# value of $\gamma$ in the H-infinity norm

Suppose I have the system:

$$\dot{x} = Ax+Bu$$

$$y=Cx+Du$$

and the following Hamiltonian matrix:

$$H=\begin{pmatrix} A & \frac{1}{2}B^TB\\ -CC^T&-A \end{pmatrix}$$

I want to find the value of $$\gamma$$ which is the bound of the $$H_{\infty}$$ norm, so it is the value such that $$\left |T(j\omega) \right |_{\infty }<\gamma$$.

I know that for the bounded real lemma, if the eigenvalues of $$A$$ have negative real part, and $$I\gamma^2-DD^T>0$$, then the Hamiltonian have no eigenvalues on the imaginary axis. I also know that if the eigenvalues of $$A$$ have negative real part, then $$\left |T(j\omega) \right |_{\infty }<\gamma$$ holds.

But my question is : how do I find the value of $$\gamma$$?

I have been told the result is $$\gamma=0.5$$ but I really can't get to this result. I have tried using the Shur's complement to see if this matrix is negative definite (so before doing that i switched sign to the Hamiltonian). In this way i thiught that if the A matrix is negative definite, it has all eigenvalues with negative real part, so the resulting value of $$\gamma$$ form the computation would have been the searched value. But I don't find the desired result. Maybe I am missing a point and doing something wrong, or maybe I am complitely on the wrong path.

• You may have better luck if you migrate this question to the computational science group scicomp.stackexchange.com or the maths stack. Yours is essentially a linear algebra problem and there may be experts in those groups. Also, you may want to define the operator $T$. Sep 25, 2019 at 20:32
I believe you should take a look of the real bounded lemma. Because the hamiltonian matrix has another shape. I quote the book from Zhou - Essentials of Robust Control p.238. $$H=\begin{bmatrix} A+BR^{-1}D^{\top}C & BR^{-1}B^{\top} \\ -C^{\top}(I+DR^{-1}D^{\top})C & -(A+BR^{-1}D^{\top}C)^{\top} \end{bmatrix} \$$ where $$R=\gamma^2I-D^{\top}D$$.
I would compare both hamiltionian matrixes and try to find $$\gamma$$.