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'm learning multivariate analysis. Cauchy-Schwarz Inequality plays an important role in several multivariate techniques.

  1. Cauchy-Schwarz Inequality:Let b and d be any two p $\times$ 1 vectors. Then $$(b'd)^2\leq(b'b)(d'd)$$

  2. Extended Cauchy-Schwarz Inequality:Let b and d be any two p $\times$ 1 vectors and B be a p $\times$ p positive definite matrix. Then $$(b'd)^2\leq(b'Bb)(d'B^{-1}d)$$

It is not that difficult to prove. I'm NOT asking how to prove it.

My question:

Consider there is a p $\times$ p identity matrix in the right hand of the Cauchy-Schwarz Inequality, that is, (b'Ib)(d'Id). Why can we turn I into a positive definite matrix so that the Inequality still remains? How to understand this fact intuitively?

share|improve this question
    
Is the LHS of 2. correct? Shouldn't there be a B there also? –  Berci Nov 17 '12 at 15:10
1  
@Berci No. It's just 1. applied to $B^{1/2}b$ and $B^{-1/2}d$. –  martini Nov 17 '12 at 15:13
    
@Berci I copied 1 and 2 from P78-P79 of Applied Multivariate Statistical Analysis written by Richard A. Johnson. I'm sure there is no typo. –  J.A.F Nov 17 '12 at 15:13
add comment

1 Answer

up vote 2 down vote accepted

I would say, it's about the possible scalar products one can impose on a given vector space.

If $B$ is a positive definite (symmetric) matrix, then $(u,v)\mapsto u'Bv$ just defines a scalar product. Cauchy-Schwartz inequality holds for any scalar product.

share|improve this answer
    
In fact any positive definite $B$ gives a scalar product, the symmetric $B$ is just the one that is canonically associated to it (in a given basis). –  Max Nov 17 '12 at 16:53
    
Well, the mapping above is symmetric only if $B=B'$. But, we can take $\displaystyle\frac{B+B'}2$ for any given positive definite $B$. –  Berci Nov 17 '12 at 17:31
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.