Prove existence symmetrical solution of $AX=0$ if $\operatorname{rank}A\leq\frac{n}{2}$

Question

Question: Given matrix $A\in M_n\left(K\right)$, $\operatorname{rank}(A)\leq\frac{n}{2}$. Prove that there exists $X\in M_n\left(K\right)$ such that $X$ is symmetrical and $X$ satisfies $AX=0$.

My thoughts. I know that $\operatorname{rank}(X)\geq\frac{n}{2}$ as $\operatorname{rank}(X)=n-\operatorname{rank}$. And than I stuck.

EDIT

$X$ is non-zero

Symmetrical means $X=X^T$

Answer 1

The solution is simple if $K=\mathbb{R}$. Since $\operatorname{rank}(A) < n$. There is a $v \neq 0$ such that $Av=0$. Clearly $X=vv^T \neq 0$ (since $\operatorname{trace}(X) = \|v\|^2 > 0$) and is symmetric, and $AX=0$.

Answer 2

Let, $$ A = USU^{-1} $$

denote the eigen-decomposition of the matrix $A$. Since the rank of $A$ is less than $\frac{n}{2}$ it follows that there are at least $\frac{n}{2}$ zero eigen-values.

If we let, $$ \hat{U} = [\textbf{u}_0, \ldots,\textbf{u}_k, \ldots \textbf{u}_K] $$ denote the eigen vectors corresopnding to the zero eigen-values, i.e, $$ A \textbf{u}_k = 0 $$ then, $$ A \hat{U} = 0 $$ It follows that, $$ A \hat{U}\hat{U}^T = 0 $$ and so we can let $X = \hat{U}\hat{U}^T$.