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I am having a linear algebra problem here. I will be grateful if someone can help me.

Let $A\in \mathbb{R}^{n\times n}$ be Hurwitz and diagonizable, and let $B$ be a diagonal matrix whose diagonal elements are non-negative. Is $A-B$ still Hurwitz?

I know that if $B=cI$, where $c$ is a positive scalar, $A-B$ is a Hurwitz matrix. However, I am not sure whether $A-B$ is still a Hurwitz matrix when some diagonal elements in $B$ are zero and the others are positive. Are there any general results on the similar topic?

Thanks in advance!

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  • $\begingroup$ Yes. Sorry about the unclear statement. I mean that all the eigenvalues of $A$ have negative real part. $\endgroup$ – MLearner Jul 17 '16 at 1:32
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No. Random counterexample: $$ A=\pmatrix{-4&3\\ -2&1},\ B=\pmatrix{3&0\\ 0&0},\ A-B=\pmatrix{-7&3\\ -2&1} $$ The eigenvalues of $A$ are $-2$ and $-1$; hence $A$ is Hurwitz and diagonalisable over $\mathbb R$. However, $A-B$ is a $2\times2$ real matrix with negative determinant, so it has exactly one positive eigenvalue and one negative eigenvalue.

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  • $\begingroup$ Thanks for your help! For your example, if $B=\left[\begin{array}{cc}0& 0\\0 & b\end{array}\right]$, where $b>-0.5$, $A-B$ will always be Hurwitz stable. I suppose for other cases there exist similar conditions to ensure Hurwitz. It seems that even if no general condition exists but a relaxed version of the statement under study can be explored case by case. $\endgroup$ – MLearner Jul 17 '16 at 13:56
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If $A$ is not only Hurwitz, but also symmetric, then it is negative definite and, thus, $-A$ is positive definite. Let

$$D := \mbox{diag} (d_1, d_2, \dots, d_n)$$

where $d_i \geq 0$, be a positive semidefinite diagonal matrix. Hence, $-(A-D) = -A + D \succ 0$ and, thus, $A - D \prec 0$. As $A-D$ is negative definite, it is also Hurwitz. We conclude that Hurwitz-ness is preserved when $A$ is symmetric.

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  • $\begingroup$ Thanks! This also gives a good perspective of the problem. Another interesting problem is to find conditions to make $A+D\prec 0$. Let $\overline{\delta}$ be the supremum of the eigenvalues of $A$, and $\overline{d}$ be the supremum of all the elements of $D$, then if $\overline{\delta}+\overline{d}<0$, the statement holds. $\endgroup$ – MLearner Jul 17 '16 at 21:40

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