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Consider a sparse $N \times N$ matrix $P$ with at most $2N$ nonzero entries and $\text{diag}(P) = 1$. Define the norm $|P|_1$ as the sum of the absolute values of all entries in $P$, i.e., $|P|_1=\sum_{i, j}\left|P_{i j}\right|$.

For which matrices $P$ does the inequality $|UP|_1 \geq |P|_1$ hold for every unitary matrix $U$, while ensuring that $\text{diag}(UP) = 1$?

Here, $|UP|_1$ is calculated as $\sum_{i, j}\left|(UP)_{i j}\right|$, where $(UP)_{i j}$ is the $(i, j)$ entry of the matrix product $UP$, and can also be expressed as $\sum_{i, j}\left|u_i^T p^j\right|$. (For a matrix $X$, let $x_i$ and $x^j$ denote the $i$-th row and $j$-th column of $X$, respectively.)

Background:

In my research, I developed an algorithm to estimate $P$, and it generally converges correctly (I'm sorry details and code cannot be shared). And if my algorithm can find the correct solution, this problem is a corollary. The algorithm's effectiveness suggests that there might be a broad class of matrices $P$ for which this condition holds.

Empirically, when $P$ is initialized with $N$ off-diagnal elements randomly chosen within the range from -1 to 1 , convergence is fast. However, when the elements are all set to 1 , convergence typically fails. I'm looking to understand what specific properties or characteristics of $P$ ensure the condition $\|U P\|_1 \geq\|P|_1$ is always met.

Update:

Considering a simplified scenario where all entries of $P$ are positive might provide a useful starting point. When analyzing $\|UP\|_1 \geq \|UP_1\|_1 = \|Up\|_1$, with $P_1$ defined as a column vector of all ones, we examine $\|Up\|_1$ by evaluating the sum of the magnitudes of the projections of $U$'s row vectors onto $p$. Here, $p = P_{1,1}$ can be any column vector, and we express:

$$ \|Up\|_1 = \sum_i |u_i p| = \sum_i \|u_i\|_2 \|p\|_2 \cos \theta_i, $$

where $\cos \theta_i$ is the angle between $p$ and $u_i$. This formulation simplifies to the minimum at $\|p\|_2$, but unfortunately, $\|p\|_2$ is not necessarily greater than $\|P\|_1$. This observation suggests that the initial diagonal and sparsity constraints might need to be revisited or revised in the theoretical model.

Update 8.2

To confirm the correctness of the problem, I have developed a program. Dealing with the constraint of unitarity was challenging. My approach involved projecting a non-unitary matrix onto the space of unitary matrices, which is an application of the Orthogonal Procrustes problem.

The code is iterative, ensuring that matrix $U$ ultimately remains unitary while also maintaining that $\operatorname{diag}(UP) = 1$.

For those interested in reviewing or utilizing the methodology, the code can be accessed through the following link: Code.

I welcome any feedback or suggestions on the implementation and the theoretical approach.

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  • $\begingroup$ Some new update. diag(UP)=1 might be redundant $\endgroup$
    – silver
    Commented Aug 1 at 23:32
  • $\begingroup$ Are there results in literature for $\min_{U \in U(n)}|UP|_1$? Have you tried solving for the critical points of the function $f(U) = |UP|_{1}$ on the manifold $U(n)$? You can get some equations for setting derivative to $0$ that may help. At worst, you can do numerical optimization. $\endgroup$
    – Mason
    Commented Aug 2 at 1:12
  • $\begingroup$ @Mason I haven't explored $\min _{U \in U(n)}|U P|_1$. Actually, I don't know how to solve for the critical points you mentioned. Can you provide more details? Yes, that's what I'm currently working on. I am mainly doing numerical optimization, and my program converges to the correct results in almost all cases, so I guess there must be some broader mathematical conclusion. $\endgroup$
    – silver
    Commented Aug 2 at 1:28
  • $\begingroup$ Let $U$ be a minimizer. The critical point equation is $df(U) = 0$, which here means $\sum_{j, k = 1}^{N}\text{sgn}(|(UP)_{j,k}|)|(AP)_{j, k}| \ni 0$ for all $A$ with $A^T = -A$. Oh this is assuming you are working with real matrices only. For complex matrices it will be different. $\endgroup$
    – Mason
    Commented Aug 2 at 1:51
  • $\begingroup$ @Mason Thanks for your help. Yes. I'm only working with real matrices. Do you mean this $\sum_{j, k=1}^N \operatorname{sgn}\left((U P)_{j, k}\right)\left|(A P)_{j, k}\right| \ni 0$? what does $\ni$ here mean? Do you mean there exist A to let LHS=0 $\endgroup$
    – silver
    Commented Aug 2 at 2:09

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