TITLE: How to show, that a Hermitian matrix is positive definite, if all eigenvalues are positive.
QUESTION [8 upvotes]: I want to show, that a Hermitian matrix is positive definite, if all eigenvalues of the matrix are positive. And the other way round.
Also I wonder, if every Hermitian, strict diagonally dominant matrix is positive definite.
Thank you guys.

REPLY [3 votes]: If $A$ is HPD and $\lambda\in\mathbb{R}$ is its eigenvalue such that $Ax=\lambda x$ for some $x$ such that $x^*x=1$, then $0<x^*Ax=\lambda x^*x=\lambda$. Hence $\lambda$ is positive.
The other way around works with the spectral theorem for Hermitian matrices. Every Hermitian matrix $A$ can be unitarily transformed to a diagonal matrix $\Lambda$. It is not hard to check that if $\Lambda$ is HPD (that is, $A$ has positive eigenvalues) then $A$ is HPD (see the other answer).
Every Hermitian and strictly diagonally dominant matrix is positive definite. This is a consequence of the Gershgorin theorem which says that the spectrum is contained in the union of closed circles in the complex plane, where each circle is centered in the diagonal element and its radius given by the sum of the absolute values of the off-diagonal elements of the row. For $A=(a_{ij})_{i,j=1}^n$, it means that
$$
\Lambda(A)\subset\bigcup_{i=1}^n D_i,
$$
where
$$
D_i=\{z\in\mathbb{C}: |a_{ii}-z|\leq\sum_{j=1}^n|a_{ij}|\}.
$$
Since $A$ is diagonally dominant, all $D_i$'s are centered on the positive part of the real line and do not contain $0$. That means (since Hermitian matrices have real eigenvalues), that no eigenvalue of $A$ is $0$ or negative. Hence $A$ is HPD.
This also works for irreducible weakly diagonally dominant matrices, it's just not that easy.