Condition for positive Determinant

Question

If we have that for any non-zero vector $x$ that $Ax\cdot x$ is a positive, why is determinant of $A$ positive?

Answer 1

Proof for the complex case (where the hypothesis applies to any non-zero vector, real or complex): the determinant is the product of the eigen values. If $\lambda$ is an eigen value then there exists a non-zero vector $x$ such that $Ax=\lambda x$ This gives $Ax.x=\lambda \|x\|^{2}$. LHS is positive by hypothesis and $\|x\|^{2}>0$. Hence $\lambda >0$. Hence determinant of $A$ is a product of positive numbers.

Real case: if $A$ is a real matrix then eigen values can only occur in pairs $\lambda, \overline {\lambda}$ and the product of these is $|\lambda|^{2}$. Hence it is enough to see that the product of real eigen values is positive which follows by above argument since any eigen vector leads to a real eigen vector by taking real parts of its components.