User ziv goldfeld - MathOverflow

Convexity of a Certain Set of Covariance Matrices

2013-01-19T17:59:48Z

Hello,

My question is about a certain set of matrices being convex or not. I'll start with some preliminaries in order to define myself properly. Let $X_1,U,X_2$ be three zero-mean Gaussian random vectors (RVs) of dimension $N\times 1$, that admit the Markov relation $X_1-U-X_2$. Let us use the notations: $\Sigma_i=\mathbb{E}[X_iX_i^T]$, for $i=1,2$, $\Sigma_U=\mathbb{E}[UU^T]$, $\Sigma_{iU}=\mathbb{E}[X_iU^T]$, for $i=1,2$ and finally $\Sigma_{12}=\mathbb{E}[X_1X_2^T]$. The Markov relation is equivalent to the fact that the auto- and cross- covariance matrices of the RV satisfy: $\Sigma_{12}=\Sigma_{1U}\Sigma_U^{-1}\Sigma_{U2}$.

Let us define the matrix:

$\Sigma=\left( \begin{array}{ccc} \Sigma_1 & \Sigma_{1U} & A\\ \Sigma_{1U}^T & \Sigma_U & \Sigma_{2U}^T\\ A^T & \Sigma_{2U} & \Sigma_{2}\end{array}\right) $.

where $A=\Sigma_{1U}\Sigma_{U}^{-1}\Sigma_{U2}$. My question regards the set of all legitimate matrices $\Sigma$, is this set convex? How can one check this?

Thank you all in advance,

Best regards!

Schur complement and negative definite matrices

2013-01-04T13:05:34Z

Hello,

My question regards to the Schur complement lemma. Consider the matrix $M=\left( \begin{array}{cc} A & B\\ B^T & C \end{array}\right) $.

According to the lemma $M\geq0$ iff $C>0$ and $A-BC^{-1}B^T\geq 0$.

In my current research I'm working on an optimization problem over a domain of matrices; I'm trying to convert this optimization problem into it's convex form. In order to do so I need a similar relation for negative definite matrices. Can the Schur complement lemma be extended to the case of negative definite matrices? And if so, how? Namely, is it true that for a matrix $M$ of the same structure we have $M\leq 0$ iff $C<0$ and $A-BC^{-1}B^T\leq 0$?

Another similar but different problem I have regards to the following non-convex (nor linear) constraint: $A-BC^{-1}B^T\leq 0$ and $A\geq 0$. Is there some way such constraints can be converted to an equivalent constraint which is linear in these variables? For example the two constraints $A-BC^{-1}B^T\geq 0$ and $C> 0$ can be simply converted to $M\geq 0$ using the Schur complement lemma. The new constraint $M\geq 0$ is equivalent to the two old ones and is indeed linear in the matrices $A,B,C$. I'm looking for a way to do something similar to this for my case.

Thank you all in advance,

Best regards!

Geometric Mean of Positive Matrices

2012-11-19T13:28:22Z

Hello all,

My question regards the geometric mean (GM) of two positive matrices. The definition of the GM for two positive matrices $(A,B)$ is given by: $M_0(A,B)=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\frac{1}{2}}A^{\frac{1}{2}}$. Moreover, when $A$ and $B$ commute, this definition reduces to $M_0(A,B)=A^{\frac{1}{2}}B^{\frac{1}{2}}$. The GM is known to be jointly concave in the pair $(A,B)$.

My question regards the reduced structure, namely $M_0(A,B)=A^{\frac{1}{2}}B^{\frac{1}{2}}$, for the general case where $A$ and $B$ do no necessarily commute. Is $A^{\frac{1}{2}}B^{\frac{1}{2}}$ jointly concave in the pair $(A,B)$ for any two positive matrices? If so, how can one proof this? In not? Which additional conditions on $(A,B)$ one should assume (without assuming commutativity) in order for it to be jointly concave in the pair?

Thank you very much in advance!

Comment by Ziv Goldfeld

2013-01-20T10:03:13Z

I'm sorry, but I didn't understand your answer fully. When you say that the data are independent apart from some inequalities do you mean the following: $\Sigma_i-\Sigma_{iU}\Sigma_U^{-1}\Sigma_{Ui}\geq 0$, and the non-negativety constraints, i.e., $\Sigma_i\geq 0$ and $\Sigma_U\geq 0$, where $i=1,2$. Are there any additional relations between the data matrices? What do you mean by "leaves room for an open set"? And finally, why $A$ must be linear if the set is to be convex? There are convex sets that are not linear. I'd appreciate if you could clarify this for me. Thank you in advance!

Comment by Ziv Goldfeld

2013-01-05T21:58:35Z

Thank you. By the way, why is it important to have $C>0$ (or $C<0$ for the negative version) in order to use the lemma? Why isn't $C\neq 0$ enough? I ask it since we do not demand anything regarding the matrices $A$ and $B$ as far as non-negativeness, but we do demand it for $C$. Why is that?

Comment by Ziv Goldfeld

2013-01-04T20:34:37Z

Ohh I'm sorry, I accidentally mixed the two version of the lemma. I've edited the original question and now I think it's fine; thanks for the remark! So you say that $M\leq 0$ iff $C<0$ and $A-BC^{-1}B^T$ holds? Any ideas regarding the second question? Thanks again!

Comment by Ziv Goldfeld

2012-11-19T20:58:01Z

I saw an extension to the definition of positive definite matrices to the asymmetrical case. Namely, we say that a real (unnecessary symmetric) matrix $M$ is positive definite in the wide sense if and only if $z^TMz>0$ for all non-zero vectors $z$. This is equivalent to that the symmetrical part of $M$, which is $\frac{M+M^T}{2}$, is positive definite in the narrow sense. Can we consider an order on matrices using the wider definition of positive definiteness? If so, does the image of the degraded geometric mean operator $A^{\frac{1}{2}}B^{\frac{1}{2}}$ is concave using that order?

Comment by Ziv Goldfeld

2012-11-19T14:58:57Z

Ohh I see. However if the matrices are assumed to be real then the problem of the definition is solved, isn't it? If so, does my first question has an answer for this case? Thank you!

Comment by Ziv Goldfeld

2012-11-19T14:04:43Z

Sorry for the ignorance but why is it necessary that the operator maps to an Hermitian space in order to define joint concavity?