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.

Simple question: do we really need the conjugate in the inequality?

$$ |\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2 $$

share|improve this question
1  
The answer is No. –  Tom Nov 17 '12 at 12:31

2 Answers 2

up vote 4 down vote accepted

The Cauchy-Schwarz inequality says that $|\langle a,b \rangle| \leq \| a \| \|b\|$. In $\mathbb C^n$, the inner product is \begin{equation} \langle a, b \rangle = \sum_{j=1}^n a_j \bar{b_j}. \end{equation}

That's why we the Cauchy-Schwarz inequality in $\mathbb C^n$ has conjugates in it.

While it is true that you could omit the conjugates in your inequality and still have a true statement, that would only take us further away from the nice statement that $|\langle a,b \rangle| \leq \| a \| \|b\|$.

share|improve this answer
1  
Ok thanks for the answer. Haven't read about the inner product for complex numbers yet –  Dan Nov 17 '12 at 12:44

Simple answer, yes. Look back to the definition of the inner product (for complex numbers) and note the complex conjugate symmetry (as opposed to pure symmetry)

share|improve this answer
    
If you replace $b_j$ by $\overline{b_j}$ on both sides, you have no more $\overline{.}$ on the left and on the right, you don't have any either since you have $|b_j| = |\overline{b_j}|$... –  xavierm02 Nov 17 '12 at 12:33

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.