$\newcommand{\lg}{\langle}$ $\newcommand{\rg}{\rangle}$ A matrix $A \in \mathbb{R}^{n \times n}$ is semi positive defined if $\langle x, Ax \rangle = x^{T}Ax = \sum_{i,j=1}^{n} x_{i}A_{ij}x_{j} \ge 0$ for all $x \in \mathbb{R}^{n}$. Prove or give a counterexample
a)Let $v_{1},v_{2}, \ldots, v_{n}$ a basis of $R^{n}$. If $v_{j}^{T}Av_{j} \ge 0$ for all $1\le j \le n$ then $A$ is semi positive defined.
b) If $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times n}$ and satisfy $x^{T}Ax=x^{T}Bx$ for all $x\in \mathbb{R}^{n}$ then $A=B$.
c)If $A\in \mathbb{R}^{n \times n}$ is semi positive defined and symmetric then $\sqrt{A_{ii}A_{jj}} \ge |A_{ij}|$ for any $1\le i, j \le n$.
For a), is false. In $\mathbb{R}^{2}$, let $v_{1}=(1,0)$ and $v_{2}=(0,1)$ the canonical basis, and let $$A= \begin{equation} \begin{pmatrix} 0 & 2 \\ -3 & 0 \end{pmatrix} \end{equation} $$ Then $v_{1}^{T} A v_{1}=0$ and $v_{2}^{T}Av_{2}=0$, but $(1,1) A \begin{equation} \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix} \end{equation} = -1 <0$.
b) false In $\mathbb{R}^{2}$, let $A=\begin{equation} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \end{equation}$ and $B=\begin{equation} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \end{equation}$, is easy to see thar for all $x\in \mathbb{R}^{2}$, $A$ and $B$ satisfy $x^{T}Ax=x^{T}Bx$ but $A \neq B$.
c) I think is true, but I don't know how to prove it, it seems kind obvious but I don't see it.