I want to solve a matrix $\Omega$ from a equation $\sum_k (\Omega + \Theta_k)^{-1} = Q$. The $Q$ and $\Theta, \forall k=1...K$ are known, and are positive definite matrices. $\Omega$ also has to be positive definite. all matrices are large (a few thousands of columns and rows). My questions are:

(1) Is there a closed-form solution? How do I simplify the sum of the inverse of two matrix sum?

(2) I'm OK to go for a numerical solution. But how do I define this problem? An optimization problem to minimize something like $f(\Omega) = ||\sum_k (\Omega + \Theta_k)^{-1} - Q||$? Do I need to minimize the frobenius norm, (just like minimizing the L-2 norm in a least square problem)? Considering the constraint that $\Omega$ is positive definite, can I solve it by semi-definite programming? How do I redefine the problem in a linear/semi-definite programming? I don't have much knowledge of linear programming. I would prefer a general gradient descent rather than LP. But I'm OK to use LP if I know how to do.

This problem comes from the estimation of inverse covariance matrix of multi-variate Gaussian distribution.

EDIT: Both $\Theta_k$ and $\Omega$ are sparse, if that helps.

  • What is known about $\Theta_{k}$? Are they, perhaps, of low rank? – Felix Goldberg Nov 27 '12 at 0:03
  • There might not always be a positive definite solution; are you ok with negative solutions? – Suvrit Nov 27 '12 at 1:00
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    How closed form would you want it to be? Even in the $1\times1$ case of real variables, this reduces to finding roots of polynomials, for which there is no closed form past degree 5. – Ralph Furmaniak Nov 27 '12 at 1:43
  • Felix, $\Theta_k$ are not necessarily low rank, but they are sparse matrices. And I just edited the question for this information. – Wei Liu Nov 27 '12 at 3:44
  • Hi Suvrit, by definition, $\Omega$ is the inverse covariance matrix of multivariate Gaussian. So I need it to be positive definite. – Wei Liu Nov 27 '12 at 3:47
up vote 5 down vote accepted

Here is a partial solution to the first question in the original post. Let's look at the equation \begin{equation}\tag{1} \sum\nolimits_{i=1}^m (X+ \Theta_i)^{-1} = Q. \end{equation}

Lemma (Existence). If all $\Theta_i$ are (strictly) positive definite, then (1) has a positive semidefinite solution only if $Q \preceq \sum_i \Theta_i^{-1}$.

Proof. Suppose $Q=\sum_i \Theta_i^{-1}$, then clearly $X=0$ is the solution. Since, $(X+\Theta_i)^{-1} \preceq \Theta_i^{-1}$ for any $X \succeq 0$, on summing up we see that $Q \preceq \sum_i \Theta_i^{-1}$ must hold. Moreover, in this case if there is a solution, then it must be strictly positive definite. A little extra argument shows that in this case, there must exist a unique positive definite solution.

This lemma shows that in case $Q$ does not satisfy the requirement, the original equation has no solution, and it might be preferable to minimize $\|\sum_i (X+\Theta_i)^{-1}-Q\|_F^2$ instead.

Lemma (Bounds). Any feasible solution to (1) must lie in the set $\Omega := [0, mQ^{-1}]$.

Proof. The lower bound $X \succeq 0$ is obvious. Following an argument similar to the previous lemma, we see that $Q=\sum_i (X+\Theta_i)^{-1} \preceq \sum_i X^{-1}$, which implies that $m X^{-1} \succeq Q$, or equivalently, $X \preceq m Q^{-1}$.

Idea Now that we have a compact set $\Omega$, we just need to setup a strictly contractive nonlinear map $G : \Omega \to \Omega$. I have not proved strict contraction of the map below, but numerically it seems to work. As one might suspect from the above lemmas, the rate of convergence depends on $\|Q-\sum_i \Theta_i^{-1}\|$, so that for small values of this quantity, the iteration converges more slowly.

Suppose, that $X \succ 0$. Denote by $S^{++}$ the set of $n\times n$ strictly positive definite matrices. Then, define the nonlinear map $\mathcal{G} : S^{++} \to S^{++}$ as \begin{equation*} \mathcal{G} = X \mapsto X^{1/2}\left(\sum\nolimits_{i=1}^m Q^{-1/2}(X+\Theta_i)^{-1}Q^{-1/2}\right)X^{1/2}. \end{equation*}

TODO If I get time, I might think about proving that the above map generates convergent solutions. Or one can come up with some other fixed point iteration.

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    I think that the technique in doi:10.1088/0951-7715/21/4/011 and doi:10.1016/j.laa.2008.10.034 can be applied to this problem. Short summary: composing sums, inversions and similarities should yield a contraction in the Finsler logarithmic metric on positive definite matrices; therefore, your matrix equation can be rewritten as $X=f(X)$ with $f$ a Finsler contraction; bingo. – Federico Poloni Nov 27 '12 at 9:08
  • I would have written a Thompson metric contraction, but the presence of the $X^{1/2}$ in my iteration prevents that. Perhaps the papers that you cite still have a way around that (Also, it seems to not be the case that my iteration actually maps $\Omega \to \Omega$, but that is not so problematic, as we can always increase $\Omega$ in size until this gets ensured). But nevertheless, some version of the above idea can be made to work without too much difficulty (from its appearance it might be a contraction in some other less structured metric than the Finslerian class) – Suvrit Nov 27 '12 at 18:06
  • Thanks Suvrit. The answer is so helpful. I'm not sure the Q satisfy the condition you gave, but at least the fixed point iteration gives me a starting point. It looks there is a 'nonlinear semidefinite problem' but it is far less explored. I'll give your solution a try, with the sparsity property of my matrices in mind. – Wei Liu Dec 7 '12 at 17:44
  • @Wei -- as I showed above, if the $Q$ does not satisfy the conditions I mentioned, then your nonlinear equation has no solution, and you should rather solve a minimization problem---perhaps a technique similar to what I wrote above applies even here! – Suvrit Dec 10 '12 at 5:28
  • @Suvrit -- I would like to ask you if you eventually managed to prove that your proposed nonlinear mapping $\mathcal{G}$ is contractive w.r.t. a suitable metric. I ask you this because I found a kind of similarity with another nonlinear matrix mapping (see here) which I have been struggling to prove convergence for quite a long time. Thanks in advance for your help. – Ludwig Jan 6 '16 at 12:01

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