4

Let

$$ A=\begin{bmatrix} P & \alpha x\\ -y^\top & 0\end{bmatrix}$$

where $P \in \mathbb{R}^{n \times n}$ is Hurwitz (the eigenvalues of $P$ have strictly negative real parts), $x, y \in \mathbb{R}^{n}$, and $\alpha$ is a real positive scalar. Find the condition for $A$ to be Hurwitz for sufficiently small $\alpha > 0$.

I have analyzed that this is true whenever $y^\top P^{-1} x < 0$ holds. However, I am not able to prove this fact. Help in this regard would be appreciated.

Rodrigo de Azevedo
  • 23,223
neelarnab
  • 189
  • If what you state is true, then scaling $y$ by a positive scalar $\beta$ should also still give a Hurwitz matrix. From this is can also be noted that this still holds if $\alpha,\beta>0$, so when $\alpha$ and $\beta$ both are negative (so in general they have the same sign). – Kwin van der Veen Sep 25 '17 at 13:05
  • Yes. I hope that will work too. But there will be some upper bound on the product $\alpha \beta$ – neelarnab Sep 25 '17 at 13:30
  • For the scalar case it indeed does always hold, but I have yet no idea for larger $n$. Maybe a conveniently chosen similarity transformation could help. – Kwin van der Veen Sep 26 '17 at 12:29

1 Answers1

0

Since $P$ is Hurwitz it is invertible. Hence

$$ A \textrm{ is Hurwitz.} \iff \underbrace{\begin{bmatrix} I& 0 \\y^TP^{-1} & 1\end{bmatrix}}_{\det(T)\textrm{ is 1}} A = TA = \begin{bmatrix} P & \alpha x\\ 0 & \alpha y^TP^{-1}x \end{bmatrix} \textrm{ is Hurwitz.} $$ since $\alpha>0$ this leads to the condition you have given.

percusse
  • 562
  • sorry, I couldn't follow the logic in your answer. Are you finally concluding that $A$ will be Hurwitz for any positive $\alpha$ if $y^\top P^{-1}x<0$ holds? – neelarnab Dec 12 '17 at 14:22