Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let matrix $X$ satisfy a differential equation $$ \dot X = f(t,X) $$ where right side is real and symmetric. Let $X(0) = M = M^{T} \succeq 0$ and matrix $Y$ satisfy differential inequality $$ \dot Y \succeq f(t,Y), \;\;\; Y(0) = M $$ where $A \succeq B$ means that for any vector $v$ we have $v^{T}Av \geq v^{T}Bv$. Is it true that $$ Y(t) \succeq X(t) $$ for $t> 0$? Function $f$ is sufficiently smooth.

share|cite|improve this question
Is $Y$ also symmetric? If so, I'd suggest the notation $A \succeq B$ instead of $A \geq B$, which is commonly used in describing semidefinite matrices. – Erick Wong Jun 12 '12 at 23:25
@ErickWong It seems to me that Nimza's is consisten with this. Using the definition given, $A \geq B$ means exactly that $A-B$ is positive semi-definite. – user12014 Jun 12 '12 at 23:54
@PZZ: I think Erick meant that $A \succeq B$ is commonly used in describing semidefinite matrices. I'll grant that it would seem an unusual parsing of his statement, if not for the fact that I have always seen $A \succeq B$ used to denote the positive semidefiniteness of $A - B$ in optimization texts. – Rahul Jun 12 '12 at 23:58
@RahulNarain Ah, I see how it could also be interpreted that way... – user12014 Jun 13 '12 at 0:01
Sorry for the confusion: the misplaced modifier "which is..." was indeed intended for $A \succeq B$. – Erick Wong Jun 13 '12 at 0:23
up vote 4 down vote accepted

I think this is a counterexample, even with a time-independent equation in which the matrices are diagonal. Let $f((a,b))=((0,-2a))$ where $((a,b))$ means the 2 by 2 diagonal matrix with diagonal entries $a,b$. Starting with $M=((0,0))$, we have $X=((0,0))$ for all times. On the other hand, $Y(t)=((t,-t^2))$ satisfies $\dot{Y}(t)=((1,-2t))\ge ((0,-2t))=f(Y)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.