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Matrices I discuss are all $N\times N$ hermitian matrices. Define two positive (semi)definite matrices $H_1$ and $H_2$. Define the following matrices

\begin{align} P_1&=H_1+(I+H_2)^{-1} \\\ P_2&=(I+H_1)^{-1}+H_2 \end{align}

I was just curious if there are any connections between them. It can be from any perspective, eigenvalues, rank, eigenbasis, simultaneous diagonalization or any such concept. Even special cases are welcome, for instance, say they are rank-one matrices, Does it make any difference?

Context: This arises as a (very) special case of problem formulation in multiuser wireless communication. $H_1$ is a kind of signal Covariance matrix and $(I+H_2)$ is a kind of interfering signal's covariance matrix. if you take every parameter in the system to be one, you get $P_1$ at user1 receiver and you get $P_2$ at user2 receiver. Then this will become the constraint matrices in a homogenous convex quadratic programming problem with two non-convex homogenous quadratic constraints.

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Is there any chance you could pose a more precise question, rather than a catch-all "can anyone tell me loads about X"? –  Yemon Choi Dec 2 '12 at 7:29
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it was more of a question out of curiosity rather than necessity, so I don't know what to point out for me to ask. The question was more like "is there anything interesting about this two matrices". And the context was explained to make the point that it was not only curiosity but there was a reason that matrices had that form. –  dineshdileep Dec 2 '12 at 9:35
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Personally, I prefer MO questions that are more focused. "Is there anything interesting about X?" seems to me like idle curiosity. –  Yemon Choi Dec 2 '12 at 10:03
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Cf. the dictum "MO is not intended for requests for someone to write an encyclopaedia entry" –  Yemon Choi Dec 2 '12 at 10:04
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the two matrices $(1+P_1)(1+P_2)^{-1}$ and $(1+H_1)(1+H_2)^{-1}$ have the same eigenvalues. –  Carlo Beenakker Dec 2 '12 at 19:32
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1 Answer

up vote 10 down vote accepted

I assume that the problem is to try to determine which pairs $(P_1,P_2)$ of positive definite Hermitian symmetric $N$-by-$N$ matrices can be written in the above form for some pair $(H_1,H_2)$ of positive semi-definite Hermitian symmetric $N$-by-$N$ matrices.

Let $\mathcal{H}_N$ denote the set of Hermitian symmetric $N$-by-$N$ matrices, and let $\mathcal{P}_N\subset \mathcal{H}_N$ denote the open convex cone consisting of the positive definite ones, with $\mathcal{K}_N = \overline{\mathcal{P}_N}$ denoting its closure, i.e., the set of positive semidefinite ones. Finally, let $\mathcal{J}_N\subset\mathcal{H}_N$ denote the convex open set consisting of the elements $H\in \mathcal{H}_N$ such that $I{+}H$ lies in $\mathcal{P}_N$. Note that $\mathcal{P}_N\subset \mathcal{K}_N\subset \mathcal{J}_N$. Consider the smooth map $F:\mathcal{J}_N\times \mathcal{J}_N\to \mathcal{P}_N\times \mathcal{P}_N$ defined by $$ F(H_1,H_2) = \bigl(H_1 + (I{+}H_2)^{-1}, H_2 + (I{+}H_1)^{-1}\bigr). $$ A straightforward computation shows that if $(H_1,H_2)\in\mathcal{K}_N\times\mathcal{K}_N$ satisfies $H_1{+}H_2\in \mathcal{P}_N$, then $F'(H_1,H_2):\mathcal{H}_N{\oplus}\mathcal{H}_N\to \mathcal{H}_N{\oplus}\mathcal{H}_N$ is an isomorphism, so, in particular, $F$ is a local diffeomorphism on a neighborhood of $(H_1,H_2)$. In particular, the mapping $$ F:\mathcal{P}_N\times \mathcal{P}_N\to \mathcal{P}_N\times \mathcal{P}_N $$ is a local diffeomorphism, so $F\bigl(\mathcal{P}_N\times\mathcal{P}_N\bigr)$ is an open subset of $\mathcal{P}_N\times \mathcal{P}_N$. Thus, there cannot be any algebraic relations implied between $P_1$ and $P_2$ in order that there exist an $(H_1,H_2)\in\mathcal{K}_N\times\mathcal{K}_N$ so that $F(H_1,H_2) = (P_1,P_2)$. The most one can hope for is inequalities relating $P_1$ and $P_2$ that might characterize the image $F\bigl(\mathcal{K}_N\times\mathcal{K}_N\bigr)$ in $\mathcal{P}_N\times\mathcal{P}_N$.

One inequality is very easy, and turns out to be quite useful: Note that $$ \begin{align} P_1 &= H_1 + (I{+}H_2)^{-1} = I + H_1 - H_2 + {H_2}^2(I{+}H_2)^{-1}\\\\ P_2 &= H_2 + (I{+}H_1)^{-1} = I + H_2 - H_1 + {H_1}^2(I{+}H_1)^{-1}, \end{align} $$ so, since ${H_i}^2(I{+}H_i)^{-1}$ is positive semidefinite for $i=1,2$, we have $$ P_1{+}P_2 = 2I + {H_1}^2(I{+}H_1)^{-1} + {H_2}^2(I{+}H_2)^{-1} \ge 2I. $$ In particular, $P_1{+}P_2{-}2I\ge 0$, so this is a necessary inequality. Moreover, if $K\subset\mathbb{C}^N$ is the kernel of $P_1{+}P_2{-}2I$, then $K$ is also the set of vectors annihilated by both $H_1$ and $H_2$, so $P_1{-}I$ and $P_2{-}I$ must also vanish on $K$. In particular, if $W\subset\mathbb{C}^N$ is the orthogonal complement to $K$, then each $P_i$ preserves $W$, and we can restrict everything to $W\simeq\mathbb{C}^n$ for some $n\le N$, which reduces us to the case in which $P_1{+}P_2{-}2I > 0$, which, in turn, implies that $H_1{+}H_2>0$, so I will assume this from now on.

In fact, one can almost solve for $H_1$ and $H_2$. Set $M=(I{+}H_1)(I{+}H_2)$ and note that the equations imply $$ (I{+}P_1)(I{+}P_2) = 2I + M + M^{-1}. $$ Setting $Q = \tfrac12(P_1P_2{+}P_1{+}P_2-I)$, one gets $M{+}M^{-1} = 2Q$, so that $Q$ must commute with $M$, which allows the above equation to be written in the form $$ (M{-}Q)^2 = Q^2 - I. $$ In particular, $Q^2-I$ must be a square, which is a nontrivial condition on $Q$. (By the way, neither $M$ nor $Q$ is Hermitian symmetric, a priori.) If $Q^2-I$ has $n$ distinct nonzero eigenvalues (which is 'generic'), then it has exactly $2^n$ 'square roots'. If $Q^2-I$ has multiple eigenvalues, then it may have positive dimensional families of square roots. In any event, let $(Q^2{-}I)^{1/2}$ denote one of the square roots of $Q^2{-}I$. Then $M$ must satisfy $M = Q + (Q^2{-}I)^{1/2}$ for some one of these square roots. One then has $$ (I{+}P_1)(I{+}H_2) = M + I = I + Q + (Q^2{-}I)^{1/2}, $$ so $$ H_2 = (I{+}P_1)^{-1}\bigl(I + Q + (Q^2{-}I)^{1/2}\bigr) - I = \tfrac12(P_2{-}I) + (I{+}P_1)^{-1}(Q^2{-}I)^{1/2}\ , $$ Now we see a necessary condition on the pair $(P_1,P_2)$, which is that the expression on the right hand side of the above equation must be a positive semidefinite matrix for some choice of the square root. One gets a similar condition when one solves for $H_1$. These are the necessary and sufficient conditions that characterize the image $F\bigl(\mathcal{K}_N\times\mathcal{K}_N\bigr)$.

Remark about square roots: I thought a bit about how to make the solution more algorithmic, and remembered that, if $\mathcal{S}_N$ is the connected open set consisting of those $N$-by-$N$ matrices without a real, nonpositive eigenvalue, then there is a canonical square root function $\sigma:\mathcal{S}_N\to \mathcal{M}_N$ ($=$ all $N$-by-$N$ complex matrices) with the property that $\sigma(A)^2 = A$ for all $A\in \mathcal{S}_N$ and that all of the eigenvalues of $\sigma(A)$ have positive real part. Using $\sigma$, we have an explicit formula for $(H_1,H_2)$ in terms of $(P_1,P_2)$, when $(P_1,P_2)$ satisfy the open condition that $Q^2{-}I$ lie in $\mathcal{S}_N$. For example, we have $$ H_2 = \tfrac12(P_2{-}I) + (I{+}P_1)^{-1}\sigma(Q^2{-}I)\ , $$ in this case. What's not completely obvious is that the expression on the right hand side is always an Hermitian symmetric matrix, but, because the second term $(I{+}P_1)^{-1}\sigma(Q^2{-}I)$ is an analytic expression in $(P_1,P_2)$ on the (nonempty) open set in $\mathcal{P}_N{\times}\mathcal{P}_N$ for which $Q^2{-}I$ lies in $\mathcal{S}_N$ and because this open set clearly has nontrivial intersection with the set $F\bigl(\mathcal{P}_N{\times}\mathcal{P}_N\bigr)$ where this expression is Hermitian symmetric, it follows that it must be Hermitian symmetric whenever $Q^2{-}I$ lies in $\mathcal{S}_N$. Thus, I believe this gives us an 'explicit' inversion of $F$ on a large open set in the image of $F$. (I put 'explicit' in quotes because you may not feel that $\sigma$ has been 'explicitly' defined.)

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it took me two days to figure out what you have written down!!. Thanks for the excellent answer. It gave me different perspectives to attack the problem. –  dineshdileep Dec 6 '12 at 11:59
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