Linear Algebra – Prove Positive Definite Matrix is Non-Singular

Question

First, going through what it means to be positive definite and non-singular:

Positive definite implies

• $\det(A) > 0$
• All eigenvalues of $A$ are positive, and so $0$ is not an eigenvalue of $A$

Nonsingular implies

• $\det(A) \neq 0$
• All eigenvalues of A are nonzero
• The product of eigenvalues of $A$ $= \det(A)$

It seems as though these two characterizations go hand in hand, though I assume negative eigenvalues could form a non-singular matrix but not a positive definite matrix. Can this be proven directly, or do I need to figure out how to prove by contradiction?

Thanks!

Answer 1

Here's a one-line proof by contraposition:

Suppose there is nonzero $x$ such that $Ax = 0$. Then $\langle x, Ax \rangle = 0$, and so $A$ is not positive definite.