求对角矩阵与对称矩阵乘积的特征值边界
Great question! Let's break this down step by step since there's a nice connection here that simplifies things a lot.
First, let's restate the setup clearly to avoid confusion:
- $A$ is an $n \times n$ symmetric matrix with eigenvalues $\lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n$
- $D$ is an $n \times n$ diagonal matrix with positive diagonal entries $d_1, d_2, \dots, d_n$ (so $D$ is positive definite)
- $J = DA$
Key Observation: J's eigenvalues are real
Since $D$ is positive definite, we can take its square root $D^{1/2}$ (another positive definite diagonal matrix with entries $\sqrt{d_i}$). Notice that:
$$D^{-1/2} J D^{1/2} = D^{-1/2} D A D^{1/2} = D^{1/2} A D^{1/2}$$
The matrix $S = D^{1/2} A D^{1/2}$ is symmetric (because $A$ is symmetric, and $D^{1/2}$ is symmetric). Similar matrices have identical eigenvalues, so all eigenvalues of $J$ are exactly the eigenvalues of $S$—which are real numbers. That's a huge simplification!
Global Eigenvalue Bounds
Now we can derive bounds for the eigenvalues of $S$ (and thus $J$) using the known eigenvalues of $A$ and the diagonal entries of $D$.
Let $m = \min{d_1, d_2, \dots, d_n}$ (smallest diagonal entry of $D$) and $M = \max{d_1, d_2, \dots, d_n}$ (largest diagonal entry of $D$).
For any eigenvalue $\nu$ of $J$ (and $S$), the following holds:
$$\min_{1 \leq i,j \leq n} (\lambda_i d_j) \leq \nu \leq \max_{1 \leq i,j \leq n} (\lambda_i d_j)$$
Simplifying the Bound
We can rewrite this interval based on the sign of $A$'s eigenvalues:
If $A$ is positive semi-definite ($\lambda_1 \geq 0$):
The minimum bound becomes $\lambda_1 \cdot m$ (smallest eigenvalue of $A$ times smallest diagonal entry of $D$), and the maximum bound becomes $\lambda_n \cdot M$ (largest eigenvalue of $A$ times largest diagonal entry of $D$).
$$\lambda_1 m \leq \nu \leq \lambda_n M$$If $A$ is negative semi-definite ($\lambda_n \leq 0$):
The minimum bound becomes $\lambda_1 \cdot M$ (most negative eigenvalue of $A$ times largest diagonal entry of $D$), and the maximum bound becomes $\lambda_n \cdot m$ (least negative eigenvalue of $A$ times smallest diagonal entry of $D$).
$$\lambda_1 M \leq \nu \leq \lambda_n m$$If $A$ has both positive and negative eigenvalues:
The minimum bound is $\lambda_1 \cdot M$ (most negative eigenvalue times largest $d_j$) and the maximum bound is $\lambda_n \cdot M$ (most positive eigenvalue times largest $d_j$).
$$\lambda_1 M \leq \nu \leq \lambda_n M$$
Why This Bound Holds
To see why this works, consider the Rayleigh quotient for $S$: for any unit vector $x$,
$$\nu = x^T S x = x^T D^{1/2} A D^{1/2} x = y^T A y$$
where $y = D^{1/2}x$. The value $y^T A y$ is a weighted average of $A$'s eigenvalues, with weights proportional to the squared components of $y$ relative to $A$'s eigenvectors.
Since $|y|^2 = x^T D x$, which ranges between $m$ (when $x$ is aligned with the smallest $d_i$) and $M$ (when $x$ is aligned with the largest $d_i$), the weighted average $y^T A y$ can't be smaller than the smallest possible combination of $\lambda_i$ and $d_j$, nor larger than the largest possible combination.
Tightness of the Bound
This bound is tight when $A$ and $D$ commute (i.e., $A$ is also a diagonal matrix). In that case, $J = DA$ is diagonal, and its eigenvalues are exactly $\lambda_i d_i$—so the minimum and maximum of these values match the bounds we derived.
For non-commuting $A$ and $D$, the eigenvalues of $J$ will lie strictly inside the interval (unless there's some alignment between $A$'s eigenvectors and $D$'s diagonal entries).
内容的提问来源于stack exchange，提问作者Peaceful
