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Suppose the hermitian part $H$ of a complex matrix $A$ be defined by $H=\frac{A+A^\ast}{2}$ and the skew hermitian part $S$ by $S=\frac{A-A^\ast}{2}$. If the hermitian part $H$ of $A$ is negative definite, prove that $A$ is stable. Is the converse also true?

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Suppose matrix $A$ has a negative definite hermitian part $H$. For any eigenvalue $\lambda$ of $A$, there exists a non-zero eigen-vector $x$ such that:

$$A x = \lambda x \implies x^{*} A x = \lambda x^{*} x \implies x^{*} A^{*} x = ( x^{*} A x )^{*} = \lambda^{*} x^{*} x $$ Adding the $2^{nd}$ and $3^{rd}$ equalities and divide by 2, we get:

$$x^{*} H x = x^{*}(\frac{A + A^{*}}{2} )x = \frac{\lambda + \lambda^{*}}{2} x^{*} x = \Re(\lambda) x^{*}x$$

Since $H$ is negative definite, $x^{*} H x < 0$ and hence $\Re(\lambda) < 0$. This means the real parts of eigenvalues of $A$ are all negative and hence $A$ is stable.

About the converse, that is false. Consider the matrix $A$ and its Hermitian part $H$:

$$A = \begin{pmatrix}-1 & 2 & 0\\0 & -1 & 2\\0 & 0 & -1\end{pmatrix} \quad\text{ and }\quad H = \frac12(A + A^{*}) = \begin{pmatrix}-1 & 1 & 0\\1 & -1 & 1\\0 & 1 & -1\end{pmatrix}$$

$A$ is stable because its has only one eigenvalue $-1$. However, $H$ is not negative definite because its characteristic polynomial $\det(tI_3 - H) = t^3 + 3 t^2 + t - 1 $ has a positive root $\sqrt{2} - 1$.

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Thank you a lot... I am highly grateful to you..thank you again. – Sujeet Apr 23 '13 at 17:46

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