What Is the Correct Form of the Cauchy-Schwarz Inequality?

[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

 

[jpreed]

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

 

[jgens]

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

 

[boboYO]

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

 

[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.

 

[EngWiPy]

So, can we write:

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

 

[HallsofIvy]

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

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.

 

[EngWiPy]

Ok, thank you all guys.

 

[Rasalhague]

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.

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

 

[slider142]

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

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 .

 

[HallsofIvy]

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

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