Positive Semidefinite/Definite Matrix

The Problem

A real symmetric matrix $A\in {\mathbb{R}}^{n\times n}$ is called positive semidefinite (PSD), written $A⪰0$, if ${\vec{x}}^{T}A\vec{x}\ge 0$ for all $\vec{x}\in {\mathbb{R}}^n$. It is positive definite (PD), $A\succ 0$, if the inequality is strict for all $\vec{x}\ne 0$.

Prove the following equivalent characterisations:

1. $A⪰0$ if and only if all eigenvalues of $A$ are non-negative.
2. $A⪰0$ if and only if $A=B^{T}B$ for some matrix $B$.
3. If $A\succ 0$ then $A$ has a unique PD square root: there exists a unique PD matrix $S$ with $S^2=A$.

Why does this matter? Covariance matrices are PSD; this is what makes portfolio optimisation, Kalman filtering, and Gaussian processes all well-posed.