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I am working on a matrix optimization problem, and the constraints are difficult to handle. The constraints are in the following form, \begin{align} \text{Given: } &b \in \mathbb{R}^n \text{ , and } t \in \mathbb{R} \\\\ \text{Constraints: } & (b^T X^{-1} b)^2 \le t^2 \\\\ \text{Variable: } &X \in \mathbb{S}^n \text{ (a real non-singular symmetric matrix, but not necessarily positive semidefinite)} \end{align} I have attempted to apply the properties of the Schur complements and formulate it as semidefinite programming. \begin{align} X \succ 0,\ b^T X^{-1} b < t \iff \begin{bmatrix} t & b^T \\ b & X \end{bmatrix} \succ 0 \end{align} However, even if $X^2≻0$ , I found it hard to generalize this idea to my original problem, since \begin{align} (b^T X^{-1} b)^2 = b^T \underbrace{(X^{-1} bb^TX^{-1})}_{\text{singular}} b < t^2. \end{align}

I wonder if there is some similar technique that can convert the original problem into an equivalent semidefinite programming, or if there are other approaches to solve the matrix optimization problem.

I appreciate any ideas or suggestions. Thanks in advance.

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  • $\begingroup$ Do you have a response to the answer below? $\endgroup$ Commented Oct 15 at 20:42
  • $\begingroup$ Thanks. I will think about it. However, $X$ is my optimization variable. If I want to add some constraint about $X$, can I still write $X ^{-1} = Y_1 - Y_2$? For example, let $X = \sum_{i=1}^n \alpha_i X_i$, and $X_i$ are given symmetric (but not positive semidefinite) matrices, and $\alpha_i$ are real coefficients, is your idea still working? Thanks for your kind suggestion. $\endgroup$
    – zycai
    Commented Oct 16 at 4:27
  • $\begingroup$ (i) For any nonsingular symmetric matrix $X$, one can write $X^{-1}$ as $Y_1-Y_2$, where $Y_1$ and $Y_2$ are positive-semidefinite symmetric matrices; I have now added a detail on this. (ii) The only constraint on the nonsingular symmetric matrix $X$ stated in your post is $(b^T X^{-1} b)^2 \le t^2$. If you impose additional constraints on $X$, that would be a completely different problem, which you might want to post separately. $\endgroup$ Commented Oct 16 at 12:54
  • $\begingroup$ Do you have a further response to my answer and comment? $\endgroup$ Commented Oct 17 at 19:00
  • $\begingroup$ Got it. Thanks. $\endgroup$
    – zycai
    Commented Oct 21 at 4:20

1 Answer 1

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$\newcommand\diag{\operatorname{diag}}$Write $X^{-1}$ as $Y_1-Y_2$, where $Y_1$ and $Y_2$ are positive-semidefinite symmetric matrices.

Detail: This can be done for any nonsingular symmetric matrix $X$. Indeed, one has the spectral decomposition $X=QDQ^T$ for some orthogonal matrix $Q$ and some diagonal matrix $D=\diag(d_1,\dots,d_n)$ with nonzero real diagonal entries $d_1,\dots,d_n$, so that $X^{-1}=QD^{-1}Q^T$. Letting now $$Y_1:=Q\diag\Big(\frac1{d_1^+},\dots,\frac1{d_n^+}\Big)Q^T,\quad Y_2:=Q\diag\Big(\frac1{d_1^-},\dots,\frac1{d_n^-}\Big)Q^T,$$ where $u^\pm:=\max(0,\pm u)$ for real $u$, one has $X^{-1}=Y_1-Y_2$.

Then $$b^T X^{-1}b=B\cdot Y,$$ where $$B:=\begin{bmatrix}bb^T&0\\0&-bb^T\end{bmatrix},\quad Y:=\begin{bmatrix}Y_1&0\\0&Y_2\end{bmatrix},$$ and $U\cdot V:=\text{tr}(UV^T)$, the trace of $UV^T$.

So, the restriction $(b^T X^{-1}b)^2\le t^2$ can be rewritten in the following semidefinite programming way: $$Y\text{ is a $2n\times2n$ positive-semidefinite symmetric matrix,}$$ $$ B\cdot Y\le|t|,\quad(-B)\cdot Y\le|t|,$$ $$E_{i,j}\cdot Y=0\text{ if }i\in[n]\ \&\ j\in[2n]\setminus[n],$$ where $[n]:=\{1,\dots,n\}$.

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