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 playing with some quadratic form for my research.

  • In my setting, $\mathbf A$ is an $n\times n$ real symmetric matrix with only two types of eigenvalues: they are either $\frac{1}{n-x}$ with algebraic multiplicity $q$ or $\frac{1}{-x}$ with algebraic multiplicity $n-q$. (You may assume that $x > 0$).
  • And let $\mathbf v$ be a column vector of $\pm 1$s.

In my work, something interesting happens when the quadratic form $\mathbf v^T\mathbf A\mathbf v$ becomes $-n/x$. And another interesting will happen if this value is the minimum.


(Updated) My Conjecture: When $x> 0$ $$-n/x = \min_{\mathbf v\in\{\pm 1\}^{n\times 1}} \mathbf v^T\mathbf A\mathbf v$$ where $\mathbf A$ admits an eigendecomposition as follows: $$\mathbf A = \mathbf U \begin{bmatrix} \frac{1}{n-x}\mathbf I_{q\times q} & \mathbf {0}_{q\times(n-q)} \\ \mathbf{0}_{(n-q)\times q} & \frac{1}{-x}\mathbf I_{(n-q)\times (n-q)} \end{bmatrix} \mathbf U^T$$


I ran a large set of experiments and it seems that my conjecture to be true. But I cannot prove it. Is my conjecture true? If it is, is there any way to prove it?

share|improve this question
    
What conjecture????? –  Will Jagy Jun 30 '13 at 18:54
add comment

1 Answer

It is not true. Here is a counterexample: $n=3,\ x=\frac12,\ q=2$, $$ A=\pmatrix{ \tfrac{1}{\sqrt{2}}&\tfrac{1}{\sqrt{3}}&\tfrac{1}{\sqrt{6}}\\ \tfrac{1}{\sqrt{2}}&\tfrac{-1}{\sqrt{3}}&\tfrac{-1}{\sqrt{6}}\\ 0&\tfrac{-1}{\sqrt{3}}&\tfrac{2}{\sqrt{6}}\\ } \pmatrix{\tfrac25\\ &\tfrac25\\ &&-2} \pmatrix{ \tfrac{1}{\sqrt{2}}&\tfrac{1}{\sqrt{3}}&\tfrac{1}{\sqrt{6}}\\ \tfrac{1}{\sqrt{2}}&\tfrac{-1}{\sqrt{3}}&\tfrac{-1}{\sqrt{6}}\\ 0&\tfrac{-1}{\sqrt{3}}&\tfrac{2}{\sqrt{6}}\\ }^T = \frac25\pmatrix{0&1&-2\\ 1&0&2\\ -2&2&-3}. $$ For $v\in\{-1,1\}^3$, we have $v^TAv=\frac25(2v_1v_2-4v_1v_3+4v_2v_3-3v_3^2)$. Its minimum occurs at $v=(1,-1,1)$, where $v^TAv=\frac25(-2-4-4-3)=\frac{-26}{5}\neq-6=\frac{-n}{x}$.

I have also run a small set of numerical experiments. Contrary to your claim, I found that your conjecture is almost always false.

share|improve this answer
    
Hmm.. Interesting. In my setting, $n$ is always a multiple of 4. And $\mathbf A$ is somewhat specially constructed in an iterative manner which I omitted. Maybe that's the reason why your experiment mostly disproved my conjecture. Thanks anyway. –  Federico Magallanez Jun 30 '13 at 21:48
    
@FedericoMagallanez I tried $n=4$ and your conjecture is true only in about 3% of all cases. So, it is not $n$ but your iterative construction that makes your conjecture seems true. –  user1551 Jul 1 '13 at 4:02
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.