# Umm, cauchy-schwarz

1. Sep 6, 2005

### misogynisticfeminist

I'm self taught though, so please bear with any questions i have. One side of the cauchy schwarz innequality (CSI, nice acronym) is

l u.v l

Firstly what's the difference between ll u ll and l u l . I thought the norm was the length.

Also, what does it mean by the length of the dot product of u and v? I thought the dot product was a number itself, and not a tuple or something.

Last edited: Sep 6, 2005
2. Sep 6, 2005

### quasar987

llull = norm/lenght of the vector u.

|u| = ABSOLUTE VALUE of the scalar u.

3. Sep 7, 2005

### TD

$$\left| {\vec x \cdot \vec y} \right| \leqslant \left\| {\vec x} \right\|\left\| {\vec y} \right\|$$

As quasar987 said, " | | " is for the absolute value. So as you said correctly, the inner product is a scalar so the inequality states that the absolute value of the inner product of two vectors is always less than (or equal too) the product of the norms of both vectors.

4. Sep 7, 2005

### matt grime

However, many people use |x| and ||x|| interchangably since there is no real difference between them; a 1-d vector is a scalar so the scalar norm and the vector norm are the same here. And most (pure) mathematicians do not distinguish symbolically between vectors and scalars; they let the context make it clear which is meant.

5. Sep 7, 2005

### misogynisticfeminist

ahh, that cleared the doubts. Thanks alot.

6. Sep 12, 2005

### Dr Avalanchez

I do not agree on that. ||.|| is a norm and | . | is the absolute value. norm() applies to vectors, abs() applies to scalars. abs() is a well defined function for real numbers (and the complex analogon), whereas norm() is not. A function is a norm in some sort of vector space iff it satisfies the parallellogram identity. The well-known frobeniusnorm is just an example, just as the 1-norm, 2-norm, maxnorm, sylvesternorm,... In fact, just like inner products on some vector space, you can "invent" a norm (as long as it satisfies the parallellogram identity).

However, if explicitly stated that the vector space is euclidean/unitarian, with standard norm, then I agree with you. But not that mathematicians do not distinguish any difference between them.

7. Sep 12, 2005

### matt grime

But we are talking about vector spaces and the euclidean norm.

In other cases this won't be true: say in the theory of elliptic curves the synmbol | | will often be taken to be the p-adic valuation.

8. Sep 12, 2005

### Hurkyl

Staff Emeritus
norm() it is too defined for the real numbers... Or are you asserting that your textbooks define the Euclidean norm specifically to exclude vector spaces of dimension 1 (and of dimension 0)?

9. Sep 14, 2005

### Dr Avalanchez

No, I was just pointing out that you can have all sorts of norms, so it should be stated what vector space we're talking about, and which norm. It was a critique on Matt that mathematicians do make a distinction (unless it's clear what we're talking about, like here).