# What's the relationship between Cauchy-Schwarz Inequality and Extended Cauchy-Schwarz Inequality?

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?

• Is the LHS of 2. correct? Shouldn't there be a B there also? Nov 17 '12 at 15:10
• @Berci No. It's just 1. applied to $B^{1/2}b$ and $B^{-1/2}d$. 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. Nov 17 '12 at 15:13

If $B$ is a positive definite (symmetric) matrix, then $(u,v)\mapsto u'Bv$ just defines a scalar product. Cauchy-Schwarz inequality holds for any scalar product.
• 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).
• Well, the mapping above is symmetric only if $B=B'$. But, we can take $\displaystyle\frac{B+B'}2$ for any given positive definite $B$. Nov 17 '12 at 17:31