I am struggling to understand a proof regarding the transformation of matrix inequalities and need your help. Thank you in advance.
I am reading the following paper: https://arxiv.org/pdf/2304.03519.pdf
In the proof of Theorem 4, in the beginning of second column of page 4, they say that:
From the inequality:
Using the dualization lemma, which is lemma 4.9 in page 107 of this lecture notes: https://www.imng.uni-stuttgart.de/mst/files/LectureNotes.pdf, they can obtain this:
I can understand the inversion of the matrix in the centre of the lefthand side, but I cannot understand how the matrix:
can turn into this matrix in the next step:
I hope you can help me understand this step-by-step. Thanks again for your help.
I assume that $[\star]^TMX$ is shorthand for $X^TMX$.
Denote $$ M = \left[\begin{array}{cc|cc} -P & 0&0&0\\ 0&P&0&0\\ \hline 0&0&\tilde Q_z&\tilde S_z\\ 0&0&\tilde S_z^\top & R_z \end{array}\right], \quad X = \left[ \begin{array}{cc} (A + B_0k^\top)^\top & B_1\\ -I & 0\\ \hline B_1^\top & 0\\ 0 & -I \end{array}\right]. $$ Note that the statement that $X^\top MX \succ 0$ is equivalent to the condition that $x^\top Mx > 0$ for all $x \neq 0$ in the column-space of $X$. Thus, the dualization lemma implies that if $Y$ is a matrix whose column-space is the orthogonal complement to the column-space of $X$, then the condition $X^TMX \succ 0$ is equivalent to requiring that $x^TM^{-1}x < 0$ for all $x \neq 0$ within the column-space of $Y$, which is to say that $Y^TM^{-1}Y \prec 0$.
Now, how can we check that the column-span of $Y$ is indeed the orthogonal complement to that of $X$? It suffices to verify the following two conditions: