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.

  • $\begingroup$ The Schur complement can be derived using the pseudoinverse. See the textbook by Boyd and Vandenberghe. $\endgroup$ – Brian Borchers Nov 29 '20 at 21:46
  • $\begingroup$ Thanks @BrianBorchers. Their treatment does not answer my question since I am trying to represent it as an LMI. I edited the question. $\endgroup$ – Morad Nov 30 '20 at 18:09
  • 1
    $\begingroup$ 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. $\endgroup$ – Brian Borchers Nov 30 '20 at 19:53
  • $\begingroup$ Thanks @BrianBorchers, but I don't know how to represent the additional $(I-AA^\dagger)B=0$ constraint as an LMI (in $A$). $\endgroup$ – Morad Dec 3 '20 at 0:39
  • $\begingroup$ The $B$ here is what you've called $Y$. What do you know about $Y$? $\endgroup$ – Brian Borchers Dec 3 '20 at 3:54

As mentioned in the comments, in the general case where $A$ is not 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}

  • $\begingroup$ A bonus fact: if A is symmetric $AA^\dagger = A^\dagger A$ $\endgroup$ – Morad Mar 2 at 20:57

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