# Semi-definite positive matrix sum up to identity in diagonal is greater than a changed form?

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!

• The inequality is related to positive definitness ? Commented Jun 8, 2022 at 6:41
• @P.Quinton Yes, the inequality $A\ge B$ means $A-B$ is semi-definite positive. Commented Jun 8, 2022 at 6:49

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.