实对角矩阵乘正交投影的特征值求解及动力系统稳定性技术问询

Alright, let's break down how to find the eigenvalues of $J = KP$ step by step—this is a common problem in stability analysis, so let's ground it in linear algebra properties we already know.

First, let's recap the given definitions to align on context:

• $K$ is an $m \times m$ real diagonal matrix.
• $P = A(A^T A)^{-1}AT$ is an orthogonal projection matrix onto the column space of $A$ (denoted $\text{Col}(A)$), where $A$ is $m \times n$ with $m \geq n$ and full column rank (a requirement for $(A^T A)^{-1}$ to exist).
• $P$ has eigenvalues $1$ (algebraic multiplicity $n$) and $0$ (algebraic multiplicity $m-n$), corresponding to $\text{Col}(A)$ and its orthogonal complement $\text{N}(A^T)$ (the null space of $A^T$), respectively.

We'll split the analysis into two clear cases: zero eigenvalues and non-zero eigenvalues.

1. Zero Eigenvalues #

Take any non-zero vector $x \in \text{N}(A^T)$. By definition of the null space, $A^T x = 0$, so $Px = A(A^T A)^{-1}AT x = 0$. Applying $J$ to $x$ gives:
$$Jx = KPx = K \cdot 0 = 0$$
This means every non-zero vector in $\text{N}(A^T)$ is an eigenvector of $J$ for the eigenvalue $0$. Since $\text{N}(A^T)$ has dimension $m-n$, the eigenvalue $0$ has geometric multiplicity $m-n$, and since we have exactly $m$ total eigenvalues (counting multiplicity), its algebraic multiplicity is also $m-n$.

2. Non-Zero Eigenvalues #

For non-zero eigenvalues $\lambda \neq 0$, there exists a non-zero vector $x$ such that $KPx = \lambda x$. Let's manipulate this equation to simplify it into a solvable form:

1. Left-multiply both sides by $P$:
   $$PKPx = \lambda Px$$
   Since $P$ is idempotent ($P^2 = P$), this reduces to:
   $$PKx = \lambda Px$$
2. Let $y = Px$. Note $y \neq 0$ (because $\lambda \neq 0$ implies $KPx \neq 0$, so $Px$ can't be zero), and $y \in \text{Col}(A)$ (since $P$ projects onto $\text{Col}(A)$). We can write $y = Az$ for some non-zero $z \in \mathbb{R}^n$.
3. Substitute $y = Az$ into $PKy = \lambda y$:
   $$P K Az = \lambda Az$$
   Replace $P$ with its definition $A(A^T A)^{-1}AT$:
   $$A(A^T A)^{-1}AT K Az = \lambda Az$$
4. Left-multiply both sides by $A^T$ (this cancels out the $A(A^T A)^{-1}$ term because $A^T A$ is invertible):
   $$A^T K Az = \lambda A^T A z$$

This is a generalized eigenvalue problem for the $n \times n$ matrices $A^T K A$ and $A^T A$. Since $A^T A$ is positive definite (thanks to $A$ having full column rank), we can convert this to a standard eigenvalue problem:

• Let $C = (A^T A)^{1/2}$ (the positive definite square root of $A^T A$).
• Substitute $z = C w$ into the equation:
  $$A^T K A C w = \lambda C^2 w$$
• Left-multiply both sides by $C^{-1}$:
  $$C^{-1} A^T K A C w = \lambda w$$

The matrix $C^{-1} A^T K A C$ is symmetric (since $K$, $A^T$, and $C$ are all symmetric), so it has $n$ real eigenvalues. These are exactly the non-zero eigenvalues of $J$, counting algebraic multiplicity.

Key Takeaways #

• $J$ has $m-n$ eigenvalues equal to $0$ (counting multiplicity).
• The remaining $n$ eigenvalues are the real solutions to the generalized eigenvalue problem $A^T K A z = \lambda A^T A z$, or equivalently the eigenvalues of the symmetric matrix $C^{-1} A^T K A C$.

内容的提问来源于stack exchange，提问作者Alexander Erlich