If $A$ is a positive definite matrix can it be concluded that the kernel of $A$ is $\{0\}$?
pf: R.T.P $\ker A = 0$. Suppose not, i.e., there exists some $x\in\ker A$ s.t $x\neq 0$, then $$Ax = 0\;\Longrightarrow x^T Ax = 0$$ which is a contradiction by definition of positive definite. Therefore $\ker A=\{0\}$.