Say that we're given the following matrices: $S\in M_{m,m}$ symmetric positive definite, $A\in M_{n,n}$, $X\in M_{n,m}$ and $Y\in M_{n,m}$.
Actually I need to use a program that doesn't accept matrix equalities and it only accepts the matrix inequality. Moreover this program does not accept the following matrix $(Y-AX)^T(Y-AX)$.
The equality that I need to set is $S=(Y-AX)^T(Y-AX)$ so I thought to reformulate the problem in the following manner:$\{S-(Y-AX)^T(Y-AX)\geq0\}\cup\{(Y-AX)^T(Y-AX)-S\geq0\}$ which means that $S-(Y-AX)^T(Y-AX)$ is at the same time positive and negative semidefinite hence a zero matrix. To avoid the use of the form $(Y-AX)^T(Y-AX)$ we can replace the two sets by the schur inequalityas follows
$\{S-(Y-AX)^T(Y-AX)\geq0\}\cup\{(Y-AX)^T(Y-AX)-S\geq0\}$ is equivalent to have
$$\left\{
\begin{bmatrix}S & (Y-AX)\\(Y-AX)^T & I_{n,n}\end{bmatrix}\geq0
\right\}\cup\left\{
\begin{bmatrix}I_{n,n} & j\times(Y-AX) \\j\times(Y-AX)^T & -S\end{bmatrix}
\geq0\right\}.$$
So my question here is as follow: are the two last sets equivalent to have $S=(Y-AX)^T(Y-AX)$. I'm wondering also if the second set remains true since we're using a complex number.
Best Answer
No. The matrix $$ B=\begin{bmatrix}I_{n,n} & j\times(Y-AX) \\j\times(Y-AX)^T & -S\end{bmatrix} $$ is not Hermitian in general (unless $Y-AX=0$). Even if it is, since $S$ is positive definite, $B$ is certainly not positive semidefinite.