eigenvalues of the sum of a diagonal matrix and a skew-symmetric matrix

Question

Suppose $A$ is a skew-symmetric matrix (i.e., $A+A^{\top}=0$) and $D$ is a diagonal matrix. Under what conditions, $A+D$ is a Hurwitz stable matrix?

Answer 1

For any $x\in\mathbb{R}^n$, $x^T(A+D)x=x^TDx$. If $x^T(A+D)x<0$ for all nonzero $x$, all eigenvalues of $A+D$ have negative real parts. Consequently, if the diagonal of $D$ is negative, $A+D$ is Hurwitz stable.

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To see that for a real matrix $B$, $x^TBx<0$ for all nonzero $x$ implies the negativity of the real part of the spectrum of $B$, consider an eigenvalue $\lambda=\alpha+i\beta$ and the associated eigenvector $x=u+iv$, where $\alpha,\beta\in\mathbb{R}$ and $u,v\in\mathbb{R}^n$. We have $$ Bx=\lambda x\quad\Leftrightarrow\quad B(u+iv)=(\alpha+i\beta)(u+iv)\quad\Leftrightarrow\quad Bu=\alpha u-\beta v, \quad Bv=\beta u+\alpha v. $$ Hence $$ u^TBu+v^TBv=\alpha u^Tu-\beta u^Tv+\beta v^Tu+\alpha v^Tv=\alpha (u^Tu+v^Tv)=\alpha\|x\|_2^2. $$ Since $u^TBu+v^TBv$ is negative (at least one of the vectors $u$ or $v$ is nonzero), we have that $\alpha<0$.

Answer 2

I would like to give an answer that relies on the Lyapunov equation.

The matrix $A+D$ satisfies the following Lyapunov equation: $$(A+D)I+I(A+D)^T=2D$$ where $I$ is the identity matrix. Suppose $\lambda\in\sigma(A+D)$ and $x^*(A+D)=\lambda x^*$. Then $$Re(\lambda)=\frac{1}{2}\frac{x^*(2D)x}{x^*x}$$

1. If $D$ is negative definite, which means all the diagonal entries are negative, then $A+D$ is stable.
2. If $D$ is negative semi-definite, which means all the diagonal entries are nonpositive, then $A+D$ is stable if and only if $(A+D, D)$ is controllable.
3. If $D$ is not definite, which means the diagonal entries can be either positive, zero, or negative, then according to the above equation the matrix $A+D$ may not be stable.

Reference: book Topics in Matrix Analysis