Take symmetric and real matrices F, f and f' such that $F \geq f$ and $F>f'$. Here $F \geq f$ means that $F-f$ is positive semi-definite, and $F>f'$ means that $F-f'$ is positive definite. I want to show that if the inequality $F \geq f$ is sharp (defined below), then this means that $f>f'$.

$F \geq f$ is sharp means that for any matrix $A$ such that $F>A$, we must have $A \ngeq f$ (I'm not 100% sure of this definition). So if we take $A=f'$, we must have that $f'\ngeq f$.

So in effect I want to show that $F \geq f$, $F>f'$ and $f'\ngeq f$ together imply that $f>f'$.

Alternatively, it would be just as good to show that $F \geq f$, $F \geq f'$ and $f'\ngeq f$ together imply that $f \geq f'$.

As anther alternative, it would be almost as good to show that $F \geq f$, $F \geq f'$ and $f'\ngeq f$ together imply that $tr(f) \geq tr(f')$ where $tr$ is the trace.

Note that these are matrix inequalities, so $f'\ngeq f$ alone does not imply that $f'< f$.


  • $\begingroup$ Hello and welcome to math.stackexchange. That's a nice question, and thank you for posting your prior work on this. As to the problem: Consider $F = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$, $f = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$ and $f' = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. Which of your assumptions are satisfied? Is the conclusion true? $\endgroup$ – Hans Engler Jul 22 '16 at 15:13
  • $\begingroup$ Hi Hans. Thanks for your quick response and for this counter example! For your 3 matrices it looks like $F \geq f$, $F \geq f'$, and $f'-f =\begin{bmatrix}-1&0\\0&1\end{bmatrix}$ which is not positive semi-definite. So all three assumptions are satisfied. However, $f \ngeq f'$, so I want to show is not true! $\endgroup$ – Paul K Jul 25 '16 at 7:26
  • $\begingroup$ From your counter example I can make a simple change to show that $tr(f) \ngeq tr(f')$ either, by using matrices $F=\begin{bmatrix}4&0\\0&4\end{bmatrix}$, $f=\begin{bmatrix}4&0\\0&1\end{bmatrix}$, and $f'=\begin{bmatrix}3&0\\0&3\end{bmatrix}$. Here the assumptions are still satisfied but $tr(f)=5 < tr(f')=6$. Thanks for your help. $\endgroup$ – Paul K Jul 25 '16 at 7:34

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