Equivilant norms of a Vector Space

In summary, we are trying to show that two norms are equivalent on a finite dimensional vector space X if and only if there exist positive real constants c_1 and c_2 such that for all x in X, the norm of x is less than or equal to c_1 times the norm of x under the first norm, and the norm of x under the first norm is less than or equal to c_2 times the norm of x under the second norm. This means that a sequence in X is cauchy under one norm if and only if it is cauchy under the other norm. The <= inequality is easy to show, but the other direction is more challenging. To prove this, we consider the unit ball in
  • #1
snoble
127
0
This is killing me that I can't see this. Why is it two norms on a finite dimensional Vector space X are equivilant if and only if there exist positive real constants [tex]c_1,\, c_2[/tex] such that [tex]\forall x\in X[/tex], [tex]\|x\|_2 \le c_1 \|x\|_1[/tex] and [tex]\|x\|_1 \le c_2 \|x\|_2[/tex].

Here equivilant means that a sequence in the VS is cauchy under one norm iff it is cauchy under the other norm.

Alright, if you assume the inequality and assume a sequence is Cauchy under one norm then it is easy to show it is Cauchy under the other norm. Just scale the [tex]\epsilon[/tex] appropriately.

Now the other direction is what has me stumped. So assuming that the v.s. norms are equivilant you can get that the norms on the underlying fields are equivilant; just take a sequence in the field that is cauchy under the first norm, multiply the sequence by any non-zero vector - that'll be cauchy in the V.S. under the first norm so cauchy under the second. Then just divide by the value of the vector under the second norm. Then it is clear that the original sequence is Cauchy in the field under the second norm.

I realize that the above is kind of fuzzy but the point is the norms on the underlying field are equivilant. And I know that if two field norms are equivilant then there exists a positive real [tex]\alpha[/tex] such that [tex]\forall k\in F[/tex], [tex]\|k\|_1^\alpha = \|k\|_2 [/tex].

But now I can't seem to get the inequality. I assume it's easiest by trying to show the converse ie if the inequality hold for no constant then you can get a sequence that is Cauchy under one norm but not under the other. But I still can't seem to see it.

Any hope would be appreciated

Thanks,
Steven
 
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  • #2
So to clarify, you're asking to show

two norms have the same cauchy sequences if and only if there are constants k and m such that |x|<k|x|' and |y|'<m|y| for all x and y. i'll use | | and | |' for the two norms.

I'm only asking because to me the second of these two (the one with the constants) is how I've usually seen the norms defined to equivalent.

as you say, the <= is trivial.

Now let's try the other way.

Let's just consider the unit ball in | |, ie the x such that |x| <=1.

If we can show this is bounded in | |' we're done, since it sufices to consider only vectors of length 1 by linearity.

If it's not bounded, then there are points x(n) such that |x(n)|' > n for each n, but the x(n) are a sequence in a closed totally bounded subset of a metric space, ie it is compact. (nb, I'm actually just assuming this, as it's the only point at which we need finite dimensionality, so it should be checked: the unit sphere in a finite dimensional normed vector space is compact)

Compact implies sequentially compact in a metric space so it must have a convergent subsequence that is cauchy, relabelling as necessary we may assume x(n) is cauchy in | |, but it must be cauchy in | |', which is a contradiction given the way they were constructed.

Note, all norms on a finite dimensional metric space are equivalent.
 
  • #3


First of all, don't worry, it's completely normal to get stuck on a math problem and have it "kill" you for a while. It's all part of the learning process and it's important to keep pushing through and seeking help when needed.

To understand why the two norms on a finite dimensional Vector space X are equivalent if and only if there exist positive real constants c_1 and c_2 satisfying the given inequality, let's break it down step by step.

First, let's define what it means for two norms to be equivalent. Two norms are equivalent if there exists a constant c such that for all x in X, \|x\|_1 \le c\|x\|_2 and \|x\|_2 \le c\|x\|_1. Essentially, this means that the two norms are "comparable" in the sense that one is always bounded above and below by a multiple of the other.

Now, let's consider the Cauchy criterion for a sequence in a vector space. A sequence (x_n) in a vector space X is Cauchy if for any \epsilon > 0, there exists an N such that for all m,n > N, \|x_m - x_n\| < \epsilon. In other words, the terms of the sequence are getting closer and closer together as n increases.

Now, let's assume that the two norms on X are equivalent, i.e. there exists a constant c such that for all x in X, \|x\|_1 \le c\|x\|_2 and \|x\|_2 \le c\|x\|_1. This means that the two norms are "comparable" in the sense that one is always bounded above and below by a multiple of the other.

Now, let's consider a sequence (x_n) in X that is Cauchy under the norm \| \cdot \|_1. By the Cauchy criterion, for any \epsilon > 0, there exists an N such that for all m,n > N, \|x_m - x_n\|_1 < \epsilon. But since the norms are equivalent, we also have \|x_m - x_n\|_2 \le c\|x_m - x_n\|_1 < c\epsilon. This means that the sequence (x_n) is also Cauchy under the
 

Related to Equivilant norms of a Vector Space

1. What are equivalent norms in a vector space?

Equivalent norms in a vector space are different ways of measuring the size or magnitude of vectors in that space. These norms may differ in their definitions, but they ultimately result in the same notion of distance or size.

2. How do I know if two norms are equivalent?

Two norms are equivalent if they satisfy the following conditions: 1) they both give a positive value for every vector except the zero vector, 2) they are both homogeneous (multiplying a vector by a constant multiplies the norm by the same constant), and 3) they are both subadditive (the norm of the sum of two vectors is less than or equal to the sum of the norms of each vector).

3. Why is it important to study equivalent norms in a vector space?

Studying equivalent norms in a vector space allows us to understand different ways of measuring the size or magnitude of vectors. This can help us compare and contrast different norms, and choose the most appropriate one for a given problem. It also allows us to prove certain properties that hold for one norm also hold for another equivalent norm.

4. Can two norms be equivalent in one vector space but not in another?

Yes, two norms can be equivalent in one vector space but not in another. Equivalent norms are dependent on the vector space and its properties, so they may differ in different vector spaces. For example, the Euclidean norm and the maximum norm may be equivalent in Rn, but not in other vector spaces.

5. How do equivalent norms affect the geometry of a vector space?

Equivalent norms do not change the geometry of a vector space, as they ultimately result in the same notion of distance or size. However, different norms may give different shapes to the unit sphere in that space, which can affect the shape of linear subspaces and the angles between vectors.

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