So my questions are:
  1. Would it be fair to say that the p-norm of a matrix is the distance of that matrix translated into p dimensions?
    For example,$\displaystyle A_{5x5}$ is in $\displaystyle R^5$, so the $\displaystyle ||l||_2$ of A is the distance described in 2 dimensions. This is how I've been conceptualizing it, but I was told the norm has nothing to do with dimensions
  2. I'm having an argument with a friend.We were trying to show that the 1-norm satisfies the properties of a norm. The first few were straightforward, but our argument is with this guy:$\displaystyle ||X +Y|| \leq ||X||+||Y|| \forall x,y \in R $ . He says that the vectors x and y must be different dimensions, and therefore in the case of the 1-norm the Cauchy Schwartz Inequality will be irrelevant as one of those vectors must = 0, so the equation above is trivially true. But my understanding is that, x and y just have to be different vectors, not necessarily different dimensions. Obviously, either one or both of us do not fully
    understand norms. So I'm hoping for some clarification.

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June 18th, 2018, 02:24 PM   #2
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1. The p-norm has nothing to do with the dimension. Different choices of $p$ define different ways of measuring distance.

An analogy (which you should not read too much into beyond some intuition) is to consider two points which lie on a cylinder. One way to measure the distance between them is to simply connect them with a line and measure the Euclidean length of that line. However, notice that this path from one point to another does not remain on the surface of the cyclinder.

Alternatively, you could measure the distance by computing the shortest path between these points which remains on the cyclinder. Neither is correct or wrong, they are just different.

In this way, different norms for vectors give different answers which are neither wrong nor correct. Actually, it turns out that for finite dimensional vector spaces, all norms are equivalent.

2. Your friend is wrong. I'm not even sure what he/she is trying to say. If $X,Y$ are vectors of different dimension, then they live in different vector spaces. So what would $X+Y$ even mean? Cauchy-Schwarz is also not a good idea since it applies to inner product spaces only which is less general than a normed vector space. I think you should google the triangle inequality.
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June 18th, 2018, 11:40 PM   #3
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For the 1-norms, this is less relevant, but for p-norms with $1<p<+\infty$, there is the Holder inequality which is a generalization of Cauchy-Schwartz.
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June 20th, 2018, 02:42 PM   #4
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Thanks a lot, that helps to clarify things a bit. I think my issue might be that I need to take a pure linear algebra course, as opposed to courses that apply linear algebra...It might help with these conceptual issues.
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