# The comparison theorem for matrices

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.

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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

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 $\dot{Y}(t)=((1,-2t))\ge ((0,-2t))=f(Y)$.