When the number of matrices become infinity in a matrix inequality?

Question

I will first show my matrix inequality, and state my confusion after it. Suppose we have $n$ positive semi-definite matrices $M_i$ which further satisfy $\sum_{i=1}^n{M_i}=I$. The dimension of $M_i$ is $d$, a finite number. Then I have the following matrix inequality $$\left( \begin{matrix} M_1& 0& 0\\ 0& ...& 0\\ 0& 0& M_n\\ \end{matrix} \right) \succcurlyeq \left( \begin{array}{c} M_1\\ ...\\ M_n\\ \end{array} \right) \left( \begin{matrix} M_1& ...& M_n\\ \end{matrix} \right) .$$ To see this, we only need to mind that $$\left( \begin{array}{c} M_1\\ ...\\ M_n\\ \end{array} \right) \left( \begin{matrix} M_1& ...& M_n\\ \end{matrix} \right) =\left( \begin{matrix} \sqrt{M_1}& 0& 0\\ 0& ...& 0\\ 0& 0& \sqrt{M_n}\\ \end{matrix} \right) \left( \begin{array}{c} \sqrt{M_1}\\ ...\\ \sqrt{M_n}\\ \end{array} \right) \left( \begin{matrix} \sqrt{M_1}& ...& \sqrt{M_n}\\ \end{matrix} \right) \left( \begin{matrix} \sqrt{M_1}& 0& 0\\ 0& ...& 0\\ 0& 0& \sqrt{M_n}\\ \end{matrix} \right) \\ \preceq \left( \begin{matrix} M_1& 0& 0\\ 0& ...& 0\\ 0& 0& M_n\\ \end{matrix} \right) $$ where I have used the fact that $$\left( \begin{array}{c} \sqrt{M_1}\\ ...\\ \sqrt{M_n}\\ \end{array} \right) \left( \begin{matrix} \sqrt{M_1}& ...& \sqrt{M_n}\\ \end{matrix} \right) \left( \begin{array}{c} \sqrt{M_1}\\ ...\\ \sqrt{M_n}\\ \end{array} \right) \left( \begin{matrix} \sqrt{M_1}& ...& \sqrt{M_n}\\ \end{matrix} \right) =\left( \begin{array}{c} \sqrt{M_1}\\ ...\\ \sqrt{M_n}\\ \end{array} \right) \left( \begin{matrix} \sqrt{M_1}& ...& \sqrt{M_n}\\ \end{matrix} \right) $$ that is, $\left( \begin{array}{c} \sqrt{M_1}\\ ...\\ \sqrt{M_n}\\ \end{array} \right) \left( \begin{matrix} \sqrt{M_1}& ...& \sqrt{M_n}\\ \end{matrix} \right) $ is a projection. I wonder when $n$ tends to infinity, will the matrix inequality still exist?

Answer 1

I think the question is really how to interpret your inequality for $n = \infty$. It should make sense as a statement about linear operators on an infinite-dimensional separable Hilbert space. Let Hilbert space $\mathcal H = \mathcal K \oplus \mathcal K \oplus \ldots$ be the direct sum of countably many copies of Hilbert space $\mathcal K$; for each natural number $i$ let $M_i$ be a positive definite bounded linear operator on $\mathcal K$, with $\sum_{i=1}^\infty M_i = I$. The left side of your inequality is the operator $L: \mathcal H \to \mathcal H$ defined by $$L(x_1,x_2, \ldots) = (M_1 x_1, M_2 x_2, \ldots)$$ and the right side is $AB$ where $B: \mathcal H \to \mathcal K$ is defined by $$ B(x_1, x_2, \ldots) = \sum_{i=1}^\infty M_i x_i $$ and $A: \mathcal K \to \mathcal H$ is defined by $$ A x = (M_1 x, M_2 x, \ldots)$$ and the inequality says $L - AB$ is positive definite. That is, for every $x = (x_1,x_2, \ldots) \in \mathcal H$, $$ \langle x, (L - AB) x \rangle \ge 0$$ i.e. $$ \sum_{i=1}^\infty \langle x_i, M_i x_i \rangle \ge \sum_{i=1}^\infty \sum_{j=1}^\infty \langle M_i x_i, M_j x_j \rangle $$

[EDIT] The proof can be adapted from the matrix case. Positive definite bounded linear operators have unique positive definite square roots, which I'll denote using $\sqrt{\ }$. If $C: \mathcal K \to \mathcal H$ and $D: \mathcal H \to \mathcal K$ are defined by $$\eqalign{ C x &= (\sqrt{M_1} x, \sqrt{M_2} x, \ldots)\cr D(x_1, x_2, \ldots) &= \sum_{i=1}^\infty \sqrt{M_i} x_i \cr}$$ then $DC = I$ so $CD$ is a projection. Note that $C^* = D$. Then $CD \preceq I$ so $$AB = \sqrt{L} C D \sqrt{L} \preceq \sqrt{L} \sqrt{L} = L$$