[Math] Minors of a positive definite matrix are positive definite

Question

All main minors of a positive definite matrix are positive definite as
well and therefore $A$ is strictly invertible.

All I know about positive-definiteness is that for the symmetric matrix $A$ the following inequality holds:

$$x^TAx>0, \forall x(\neq 0) \in \mathbb R^3$$

Could you please give me some hints as to how can I prove the above theorem, specially its second part, what does positive-definiteness have to do with being invertible?

Answer 1

I think it's fairly straight-forward actually, naturally depending on how much you want to show. Do you take the following as given?

If $\mathbf{A}$ is non-invertible, then there exists a non-zero vector $\mathbf{x}$ such that $\mathbf{Ax}=\boldsymbol{0}$.

If you do, then it shouldn't be too problematic.

If $\mathbf{A}$ is not invertible, then that means there is a vector such that $\mathbf{Ax}=\boldsymbol{0}$. Thus, $\mathbf{x}^\prime \mathbf{Ax}=\mathbf{x}^\prime \boldsymbol{0}=0$. But, if $\mathbf{A}$ is positive definite, then $\mathbf{x}^\prime \mathbf{Ax}>0$. So $\mathbf{A}$ cannot be both non-invertible and positive definite. Hence, it must be invertible if it is positive definite.

Edit: This only addresses the second part.