In physics we use a set of semi-definite positive matrices to describe quantum measurement, they should satisfy $\sum_i \Pi_i=\mathbb I,\Pi_i\ge0\,\forall i$. I wonder if there are some way to show that $$\left(\begin{array}{cccc} \Pi_{1} & 0 & \ldots & 0 \\ 0 & \Pi_{2} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \Pi_{M} \end{array}\right) \geq\left(\begin{array}{c} \Pi_{1} \\ \Pi_{2} \\ \vdots \\ \Pi_{M} \end{array}\right)\left(\Pi_{1} \Pi_{2} \ldots \Pi_{M}\right).$$ where $A\ge B$ means $A-B$ is a semi-definite positive matrix.
I try to figure out when $\Pi_i$ are real positive numbers sum up to $1$, which can be proved but for $\Pi_i$ stand for semi-definite matrices, I don't know where to go.
Any suggestion or hint will be great. Thanks in advance!
Yes. Denote by $X^\ast$ the conjugate transpose of a matrix $X$. Let $$ D=\pmatrix{ \Pi_1\\ &\Pi_2\\ &&\ddots\\ &&&&\Pi_M} ,\ E=\pmatrix{I\\ I\\ \vdots\\ I}. $$ The difference between the two sides of your inequality is then $$ D-DEE^\ast D=D^{1/2}(I-P)D^{1/2}\tag{1} $$ where $P=D^{1/2}EE^\ast D^{1/2}$. Since $P^\ast=P$ and $$ P^2=D^{1/2}E(E^\ast DE)E^\ast D^{1/2}=D^{1/2}E(I)E^\ast D^{1/2}=P, $$ we see that $P$ is an orthogonal projection. Hence $I-P\ge0$ and by $(1)$, $D-DEE^\ast D$ is positive semidefinite too.