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

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LMI Condition[edit | edit source]

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 | edit source]

Proof for (3)[edit | edit source]

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 | edit source]

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

 

 

 

 

( 5-1)

 

 

 

 

(5-2)

 

 

 

 

(5-3)

Proof for (7)[edit | edit source]

Using the Schur complement lemma on (7) for

Using the property or equivalent gives


External Links[edit | edit source]