对称Hurwitz矩阵下Lyapunov方程等价改写的合理性问询

Great question—let’s break this down clearly, because the confusion here comes from mixing up two key points: the context of the equation (i.e., $R$ being the solution to the Lyapunov equation) and general matrix multiplication rules.

Short Answer #

The rewrite is valid if $R$ is the symmetric positive definite solution to the original Lyapunov equation (given $A$ is symmetric Hurwitz). However, it’s not valid for arbitrary symmetric positive definite $R$—which is likely where your "counterexample" comes from.

Detailed Explanation #

Let’s start with the basics:

• Since $A$ is symmetric, $A^T = A$, so the original equation simplifies to:
  $$AR + RA = -I$$
• Matrix multiplication is not commutative in general, so $AR \neq RA$ for most pairs of matrices. But in this specific case, the solution $R$ to the Lyapunov equation does commute with $A$, which makes the rewrite valid.

Why $R$ Commutes with $A$ (for Symmetric $A$) #

Symmetric matrices are orthogonally diagonalizable: we can write $A = Q\Lambda Q^T$, where $Q$ is an orthogonal matrix ($Q^TQ = I$) and $\Lambda$ is a diagonal matrix of $A$’s eigenvalues (all negative, since $A$ is Hurwitz).

Substitute this into the Lyapunov equation:
$$Q\Lambda Q^T R + R Q\Lambda Q^T = -I$$
Left-multiply both sides by $Q^T$ and right-multiply by $Q$:
$$\Lambda (Q^T R Q) + (Q^T R Q) \Lambda = -I$$
Let $S = Q^T R Q$. Since $R$ is symmetric, $S$ is also symmetric. The equation becomes:
$$\Lambda S + S \Lambda = -I$$

Now, since $\Lambda$ is diagonal, let’s look at the entries of this equation:

• For diagonal entries: $\lambda_i s_{ii} + s_{ii} \lambda_i = -1 \implies 2\lambda_i s_{ii} = -1 \implies s_{ii} = -\frac{1}{2\lambda_i}$ (valid because $\lambda_i < 0$, so $s_{ii} > 0$)
• For off-diagonal entries: $\lambda_i s_{ij} + s_{ij} \lambda_j = 0 \implies s_{ij}(\lambda_i + \lambda_j) = 0$. Since all $\lambda_i$ have negative real parts, $\lambda_i + \lambda_j \neq 0$, so $s_{ij} = 0$ for $i \neq j$.

This means $S$ is a diagonal matrix. Since $\Lambda$ and $S$ are both diagonal, they commute: $\Lambda S = S \Lambda$. Translating back to $R$:
$$AR = Q\Lambda Q^T Q S Q^T = Q\Lambda S Q^T = Q S \Lambda Q^T = Q S Q^T Q \Lambda Q^T = RA$$

So $AR = RA$, which means the original equation simplifies to:
$$2RA = -I$$

Why Your "Counterexample" Might Seem to Exist #

If you picked an arbitrary symmetric positive definite $R$ (not the solution to the Lyapunov equation), then $AR + RA$ will almost never equal $2RA$—because most matrices don’t commute. But the claim in question is specifically about the $R$ that satisfies the original Lyapunov equation, not any random $R$.

Example to Verify #

Take a symmetric Hurwitz matrix: $A = \begin{pmatrix} -2 & 1 \ 1 & -2 \end{pmatrix}$ (eigenvalues $-1$ and $-3$). The solution to $AR + RA = -I$ is:
$$R = \begin{pmatrix} \frac{1}{3} & \frac{1}{6} \ \frac{1}{6} & \frac{1}{3} \end{pmatrix}$$
Check $2RA$:
$$2 \begin{pmatrix} \frac{1}{3} & \frac{1}{6} \ \frac{1}{6} & \frac{1}{3} \end{pmatrix} \begin{pmatrix} -2 & 1 \ 1 & -2 \end{pmatrix} = 2 \begin{pmatrix} -\frac{1}{2} & 0 \ 0 & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix} = -I$$
Which matches the original equation.

Key Takeaway #

The rewrite is valid for the specific $R$ that solves the Lyapunov equation when $A$ is symmetric Hurwitz. The confusion arises if you test arbitrary $R$—but those $R$s don’t satisfy the original equation in the first place.

内容的提问来源于stack exchange，提问作者MrYouMath