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Consider the following homogeneous inequality

$$ A^{T}*P+PA\lt0, P > 0$$

where $A$ is a square given matrix. Since the inequality above is homogeneous, it can be rewriten as

$$A^{T}*P+PA<-I, P \geq I$$

Why is this possible?

Thanks

marianauc
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  • If the dimension is $1$ the first pair of inequalitites is the same as $2AP<0, P>0$, while the second is $2AP<-1, P\geq 1$. If $A=-1$, the set of solutions of the first pair is $P>0$, while the second is $P>1$. They don't look equivalent. Is there anything else that is known? – orole Jan 29 '18 at 18:44
  • Yes,

    1. The second inequality is non strictly, i.e. less than. \leq.

    2. This afford is done with the objective to solve the lyapunov equation A^{T}P +PA+I = 0 (I is the identity) also.

    3. This is valid not only for one dimension.

     – marianauc Jan 29 '18 at 19:09
  • Something that holds in general should be true in the particular cases. – orole Jan 29 '18 at 19:37

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