Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I see the Cauchy-Schwarz Inequality written as follows

$$|\langle u,v\rangle| \leq \lVert u\rVert \cdot\lVert v\rVert.$$

Why the is the absolute value of $\langle u,v\rangle$ specified? Surely it is apparent if the right hand side is greater than or equal to, for example, $5$, then it will be greater than or equal to $-5$?

share|improve this question
5  
The absolute value gives a stronger result, then. –  Dylan Moreland Apr 19 '12 at 14:44
add comment

2 Answers

up vote 5 down vote accepted

I assumed that we work in a real inner product space, otherwise of course we have to put the modulus.

The inequality $\langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$ is also true, but doesn't give any information if $\langle u,v\rangle\leq 0$, since in this case it's true, and just the trivial fact that a non-negative number is greater than a non-positive one. What is not trivial is that $\lVert u\rVert\lVert v\rVert$ is greater than the absolute value. But in fact the assertions $$\forall u,v \quad \langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$$ and $$\forall u,v\quad |\langle u,v\rangle|\leq \lVert u\rVert\lVert v\rVert$$ are equivalent. Indeed, the second implies the first, and consider successively $u$ and $-u$ in the first to get the second one.

share|improve this answer
add comment

What if the inner product is a complex number which can happen if $u$ and $v$ are vectors of complex numbers?

For real vectors, the Cauchy-Schwarz Inequality is better written as $$-||u||\cdot ||v|| \leq \langle u, v \rangle \leq ||u||\cdot ||v||$$ where, if $||v|| > 0$, then equality holds in the right inequality if $u = \lambda v$ with $\lambda \geq 0$ and in the left inequality if $u = \lambda v$ with $\lambda \leq 0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.