Mixing
Searching an analogues for Schur complement
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When I'm trying to solve a matrix inequality set of the following:
beginequation
beginarrayl
A-BC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix and $C$ is an unknown $qtimes q$ matrix.
At first look, I cannot say that this problem can be represented in a linear form (in the matrix variables $B,C$) and that the convex combination of a given two solutions for the inequalities is also a solution. But Schur complement provides a simple tool to change the inequalities into an equivalent form which is linear in the variables and thus also convex:
beginequation
beginbmatrix
A & B\
B^T & C
endbmatrix>0
endequation
My question is:
If there some analog transformation that produce a linear (in the variables) form of the following:
beginequation
beginarrayl
A-BC^-1DC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix, $C$ is an unknown $qtimes q$ matrix and $D$ is a given $qtimes q$ positive-definite matrix.
In some sense, if Schur complement help with 2'nd order multiplication, what can help with the 4'th order?
Thanks, Y.
lmis schur-complement
asked Jul 19 at 7:18
Yoav
63
add a comment |Â
up vote
1
down vote
favorite
When I'm trying to solve a matrix inequality set of the following:
beginequation
beginarrayl
A-BC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix and $C$ is an unknown $qtimes q$ matrix.
At first look, I cannot say that this problem can be represented in a linear form (in the matrix variables $B,C$) and that the convex combination of a given two solutions for the inequalities is also a solution. But Schur complement provides a simple tool to change the inequalities into an equivalent form which is linear in the variables and thus also convex:
beginequation
beginbmatrix
A & B\
B^T & C
endbmatrix>0
endequation
My question is:
If there some analog transformation that produce a linear (in the variables) form of the following:
beginequation
beginarrayl
A-BC^-1DC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix, $C$ is an unknown $qtimes q$ matrix and $D$ is a given $qtimes q$ positive-definite matrix.
In some sense, if Schur complement help with 2'nd order multiplication, what can help with the 4'th order?
Thanks, Y.
lmis schur-complement
asked Jul 19 at 7:18
Yoav
63
Wouldn't $B=0$ always be a solution?
– Kwin van der Veen
Jul 24 at 6:50
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When I'm trying to solve a matrix inequality set of the following:
beginequation
beginarrayl
A-BC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix and $C$ is an unknown $qtimes q$ matrix.
At first look, I cannot say that this problem can be represented in a linear form (in the matrix variables $B,C$) and that the convex combination of a given two solutions for the inequalities is also a solution. But Schur complement provides a simple tool to change the inequalities into an equivalent form which is linear in the variables and thus also convex:
beginequation
beginbmatrix
A & B\
B^T & C
endbmatrix>0
endequation
My question is:
If there some analog transformation that produce a linear (in the variables) form of the following:
beginequation
beginarrayl
A-BC^-1DC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix, $C$ is an unknown $qtimes q$ matrix and $D$ is a given $qtimes q$ positive-definite matrix.
In some sense, if Schur complement help with 2'nd order multiplication, what can help with the 4'th order?
Thanks, Y.
lmis schur-complement
asked Jul 19 at 7:18
Yoav
63
When I'm trying to solve a matrix inequality set of the following:
beginequation
beginarrayl
A-BC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix and $C$ is an unknown $qtimes q$ matrix.
At first look, I cannot say that this problem can be represented in a linear form (in the matrix variables $B,C$) and that the convex combination of a given two solutions for the inequalities is also a solution. But Schur complement provides a simple tool to change the inequalities into an equivalent form which is linear in the variables and thus also convex:
beginequation
beginbmatrix
A & B\
B^T & C
endbmatrix>0
endequation
My question is:
If there some analog transformation that produce a linear (in the variables) form of the following:
beginequation
beginarrayl
A-BC^-1DC^-1B^T>0\
C>0
endarray
endequation
Where $A$ is a given $ptimes p$ positive-definite matrix, $B$ is an unknown $ptimes q$ matrix, $C$ is an unknown $qtimes q$ matrix and $D$ is a given $qtimes q$ positive-definite matrix.
In some sense, if Schur complement help with 2'nd order multiplication, what can help with the 4'th order?
Thanks, Y.
lmis schur-complement
asked Jul 19 at 7:18
Yoav
63
edited Jul 19 at 9:02
asked Jul 19 at 7:18
Yoav
63
asked Jul 19 at 7:18
Yoav
63
asked Jul 19 at 7:18
Yoav
63
63
Wouldn't $B=0$ always be a solution?
– Kwin van der Veen
Jul 24 at 6:50
add a comment |Â
Wouldn't $B=0$ always be a solution?
– Kwin van der Veen
Jul 24 at 6:50
Wouldn't $B=0$ always be a solution?
– Kwin van der Veen
Jul 24 at 6:50
Wouldn't $B=0$ always be a solution?
– Kwin van der Veen
Jul 24 at 6:50
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.
Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^-1D,C^-1$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^-1 = C^-1D,C^-1$, so
$$
A - B,P^-1B^top succ 0 \
P succ 0
$$
and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C,D^-1C$, which is a very simplified version of a CARE, which is normally defined as
$$
A^top X + X,A - X,B,R^-1B^top X + Q = 0,
$$
but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C,D^-1C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.
Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^-1D,C^-1$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^-1 = C^-1D,C^-1$, so
$$
A - B,P^-1B^top succ 0 \
P succ 0
$$
and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C,D^-1C$, which is a very simplified version of a CARE, which is normally defined as
$$
A^top X + X,A - X,B,R^-1B^top X + Q = 0,
$$
but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C,D^-1C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
add a comment |Â
up vote
0
down vote
I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.
Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^-1D,C^-1$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^-1 = C^-1D,C^-1$, so
$$
A - B,P^-1B^top succ 0 \
P succ 0
$$
and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C,D^-1C$, which is a very simplified version of a CARE, which is normally defined as
$$
A^top X + X,A - X,B,R^-1B^top X + Q = 0,
$$
but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C,D^-1C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.
Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^-1D,C^-1$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^-1 = C^-1D,C^-1$, so
$$
A - B,P^-1B^top succ 0 \
P succ 0
$$
and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C,D^-1C$, which is a very simplified version of a CARE, which is normally defined as
$$
A^top X + X,A - X,B,R^-1B^top X + Q = 0,
$$
but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C,D^-1C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
I will assume that $B$ is given, otherwise $B = 0$ would be a trivial solution in which case $C$ can be any positive definite matrix.
Since $C$ has to be positive definite means that it and its inverse is full rank and symmetric, therefore $C^-1D,C^-1$ has to be positive definite as well if $D$ is positive definite. Therefore you could use the substitution $P^-1 = C^-1D,C^-1$, so
$$
A - B,P^-1B^top succ 0 \
P succ 0
$$
and solve this LMI for $P$. A corresponding positive definite solution for $C$ can then be found from $P$ by using the continuous time algebraic Riccati equation (CARE). Namely the relation between $P$ and $C$ can also be written as $P = C,D^-1C$, which is a very simplified version of a CARE, which is normally defined as
$$
A^top X + X,A - X,B,R^-1B^top X + Q = 0,
$$
but when using $A=0$, $X = C$, $B = I$, $R = D$ and $Q = P$ we again get $P = C,D^-1C$. This should have a solution, since a CARE requires that $R$ is positive definite and $Q$ is semi-positive definite, which is the case since $P$ and $D$ are positive definite.
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
edited Jul 30 at 12:37
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
answered Jul 24 at 7:41
Kwin van der Veen
4,3542826
4,3542826
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
add a comment |Â
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
Thanks for the Answer Kwin! But actually, the LMI I presented is only part of a larger set of LMI's so choosing $B=0$ isn't possible. I'm trying to find some transformation that keeps all degrees of freedom. I'm getting the feeling that there is no transformation but I'll keep looking. Thanks anyway, Y.
– Yoav
Jul 30 at 12:35
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
@Yoav Are the other LMI's only in $B$ or also in $C$?
– Kwin van der Veen
Jul 30 at 18:25
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
Yes, $B$ and $C$ are the only variables here, $A,D$ are given.\ The other LMI is for the state-feedback control problem if you are familiar with control theory.
– Yoav
Jul 31 at 7:08
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
@Yoav Can you maybe do some clever substitution for $B,C^-1$ or do $B$ and $C^-1$ appear separate from each other in the other LMI's? If they do then an approach that works might depend on your other LMI's.
– Kwin van der Veen
Jul 31 at 7:49
add a comment |Â
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Wouldn't $B=0$ always be a solution?
– Kwin van der Veen
Jul 24 at 6:50