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I have the following linear matrix inequality:

$$F^T P + PF < 0,$$

where $P$ is a positive definite matrix and $F$ is a matrix with appropriate dimension.

Let $Q$ be a positive definite matrix that obeys $Q \leq P$. Is this equivalence valid?

$$F^TP+PF< 0 \iff F^TQ+QF< 0$$

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Here is a counterexample to what I think is being asked.

In this counterexample, $P$, $Q$, and $P - Q$ are all symmetric positive definite, $F^TP+PF$ is negative definite, and $F^TQ+QF$ is indefinite, as seen below.

>> disp(P)
    30    15
    15    50
>> disp(Q)
     3    -2
    -2     7
>> disp(F)
    -2    -1
     1     0
>> disp(eig(P))
   21.9722
   58.0278
>> disp(eig(Q))
    2.1716
    7.8284
>> disp(eig(P-Q))
   16.2117
   53.7883
>> disp(eig(F'*P+P'*F))
  -91.6228
  -28.3772
>> disp(eig(F'*Q+Q'*F))
  -18.8062
    6.8062

Edit: Here is a counterexample to the proposed implication in the comment to this answer by the OP, i.e., with $Q - P$ being positive definite.

>> disp(P)
    3.0384    0.3833
    0.3833    2.0377
>> disp(Q)
   11.7204    3.8318
    3.8318    4.8094
>> disp(F)
   -0.6509   -0.9444
    0.2571   -1.3218
>> disp(eig(P))
    1.9078
    3.1683
>> disp(eig(Q))
    3.1051
   13.4247
>> disp(eig(Q-P))
    1.1854
   10.2684
>>  disp(eig(F'*P+P'*F))
   -8.2518
   -1.6173
>> disp(eig(F'*Q+Q'*F))
  -34.3271
    1.0879
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  • $\begingroup$ thank you so much if i inverse the condition $Q<P$ to $P<Q$ i will have the implication $F^TP+PF<0 \implies F^TQ+QF<0$ ? $\endgroup$ Commented Sep 19, 2018 at 19:50
  • $\begingroup$ @khallouq abdelmounaim I just added a counterexample to that proposed implication. $\endgroup$ Commented Sep 19, 2018 at 20:07
  • $\begingroup$ is there any solution for that implication ? under what condition can it be valid ? $\endgroup$ Commented Sep 19, 2018 at 20:45

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