# Cauchy-Schwarz Inequality

1. Aug 12, 2009

### EngWiPy

Hello,

For two n-dimensional vectors $$\mathbf{v}_1\text{ and }\mathbf{v}_2$$, what is the Cauchy-Schwarz Inequality:

1- $$|\mathbf{v}_1.\mathbf{v}_2|\leq\|\mathbf{v}_1\|\|\mathbf{v}_2\|$$, or

2- $$|\mathbf{v}_1.\mathbf{v}_2|\leq\|\mathbf{v}_1\|+\|\mathbf{v}_2\|$$

In either case, the equality holds when $$\mathbf{v}_1=a\,\mathbf{v}_2$$, where a is a positive real constant. Is there any specific way to compute a, or just pick an arbitrary positive real number?

Regards

2. Aug 12, 2009

### jpreed

I don't believe you can calculate 'a' from the inequality. You would need another route to get it.

3. Aug 12, 2009

### jgens

Your second inequality is the "triangle inequality." The Cauchy-Schwarz inequality is $| \langle \mathbf{A},\mathbf{B} \rangle | \leq \|{\mathbf{A}}\| \, \|{\mathbf{B}} \|$

4. Aug 12, 2009

### boboYO

the equality holds when v1=av2 for any scalar a.

5. Aug 13, 2009

### uart

The first is the CS inequality.

BTW the second inequality has an error (dot instead or plus). It should read :

$$|\mathbf{v}_1 + \mathbf{v}_2|\leq\|\mathbf{v}_1\|+\|\mathbf{v}_2\|$$

In which case it's the triangle inequality.

6. Aug 13, 2009

### EngWiPy

So, can we write:

$$|\mathbf{w}^H\,\mathbf{h}|^2\leq\|\mathbf{w}\|^2\,\|\mathbf{h}\|^2$$?

7. Aug 14, 2009

### HallsofIvy

What they are saying is that equality holds if an only if one vector is a multiple of the other. a could be any real number. It is not a matter of calculating a or a picking a.

8. Aug 14, 2009

### EngWiPy

Ok, thank you all guys.

9. Aug 15, 2009

### Rasalhague

Could we say, more generally, that a could be any scalar (so that this would hold for any inner product space, over any field)?

10. Aug 15, 2009

### slider142

As far as I know, Cauchy proved the inequality for complex vector spaces and Schwarz proved it for polynomial space. In my general linear algebra text, it is proven for real/complex vector spaces. It is a special case of a http://planetmath.org/encyclopedia/CauchySchwartzInequality.html [Broken].

Last edited by a moderator: May 4, 2017
11. Aug 18, 2009

### HallsofIvy

Over any ordered field, yes. That is necessary in order that we be able to say "$\le$".