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I have a $n\times n$ real symmetric indefinite non-zero matrix $A$ where diagonal elements are all the same.

Assume $x \in \{-1, 1\}^n$. There are two questions I'm tackling:

  1. Is this statement true: there exists such an A where $x^TAx = 0, \forall x$.

  2. Say I have found an $x$ such that $x^TAx = 0$. Is there an analytical (or computationally efficient) way to construct from $x$, a solution $\hat{x} \in \{-1,1\}^n$ such that $\hat{x}^T A \hat{x} \neq 0$?

There are related questions but the insights there didn't help me much: zeros of $x^*Ax$, a quadratic form Solution to a quadratic form

Any suggestions, will help. Thank you.

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    If you want to find symmetric $\rm A$ such that $\rm x^{\top} A x = 0$ for all $\rm x \in \{\pm 1\}^n$, then you have $2^n$ linear equations in $\binom{n+1}{2}$ unknowns. – 2017-02-20
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    And half of the equations are redundant. Hence, $2^{n-1}$ equations. – 2017-02-20
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    Which means that the system is *underdetermined* when $n \in \{2,3,4\}$. – 2017-02-20

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