Necessary vs sufficient conditions - matrix definiteness

Question

I'm having a little trouble understanding the wording of this theorem, and what it's implications are.

Theorem: If $A$ is an $n \times n$ symmetric matrix then for it to be positive definite it is:

• Sufficient: all leading principal minors be positive

• Necessary: all principal minors be positive

My understanding of sufficient vs necessary was that: (S) is a stronger condition than (N) in that it's possible (N) can be satisfied but A isn't Positive definite, whereas as long as (S) is satisfied, A is always positive definite.

But in this case, doesn't (N) imply (S)? If all principal minors are positive, then all the leading principal minors are positive, and hence A is PD.

I'm confused by the meaning of necessary and sufficient here.

Answer 1

You are generally right:

• “(S) is a sufficient condition for (P)” means: $(S) \implies (P)$,
• “(N) is a necessary condition for (P)” means: $(P) \implies (N)$.

The reverse implications do not necessarily hold, but may hold: in that case we have a “necessary and sufficient condition” for (P), i.e. a condition which is equivalent to (P).

That is what happens in your case: From the circular implications $$ \begin{align} &\text{All leading principal minors are positive} \\ \implies &\text{$A$ is positive definite} \\ \implies &\text{All principal minors are positive} \\ \implies &\text{All leading principal minors are positive} \end{align} $$ it follows that both your conditions are equivalent to $A$ being positive definite, i.e. both are necessary and sufficient.