If $\left( {\begin{array}{*{20}{c}} A & B \\ B & A \\ \end{array}} \right) \ge 0 \Rightarrow A \ge B$ Let $A,B\in M_n$ and $\left( {\begin{array}{*{20}{c}}
   A &amp; B  \\
   B &amp; A  \\
\end{array}} \right) \ge 0$ 

Why does $A \ge B$ ?
Let $x$ be given. Then
$$
0\le \pmatrix{x\\-x}^T\pmatrix{A&amp;B\\B&amp;A}\pmatrix{x\\-x}
 = 2x^TAx - 2x^TBx,
$$
This implies
$$
x^T(A-B)x\ge0
$$
for all $x$, hence $A\ge B$.
If you take Determinant of the above matrix, it becomes,

$A^2-B^2 \ge 0$

Square of any number is positive and to yield the number greater than zero, A should be always greater than B.

Hence $A\ge B$