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?

  • $\begingroup$ What conjecture????? $\endgroup$ – Will Jagy Jun 30 '13 at 18:54

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.

  • $\begingroup$ 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. $\endgroup$ – Federico Magallanez Jun 30 '13 at 21:48
  • $\begingroup$ @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. $\endgroup$ – user1551 Jul 1 '13 at 4:02

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