Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $a_i>0$, $x_i, y_i\in \mathbb{R}$ $i=1,\cdots, n$, such that

$\sum\limits_{i=1}^nx_iy_i=0$, $\sum\limits_{i=1}^nx_i^2=\sum\limits_{i=1}^ny_i^2=1$. Is it true $$ \left[\sum\limits_{i=1}^n\frac{1}{a_i}x_i^2\right] \left[\sum\limits_{1\le i < j \le n} a_ia_j(x_iy_j-x_jy_i)^2\right]\ge \sum\limits_{i=1}^n a_i y_i^2? $$

I tried the case $n=2$ and the special case where all $a_i$ are equal.

Motivation: To find a lower bound for difference in Lagrange identity under certain conditions.

share|improve this question
    
@Sunni, congratulations, you get an interesting inequality and a very nice proof. Here are two side observations: 1) in the case $n=2$, it is actually an identity for arbitrary $a_i$'s; 2)when all the $a_i$'s are equal, it is also an identity, for any $n$. –  Syang Chen Jul 25 '11 at 22:04
    
Right, I already observed that. –  Sunni Jul 26 '11 at 1:42
1  
Note also that from the proof of the inequality you get that equality holds if and only if the vector $(1/\sqrt{a_i}x_i)_{1\leq i\leq n}$ belongs to the linear span of $(\sqrt{a_i}x_i)$ and $(\sqrt{a_i}y_i)$. –  Mikael de la Salle Jul 26 '11 at 7:42
    
@Mikael, I could see that A is essentially the matrix matrix $\begin{pmatrix}0&\lambda \\ -\lambda&0\end{pmatrix}$ operating on a 2-dimensional subspace $X$ of $R^n$. Thus the identity $\|A\|_{op}=\frac{1}{\sqrt{2}}\|A\|_{HS}$ always holds, and $|Ax'|=\|A\|_{op}|x'|$ if and only if $x'\in X$. But I can't see why $X$ is spaned by $(\sqrt{a_{i}}x_{i})$ and $(\sqrt{a_{i}}y_{i})$. –  Syang Chen Jul 27 '11 at 2:32
    
@Xianghong, what is immediate is that the kernel of $A$ contains the orthogonal of the space $Y$ spanned by $(\sqrt{a_i} x_i)$ and $(\sqrt{a_i} y_i)$. On the other hand, since $A$ is normal, $X$ is the orthogonal to the kernel of $A$. This proves one inclusion, which is enough since $X$ and $Y$ have same dimension. –  Mikael de la Salle Jul 28 '11 at 5:56

1 Answer 1

up vote 7 down vote accepted

The answer is yes.

Edit: As pointed out in the comment below, my first answer was not correct (it was proving the inequality with a factor 2). Here is the correction.

Denote by $A$ the $n \times n$ matrix given by $A_{i,j} = \sqrt{ a_i a_j} (x_j y_i - x_i y_j)$, by $x'=(\sqrt{a_i^{-1}} x_i) \in \mathbb R^n$ and by $y'=(\sqrt{a_i} y_i) \in \mathbb R^n$. You are asking whether $|y'| \leq |x'| \| A\|_{HS}/{\sqrt 2}$. Here $|\cdot|$ denotes the usual euclidean norm on $\mathbb R^n$, and $\|\cdot\|_{HS}$ is the Hilbert-schmidt norm on matrices, ie $\|M\|_{HS} = \sqrt{Tr(M^*M)}$, where $M^*$ is the (conjugate) transpose of $M$.

By the assumptions $\langle x,y\rangle =0$ and $\langle x,x\rangle =1$, you have that $A x' = y'$. It is therefore enough to prove that $\|A\|_{op} \leq \|A\|_{HS}/\sqrt 2$, where $\|A\|_{op}$ is the operator norm acting on $\mathbb R^n$ with usual euclidean norm. But this holds because $A$ has real entries and is anti-hermitian, ie $A^* = -A$. If $\lambda$ is an eigenvalue of $A$, $\lambda \in i\mathbb R$ and $\bar \lambda$ is also an eigenvalue of $A$. This proves the desired inequality, since $\|A\|_{op}$ is $\max_\lambda |\lambda|$ and $\|A\|_{HS} = (\sum_{\lambda} |\lambda|^2)^{1/2}$, where the max and the sum run over the eigenvalues of $A$ (counted with multiplicity).

share|improve this answer
    
Is your $\otimes$ Kronecker product? I think $\sum\limits_{1\le i < j \le n} a_ia_j(x_iy_j-x_jy_i)^2=\frac{1}{2}\|A\|^2$. –  Sunni Jul 25 '11 at 13:35
    
You are correct. See my edit. –  Mikael de la Salle Jul 25 '11 at 15:42
    
+1: nice, clean proof. –  Suvrit Jul 25 '11 at 17:44

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.