Hermitian and positive eigenvalues implies positive-definite On Wikipedia it says that for Hermitian matrices, 
positive eigenvalues if and only if positive-definite.
How do you prove the forward direction?
 A: Let $A\in\mathbb{C}^{n\times n}$ be Hermitian such that $A$ has positive eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$, then we may let
$\{v_1,v_2,\ldots,v_n\}$ be an orthonormal basis for $\mathbb{C}^n$ consisting of eigenvectors of $A$ corresponding to
$\lambda_1,\lambda_2,\ldots,\lambda_n$, respectively.
Given a non-zero vector $x\in\mathbb{C}^n$, write $\displaystyle x=\sum_{i=1}^na_iv_i$
for some scalars $a_1,a_2,\ldots,a_n$ not all zero. Then
\begin{align}
x^\ast Ax
&=x^\ast A\left(\sum_{i=1}^na_iv_i\right)
=x^\ast \left(\sum_{i=1}^na_iAv_i\right)
=x^\ast \left(\sum_{i=1}^na_i\lambda_iv_i\right)\\
&=\left(\sum_{j=1}^n\bar{a}_jv_j^\ast\right)\left(\sum_{i=1}^na_i\lambda_iv_i\right)
=\sum_{j=1}^n\sum_{i=1}^n\bar{a}_ja_i\lambda_i\delta_{ji}\\
&=\sum_{j=1}^n\vert a_j\vert^2\lambda_j>0,
\end{align}
where $x^\ast$ denotes the conjugate transpose of $x$, and $\delta_{ji}$ is
the Kronecker delta function. Hence we conclude that $A$ is positive-definite.
A: Hint:Let $A=U\Lambda U^H$ be its Eigen decomposition where $U=[u_1,\dots,u_n]$ is a orthonormal matrix with columns as eigenvectors of $A$ and $\Lambda$ is a diagonal matrix with eigenvalues of $A$. Then $$x^HAx=x^HU\Lambda U^Hx=\sum_{i=1}^{N}\theta_i\lambda_i(A)$$
where $\theta_i = |x^Hu_i|^2\geq 0$ and $\sum_i \theta_i=1$. Now try to use $x^HAx>0\,,\,\forall x$ and argue why all eigenvalues should be positive.
