xA=0 sufficient condition for zero determinant?

Question

Let A be a symmetric n by n matrix and x be a 1 by n vector. If I find one x such that xA=0, does it mean A is singular?

Answer 1

A matrix is singular if and only if its kernel contains a nontrivial element, i.e. if there exists $x\in\mathbb R^n\setminus\{0\}$ such that $Ax=0$. Transpose to get $x^T A^T = x^T A = 0$ where I have chosen $x$ to be a column vector (as is usual). So yes under the additional assumption that $x\ne 0$. Notice that the symmetry is irrelevant because $A$ is invertible iff $A^T$ is invertible.