Can $X - Y A^\dagger Y^T\succ0$ be written as an LMI where $A^\dagger$ is a pseudoinverse?

Question

I have the constraint

\begin{align} X - Y A^\dagger Y^T\succ0, \end{align}

where $A^\dagger$ is the pseudoinverse of $A\succeq0$. Can we still use the Schur complement to write the constraint as an LMI?

Explicitly, can we show something like:

\begin{align} X - Y A^\dagger Y^T\succeq0 &\iff \begin{pmatrix}X&Y\\Y^T&A\end{pmatrix}\succeq0 \end{align}

Answer: It is true only when $A$ is invertible, please see answer below on the general case.

The Schur complement can be derived using the pseudoinverse. See the textbook by Boyd and Vandenberghe. — Brian Borchers, Nov 29 '20 at 21:46
Thanks @BrianBorchers. Their treatment does not answer my question since I am trying to represent it as an LMI. I edited the question. — Morad, Nov 30 '20 at 18:09
In appendix A of Boyd and Vandenberghe the characterization is given on page 651. The block matrix is PSD if and only if the A block is positive semidefinite, the Schur complement is PSD, and $(I-AA^{\dagger})B=0$. Note that you will not get that $X-YA^{\dagger}Y^{T}$ is strictly positive definite. — Brian Borchers, Nov 30 '20 at 19:53
Thanks @BrianBorchers, but I don't know how to represent the additional $(I-AA^\dagger)B=0$ constraint as an LMI (in $A$). — Morad, Dec 03 '20 at 00:39
The $B$ here is what you've called $Y$. What do you know about $Y$? — Brian Borchers, Dec 03 '20 at 03:54
It is a rectangular real matrix which is part of the optimization. @BrianBorchers — Morad, Dec 03 '20 at 06:40
Can you please provide some direction or conditions in which this condition becomes linear? @BrianBorchers — Morad, Dec 03 '20 at 23:47
Somewhat related: https://math.stackexchange.com/q/2913018/339790 — Rodrigo de Azevedo, Dec 17 '20 at 03:19

Morad · Accepted Answer · 2021-12-15T18:43:52.333

1

As mentioned in the comments, in the general case where $A\succeq0$ may not be invertible there is an orthogonality condition that reads as follows: \begin{align} X-YA^\dagger Y^T\succeq0 \ \ \text{and}\ \ Y(I - A A^\dagger)=0 \iff \begin{pmatrix} X&Y\\ Y^T &A\end{pmatrix}\succeq0. \end{align}

edited Dec 15 '21 at 18:43

answered Dec 17 '20 at 03:04

Morad

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Can $X - Y A^\dagger Y^T\succ0$ be written as an LMI where $A^\dagger$ is a pseudoinverse?

1 Answers1