If $A$ is negative-definite, then for a sufficiently big $k>0$ the eigenvalues of $M = kA + B$ are all with negative real part?

Question

I want to prove the next statement:

"If $A$ is a symmetric negative-definite matrix, then for a sufficiently big $k\in\mathbb{R}^+$, the eigenvalues of $M = kA + B$ are all with negative real part, where $B$ is just an arbitrary matrix with the appropriated dimensions. Assume further that $A$ and $B$ are real matrices."

I believe it is true since the intuition says that for a big $k$ you are making more negative the eigenvalues of $A = T kD T^{-1}$, where $T$ is just a transformation matrix and $D$ is the diagonal matrix with the eigenvalues of $A$. Since the eigenvalues are continuous functions of the elements of the matrix, $B$ can be considered as a small perturbation and then the eigenvalues of $M$ are still with negative real part.

Now I would like to prove it more rigorously and compute $k$ (it does not matter if it is conservative value). I guess one way can be looking at $M+M^T = k(A+A^T) + (B+B^T)$, if this matrix is negative-definite, then the eigenvalues of $M$ are with negative real part [1], but I do not know how to proceed, any ideas or suggestions?

Thanks in advance.

$k$ should be sufficiently big in order to make $A$ "more dominant" and see $B$ as a small perturbation. Am I right? — , May 05 '15 at 12:51
If $B$ is not symmetric, then the eigenvalues might not all be real (for any $k$) — Ben Grossmann, May 05 '15 at 12:51
If $B$ is taken to be an arbitrary (the word is arbitrary, not random) symmetric matrix, then your argument regarding $k$ is sound. — Ben Grossmann, May 05 '15 at 12:52
Thanks for your comments. I edited the question, I am asking for eigenvalues with negative real part (they can be complex indeed). — , May 05 '15 at 12:54
@Omnomnomnom, B is not symmetric, but indeed $B + B^T$ is and then we can look at $M+M^T$. What do you mean by "the argument regarding $k$ is sound?" — , May 05 '15 at 12:58
alternatively, it suffices to find a $k$ such that for real vectors $x$, $ x^TMx \leq 0$ — Ben Grossmann, May 05 '15 at 13:11
This second approach works if both $A$ and $B$ are real matrices. Is that the case here? — Ben Grossmann, May 05 '15 at 13:32
Yes, they are real matrices. Indeed Weyl's i equality provides a bound for $k$. Is the second approach less conservative? — , May 05 '15 at 13:34
It might actually provide the same bound. I'm not sure off the top of my head. — Ben Grossmann, May 05 '15 at 13:35

Ben Grossmann · Answer 1 · 2015-05-05T14:31:12.080

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We note that a real $M$ has eigenvalues with negative real part if $x^TMx < 0$ for all $x \in \Bbb R^n$ with $x \neq 0$.

We then see that $$ x^T(kA + B)x = k(x^TAx) + x^TBx $$ Let $\lambda = \lambda_{max}(A) = \min \left| \frac{x^TAx}{x^Tx} \right|$. It suffices to take $k$ so that $$ \left|k\lambda\right| > \|B\| = \sup_{x \in \Bbb R \setminus \{0\}} \frac{\|Bx\|}{\|x\|} \geq \sup_{x \in \Bbb R \setminus \{0\}} \frac{|x^TBx|}{x^Tx} $$ So, we can take $$ k > \frac{\|B\|}{|\lambda|} $$

edited May 05 '15 at 14:31

answered May 05 '15 at 13:41

Ben Grossmann

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Then, this bound for $k$ is independent of $A$? Interesting. – May 05 '15 at 13:47
Whoops! It shouldnt be. I need to fix that. – Ben Grossmann May 05 '15 at 13:48
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see my latest edit. – Ben Grossmann May 05 '15 at 13:52
Now I understand your argument. Thank you so much. I guess it should be $k |\lambda| > ||B||$ right? – May 05 '15 at 13:57
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@noether yes! fixed it again – Ben Grossmann May 05 '15 at 14:03
By the way, I cannot see it the first statement of you answer. Should it be M has eigenvalues with negative real part if $x^T (M+M^T) x < 0$ ? and then make your computations for $M + M^T$? – May 05 '15 at 14:16
Sorry, I forgot something else. The trick is that I'm only using real $x$. For any such $x$, $$ x^TMx = \frac 12 x^T(M + M^T)x $$ – Ben Grossmann May 05 '15 at 14:32
Then, the conservative bound would be $k > \frac{\lambda_{max}(B+B^T)}{ |\lambda_{max}(A+A^T)|}$. From here, we can conclude that If $\lambda_{max}(B+B^T) < 0$, then any $k > 0$ would be valid right? – May 05 '15 at 15:03
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@noether yes. More concisely, the set of (symmetric) negative-definite matrices form a cone. – Ben Grossmann May 05 '15 at 15:08

If $A$ is negative-definite, then for a sufficiently big $k>0$ the eigenvalues of $M = kA + B$ are all with negative real part?

1 Answers1