关于非对称正定矩阵场景下约束非负对角矩阵K使AᵀA+KA+AK保持正定的边界刻画方法问询

Hey Jason, great question—this kind of structured perturbation problem for positive definiteness comes up all the time in numerical linear algebra and control theory, so I totally get how frustrating it can be to hunt for concrete characterizations! Let’s break down some actionable approaches and intuition that align with your numerical observations:

首先，简化问题到对称正定框架 #

First, let’s anchor the problem in symmetric matrix terms, since your definition of positive definiteness (PD) for non-symmetric matrices relies on the symmetric part being PD. For your target expression $A^TA + KA + AK$:

• $A^TA$ is already symmetric positive definite (since $A = CX + I$ is invertible—its symmetric part is PD, so no non-zero $x$ satisfies $Ax=0$).
• The symmetric part of $KA + AK$ is $\frac{1}{2}(KA + AK + A^TK + KA^T) = K \cdot \text{sym}(A) + \text{sym}(A) \cdot K$, where $\text{sym}(A) = \frac{A+A^T}{2}$ is PD (you’ve fixed $C$ to ensure this).

So your problem reduces to ensuring the symmetric matrix:
$$S = A^TA + K \cdot \text{sym}(A) + \text{sym}(A) \cdot K$$
is PD for non-negative diagonal $K$. This is a key simplification—now we’re working entirely with symmetric matrices, which have well-understood PD conditions.

思路1：利用Rayleigh商的保守充分条件（快速验证） #

A quick but conservative sufficient condition comes from bounding the Rayleigh quotient of $S$. For any non-zero $x$:
$$x^TSx = x^TATAx + x^T(K\text{sym}(A) + \text{sym}(A)K)x$$
We know $x^TATAx \geq \lambda_{\text{min}}(A^TA) |x|^2 > 0$. For the second term, using Cauchy-Schwarz:
$$|x^T(K\text{sym}(A) + \text{sym}(A)K)x| \leq 2|K| \cdot |\text{sym}(A)| \cdot |x|^2$$
So a sufficient condition is:
$$|K| < \frac{\lambda_{\text{min}}(A^TA)}{2|\text{sym}(A)|}$$
This is easy to compute, but as you saw in experiments, it’s way too strict—it doesn’t account for the structured nature of $K$ (diagonal) or $\text{sym}(A)$.

思路2：对角占优与M矩阵性质（匹配你的数值观察） #

Your numerical result that $K$ stays PD when its entries don’t differ too much aligns perfectly with diagonal dominance arguments. Here’s why:

• $\text{sym}(A)$ is a symmetric PD M-matrix (since $X$ is a symmetric M-matrix with non-positive off-diagonal entries, $C$ is positive diagonal, so $\text{sym}(A)$ inherits non-positive off-diagonals and PD from $X$ and the identity term).
• For $S = A^TA + K\text{sym}(A) + \text{sym}(A)K$:
  • Diagonal entries: $S_{ii} = |A_{i·}|^2 + 2k_i \cdot \text{sym}(A){ii}$ (all positive, since $\text{sym}(A){ii} > 0$ and $k_i \geq 0$).
  • Off-diagonal entries: $S_{ij} = (A^TA){ij} + \text{sym}(A){ij}(k_i + k_j)$ (non-positive if $(A^TA){ij}$ isn’t large enough, since $\text{sym}(A){ij} \leq 0$).

If we enforce strict diagonal dominance on $S$ (a sufficient condition for PD), we get:
$$S_{ii} > \sum_{j \neq i} |S_{ij}|$$
Rearranging terms to isolate $k_i$ and $k_j$, this simplifies to a constraint on the ratio of $K$’s maximum and minimum entries. Let $k_{\text{max}} = \max{k_i}$, $k_{\text{min}} = \min{k_i}$:
$$\frac{k_{\text{min}}}{k_{\text{max}}} > \max_i \left( \frac{\sum_{j \neq i} |\text{sym}(A){ij}|}{2\text{sym}(A){ii} - \sum_{j \neq i} |\text{sym}(A)_{ij}|} \right)$$
This is exactly the "entries can’t be too different" pattern you observed! The right-hand side depends only on $\text{sym}(A)$’s structure, so you can precompute it for your fixed $C$ and $X$.

思路3：参数化扰动与特征值跟踪（精确边界） #

If you need tighter bounds, parameterize $K$ as $tD$ where $D$ is a fixed non-negative diagonal matrix (e.g., $D$ could be a matrix with one entry 1 and others 0, or a scaled identity). Then you can find the maximum $t_{\text{max}} \geq 0$ such that $S(t) = A^TA + t(D\text{sym}(A) + \text{sym}(A)D)$ remains PD.

Since $S(t)$’s eigenvalues are continuous in $t$, you can use numerical methods (like inverse iteration or eigenvalue solvers) to track when the smallest eigenvalue hits zero. For general $K$, you can combine results from multiple parameterized $D$ matrices to build a convex region of valid $K$ entries.

直觉与后续建议 #

• The core idea is that $A^TA$ provides a "base" of positive definiteness. Extreme $K$ (one entry much larger than others) distorts $S$ by amplifying the negative off-diagonal entries from $\text{sym}(A)$, which can cancel out the positive diagonal terms.
• For references, look into:
  • Perturbation theory for PD matrices in Matrix Computations (Golub & Van Loan)
  • M-matrix properties and diagonal dominance in Nonnegative Matrices in the Mathematical Sciences (Berman & Plemmons)
  • Papers on structured perturbations of Lyapunov/Sylvester equations (they often deal with similar PD constraints)

备注：内容来源于stack exchange，提问作者jason