The equivalence of a 2 by 2 positive semidefinite matrix and a 3 by 3 positive semidefinite matrix?

Question

I came across the following: $$\begin{bmatrix} -x^TAx-2b^Tx+c &-(Ax+b)^TR\\ -R(Ax+b) & \lambda I -RAR \end{bmatrix}\geq0 \iff \begin{bmatrix} b^TA^{-1}b+c & 0 &(x+A^{-1}b)^T\\ 0 & \lambda I &R\\ (x+A^{-1}b) & R &A^{-1} \end{bmatrix}\geq0$$ where $A\in\mathbb{S}^n_+$ is a positive semidefinite matrix, and R can be either a symmetric matrix or a scalar. This is not obvious to me. Can someone kindly teach me how is this equivalent?

Answer 1

This is not a final answer, but the work may be useful regardless. I'll denote the matrices as $M_2,M_3$ respectively. Then the Schur complement of the diagonal element $A^{-1}$ in $M_3$ is given by

\begin{align} M_3\setminus A^{-1}&= \begin{bmatrix} -b^\top A^{-1}b+c & 0\\ 0 & \lambda I\end{bmatrix}-\begin{bmatrix} (x+A^{-1}b)^T \\ R\end{bmatrix}A\begin{bmatrix} x+A^{-1}b & R\end{bmatrix}\\ &=\begin{bmatrix} -b^\top A^{-1}b+c & 0\\ 0 & \lambda I\end{bmatrix} -\begin{bmatrix}(x+A^{-1}b)^T A(x+A^{-1}b) & (x+A^{-1}b)^TAR\\ RA (x+A^{-1}b) & RAR \end{bmatrix}\\ &=\begin{bmatrix} -b^\top A^{-1}b+c-(x^\top Ax +2b^\top x+b^\top A^{-1}b) & -(Ax +b)^TR\\ -R(Ax+b) & \lambda I-RAR\end{bmatrix} \end{align}

This is altogether close to $$M_2=\begin{bmatrix} x^TAx+2b^Tx+c &(Ax+b)^TR\\ R(Ax+b) & \lambda I -RAR \end{bmatrix}.$$

But it isn't identical. One obvious difference is the sign of $R(Ax+b)$ in the corners. However, this may not be a big difference in context: If you're maximizing some $f(R)$ subject to $M_2\geq 0$, then relabeling $R\to -R$ isn't an issue. The more concerning problem is that the upper-left corner doesn't simplify the way one would want. For it to do so, one would first need to have $b^T A^{-1}b$ in the initial statement of $M_3$. But that would still only simplify to $$b^\top A^{-1}b+c-(x^T Ax +2b^\top x+b^\top A^{-1}b)=-x^T Ax-2b^T x+c$$ which is similar but not identical to the upper-left element of $M_2$.