Can I find a vector with entirely nonzero entries for the following quadratic form to evaluate to zero?

Question

I have a square, symmetric matrix $M$, of size at least $2\times 2$, with diagonal entries equal to $1$ and off-diagonal entries equal to $\pm 1$. Let the entries of $M$ be such that it is indefinite. Since it is indefinite, there is a nonzero vector $x$ such that $x^T M x$ evaluates to zero. I want to know if it is always possible to choose such a vector $x$ so that none of its entries are zero, besides satisfying $x^T M x=0$, no matter what the matrix $M$ is, as long as it satisfies the criteria mentioned beforehand. Thanks.

Answer 1

The locus of $x\in\mathbb{R}^n$ for which $x^t M x = 0$ is an unbounded quadratic surface that cannot entirely lie in a hyperplane $x_i=0$, provided that $M$ has at least two non-zero eigenvalues with opposite signs. Since $\text{Tr}(M)>0$ our matrix cannot be negative semi-definite.

Anyway, in the worst scenario where $M$ is positive semi-definite, no vector of the canonical base of $\mathbb{R}^n$ is an eigenvector of $M$, hence the quadratic surface associated with $M$ may lie on a hyperplane, but it is not a hyperplane of the form $x_i=0$, so your claim holds.

Answer 2

Take $M=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ The quadratic form associated to $M$ is $x^2+2xyy^2=(x+y)^2$. The vector $u=\begin{bmatrix}1\\-1\end{bmatrix}$ is such a vector.