LMIs in Control/Matrix and LMI Properties and Tools/Schur Complement Lemma-Based Properties

LMI Condition edit

Consider  ,  ,  ,  ,  , and  .

There exists   such that

 

 

 

 

 

(1)

if and only if

 

 

 

 

 

(2)

Any matrix   satisfying

    +     is a solution to (1)


Consider  ,  ,  ,  , ,   and  .

There exists   such that

 

 

 

 

 

(3)

if and only if

 , and  

 

 

 

 

(4)

If two inequalities in (4) hold, then a solution to (3) is given by

 

Consider  ,  ,  , and  , where  .

There exists   such that

 

 

 

 

 

(5)

if and only if

 

 

 

 

 

(6)


Consider  ,  ,  , and  , where  .

 

 

 

 

 

(7)

implies

 


Consider  ,  ,  ,  ,  ,  , and  

LMI gives

 , and  

 

 

 

 

(8)

if and only if

 

 

 

 

 

(9)


Consider  ,  ,  ,  ,  ,  , and  .

LMI gives

 , and  

 

 

 

 

(10)

are satisfied if and only if

 

 

 

 

 

(11)


Consider  ,  ,  ,  , and  ,  , where  ,  ,  , and  .

There exists  ,  ,  , and   such that

 , and  = 

 

 

 

 

(12)

if and only if

 , and     

 

 

 

 

(13)

Proof edit

Proof for (3) edit

Necessity ((3)  (4)) comes from the requirement that the submatrices corresponding to the principle minors of (3) are negative definite

Sufficiency ((4)  (3)) is shown by rewriting the matrix inequalities of (4) in the equivalent form

 , and  

Concatenating the two matrices and choosing   gives the equivalent matrix inequality

 

 

 

 

 

(3-1)

or

    

 

 

 

 

(3-2)

which is equivalent to (3) using the Schur complement lemma.

Proof for (6) edit

the LMI in (5) can be written using the Schur complement lemma as

    

 

 

 

 

( 5-1)

  

 

 

 

 

(5-2)

 

 

 

 

 

(5-3)

Proof for (7) edit

Using the Schur complement lemma on (7) for  

Using the property     or equivalent   gives

 


External Links edit