If $X^{T}\cdot A \cdot X = 0$ for some $X$, what is the condition must impose to $A$?

Question

If $X$ is a vector and is a solution of the equation $X^T\cdot A \cdot X = 0$, what is the condition must impose to $A$ to $X$ be a non-trivial solution, excluding the possibility that $A$ be a skew symmetric matrix ?

Answer 1

Guess you mean $x$ as a vector, $A$ being a skew symmetric matrix would be sufficient for all x to make $x^TAx = 0$ as the expression $x^TAx$ is actually $\Sigma_i^n \Sigma_j^n A_{ij}x_ix_j$, for general case, the condition of $A$ depends on $x$, like when $x$ is the column vector with all entries equals 1, $A$ need to be the matrix for which all of this entries sum up to 0.

Answer 2

WLOG we can assume that $A$ is symmetric (as the antisymmetric part of $A$ does not count). If $A$ is positive (or negative) definite then

$$x^T Ax =0 \ \ \Longrightarrow \ \ x=0$$

So a necessary condition is that $A$ is not definite. This condition is also sufficient. In fact let us diagonalize $A=U^T D U$ (possible since $A$ is symmetric). Call $y = Ux$.

Then

$$ x^T Ax =0 \ \ \Longrightarrow \sum_i y_i^2 d_i =0 $$

Where $d_i$ are the eigevalues of $A$. But this latter equation can be satisfied only if some $d_i$'s have opposite sign.

To summarize, a necessary and sufficient condition for the claim is that the symmetric part of $A$ is not definite.