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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.

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1 Answer 1

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.

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The lemma for the schur compliment states that if the matrix block in the upper left side, which is $I_{n,n}$ in this case, is positive definite then B is positive definite if and only if $(Y-AX)^T(Y-AX) - S$ is positive definite. And as far as I know if B is positive definite it doesn't imply that $B$ must be positive element-wise. Therefore there is no restriction to choose the matrix block in the lower right side as a negative definite matrix, i.e. $-S$. –  user2987 Feb 16 '13 at 8:35
    
@Bel Please clarify the meaning of $\ge0$ in your question. Does it mean entrywise $\ge0$ or "positive semidefinite"? –  user1551 Feb 16 '13 at 8:40
    
In addition in a hermitian matrix the main diagonal entries should be real not the antidiagonal entries. So why we should set $Y-AX=0$? –  user2987 Feb 16 '13 at 8:50
    
It means positive semidefinite and I'm trying to prove that the matrix $S-(Y-AX)^T(Y-AX)$ is at the same time positive and negative semidefinite hence zero matrix. –  user2987 Feb 16 '13 at 17:39
    
@Bel OK, then my answer remains unchanged. In that answer I am not saying that you should set $Y-AX=0$. What I'm saying is that, since $Y-AX\not=0$ in general, the matrix $B$ is not Hermitian and hence it is not positive semidefinite (all positive semidefinite matrices, be they real or complex, must be Hermitian). Also, since your question states that $S$ is positive definite, it follows that $-S$ and in turn, $B$, are not positive semidefinite. –  user1551 Feb 16 '13 at 21:51

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