Can I apply the Schur complement on both sides of an inequality












2












$begingroup$


I have the following matrix inequality that I need to express as an LMI



$ A^T Q^{-1} A - Q + sum_i A_i^T Q^{-1} A_i < 0$



($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)



which I rewrote to



$ A^T Q^{-1} A - Q < - sum_i A_i^T Q^{-1} A_i$



Now I want to use Schur's complement and express this inequality as



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
< sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
$



and therefore



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
- sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
< 0 $



which is a form my solver, solving for Q and B, should be able to handle.



Can I use the Schur complement in this way and are these two inequalities equivalent?






matrices proof-verification matrix-equations lmis schur-complement







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which variables are known and which variables do you want to solve for?
    $endgroup$
    – Kwin van der Veen
    Dec 1 '18 at 23:48










  • $begingroup$
    @KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 1:51










  • $begingroup$
    The matrices $A_i$ are also known as well?
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 2:55










  • $begingroup$
    Yes, they have the same structure, $X_i$ and $Y_i$ are known
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:38
















2












$begingroup$


I have the following matrix inequality that I need to express as an LMI



$ A^T Q^{-1} A - Q + sum_i A_i^T Q^{-1} A_i < 0$



($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)



which I rewrote to



$ A^T Q^{-1} A - Q < - sum_i A_i^T Q^{-1} A_i$



Now I want to use Schur's complement and express this inequality as



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
< sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
$



and therefore



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
- sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
< 0 $



which is a form my solver, solving for Q and B, should be able to handle.



Can I use the Schur complement in this way and are these two inequalities equivalent?






matrices proof-verification matrix-equations lmis schur-complement







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which variables are known and which variables do you want to solve for?
    $endgroup$
    – Kwin van der Veen
    Dec 1 '18 at 23:48










  • $begingroup$
    @KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 1:51










  • $begingroup$
    The matrices $A_i$ are also known as well?
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 2:55










  • $begingroup$
    Yes, they have the same structure, $X_i$ and $Y_i$ are known
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:38














2












2








2





$begingroup$


I have the following matrix inequality that I need to express as an LMI



$ A^T Q^{-1} A - Q + sum_i A_i^T Q^{-1} A_i < 0$



($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)



which I rewrote to



$ A^T Q^{-1} A - Q < - sum_i A_i^T Q^{-1} A_i$



Now I want to use Schur's complement and express this inequality as



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
< sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
$



and therefore



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
- sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
< 0 $



which is a form my solver, solving for Q and B, should be able to handle.



Can I use the Schur complement in this way and are these two inequalities equivalent?






matrices proof-verification matrix-equations lmis schur-complement







share|cite|improve this question











$endgroup$




I have the following matrix inequality that I need to express as an LMI



$ A^T Q^{-1} A - Q + sum_i A_i^T Q^{-1} A_i < 0$



($Q > 0$, $A = XQ+YB$, $A_i = X_i Q+Y_i B$)



which I rewrote to



$ A^T Q^{-1} A - Q < - sum_i A_i^T Q^{-1} A_i$



Now I want to use Schur's complement and express this inequality as



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
< sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
$



and therefore



$
begin{bmatrix}
-Q & A^T \
A & -Q
end{bmatrix}
- sum_i
begin{bmatrix}
0 & A_i^T \
A_i & Q
end{bmatrix}
< 0 $



which is a form my solver, solving for Q and B, should be able to handle.



Can I use the Schur complement in this way and are these two inequalities equivalent?






matrices proof-verification matrix-equations lmis schur-complement




matrices proof-verification matrix-equations lmis schur-complement






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 16:02







Alex bGoode

















asked Dec 1 '18 at 18:57









Alex bGoodeAlex bGoode

1257




1257












  • $begingroup$
    Which variables are known and which variables do you want to solve for?
    $endgroup$
    – Kwin van der Veen
    Dec 1 '18 at 23:48










  • $begingroup$
    @KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 1:51










  • $begingroup$
    The matrices $A_i$ are also known as well?
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 2:55










  • $begingroup$
    Yes, they have the same structure, $X_i$ and $Y_i$ are known
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:38


















  • $begingroup$
    Which variables are known and which variables do you want to solve for?
    $endgroup$
    – Kwin van der Veen
    Dec 1 '18 at 23:48










  • $begingroup$
    @KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 1:51










  • $begingroup$
    The matrices $A_i$ are also known as well?
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 2:55










  • $begingroup$
    Yes, they have the same structure, $X_i$ and $Y_i$ are known
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:38
















$begingroup$
Which variables are known and which variables do you want to solve for?
$endgroup$
– Kwin van der Veen
Dec 1 '18 at 23:48




$begingroup$
Which variables are known and which variables do you want to solve for?
$endgroup$
– Kwin van der Veen
Dec 1 '18 at 23:48












$begingroup$
@KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown
$endgroup$
– Alex bGoode
Dec 2 '18 at 1:51




$begingroup$
@KwinvanderVeen I have edited my question. X and Y are known, Q and B are unknown
$endgroup$
– Alex bGoode
Dec 2 '18 at 1:51












$begingroup$
The matrices $A_i$ are also known as well?
$endgroup$
– Kwin van der Veen
Dec 2 '18 at 2:55




$begingroup$
The matrices $A_i$ are also known as well?
$endgroup$
– Kwin van der Veen
Dec 2 '18 at 2:55












$begingroup$
Yes, they have the same structure, $X_i$ and $Y_i$ are known
$endgroup$
– Alex bGoode
Dec 2 '18 at 12:38




$begingroup$
Yes, they have the same structure, $X_i$ and $Y_i$ are known
$endgroup$
– Alex bGoode
Dec 2 '18 at 12:38










1 Answer
1






active

oldest

votes


















2












$begingroup$

You can't use Schur's complement in that way. Namely when you have the matrix inequalities



$$
A - B^top C^{-1} B succ 0, quad C succ 0,
$$



then you can combine this into the following one matrix inequality



$$
M =
begin{bmatrix}
A & B^top \ B & C
end{bmatrix} succ 0.
$$



This is because $M$ can be decomposed as follows



$$
M =
underbrace{
begin{bmatrix}
I & B^top C^{-1} \ 0 & I
end{bmatrix}
}_{T^top}
begin{bmatrix}
A - B^top C^{-1} B & 0 \ 0 & C
end{bmatrix}
underbrace{
begin{bmatrix}
I & 0 \ C^{-1} B & I
end{bmatrix}
}_{T}.
$$



Namely $M succ 0$ means that $x^top M,x > 0 forall, xneq0$. So when using $x=T,y$ one gets $x^top M,x = y^top T^top M,T,y$. It holds that $xneq0$ iff $yneq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^top M,T succ 0$.



But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that



$$
T^top left(
begin{bmatrix}
Q & -A^top \
-A & Q
end{bmatrix}
+ sum_i
begin{bmatrix}
0 & A_i^top \
A_i & Q
end{bmatrix}
right) T
$$



gives back the original matrix inequalities as blocks on the diagonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:40










  • $begingroup$
    @AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 12:50










  • $begingroup$
    Nevermind, you only have one $B$ and multiple $A$/$A_i$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 13:14











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

You can't use Schur's complement in that way. Namely when you have the matrix inequalities



$$
A - B^top C^{-1} B succ 0, quad C succ 0,
$$



then you can combine this into the following one matrix inequality



$$
M =
begin{bmatrix}
A & B^top \ B & C
end{bmatrix} succ 0.
$$



This is because $M$ can be decomposed as follows



$$
M =
underbrace{
begin{bmatrix}
I & B^top C^{-1} \ 0 & I
end{bmatrix}
}_{T^top}
begin{bmatrix}
A - B^top C^{-1} B & 0 \ 0 & C
end{bmatrix}
underbrace{
begin{bmatrix}
I & 0 \ C^{-1} B & I
end{bmatrix}
}_{T}.
$$



Namely $M succ 0$ means that $x^top M,x > 0 forall, xneq0$. So when using $x=T,y$ one gets $x^top M,x = y^top T^top M,T,y$. It holds that $xneq0$ iff $yneq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^top M,T succ 0$.



But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that



$$
T^top left(
begin{bmatrix}
Q & -A^top \
-A & Q
end{bmatrix}
+ sum_i
begin{bmatrix}
0 & A_i^top \
A_i & Q
end{bmatrix}
right) T
$$



gives back the original matrix inequalities as blocks on the diagonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:40










  • $begingroup$
    @AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 12:50










  • $begingroup$
    Nevermind, you only have one $B$ and multiple $A$/$A_i$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 13:14
















2












$begingroup$

You can't use Schur's complement in that way. Namely when you have the matrix inequalities



$$
A - B^top C^{-1} B succ 0, quad C succ 0,
$$



then you can combine this into the following one matrix inequality



$$
M =
begin{bmatrix}
A & B^top \ B & C
end{bmatrix} succ 0.
$$



This is because $M$ can be decomposed as follows



$$
M =
underbrace{
begin{bmatrix}
I & B^top C^{-1} \ 0 & I
end{bmatrix}
}_{T^top}
begin{bmatrix}
A - B^top C^{-1} B & 0 \ 0 & C
end{bmatrix}
underbrace{
begin{bmatrix}
I & 0 \ C^{-1} B & I
end{bmatrix}
}_{T}.
$$



Namely $M succ 0$ means that $x^top M,x > 0 forall, xneq0$. So when using $x=T,y$ one gets $x^top M,x = y^top T^top M,T,y$. It holds that $xneq0$ iff $yneq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^top M,T succ 0$.



But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that



$$
T^top left(
begin{bmatrix}
Q & -A^top \
-A & Q
end{bmatrix}
+ sum_i
begin{bmatrix}
0 & A_i^top \
A_i & Q
end{bmatrix}
right) T
$$



gives back the original matrix inequalities as blocks on the diagonal.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:40










  • $begingroup$
    @AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 12:50










  • $begingroup$
    Nevermind, you only have one $B$ and multiple $A$/$A_i$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 13:14














2












2








2





$begingroup$

You can't use Schur's complement in that way. Namely when you have the matrix inequalities



$$
A - B^top C^{-1} B succ 0, quad C succ 0,
$$



then you can combine this into the following one matrix inequality



$$
M =
begin{bmatrix}
A & B^top \ B & C
end{bmatrix} succ 0.
$$



This is because $M$ can be decomposed as follows



$$
M =
underbrace{
begin{bmatrix}
I & B^top C^{-1} \ 0 & I
end{bmatrix}
}_{T^top}
begin{bmatrix}
A - B^top C^{-1} B & 0 \ 0 & C
end{bmatrix}
underbrace{
begin{bmatrix}
I & 0 \ C^{-1} B & I
end{bmatrix}
}_{T}.
$$



Namely $M succ 0$ means that $x^top M,x > 0 forall, xneq0$. So when using $x=T,y$ one gets $x^top M,x = y^top T^top M,T,y$. It holds that $xneq0$ iff $yneq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^top M,T succ 0$.



But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that



$$
T^top left(
begin{bmatrix}
Q & -A^top \
-A & Q
end{bmatrix}
+ sum_i
begin{bmatrix}
0 & A_i^top \
A_i & Q
end{bmatrix}
right) T
$$



gives back the original matrix inequalities as blocks on the diagonal.






share|cite|improve this answer











$endgroup$



You can't use Schur's complement in that way. Namely when you have the matrix inequalities



$$
A - B^top C^{-1} B succ 0, quad C succ 0,
$$



then you can combine this into the following one matrix inequality



$$
M =
begin{bmatrix}
A & B^top \ B & C
end{bmatrix} succ 0.
$$



This is because $M$ can be decomposed as follows



$$
M =
underbrace{
begin{bmatrix}
I & B^top C^{-1} \ 0 & I
end{bmatrix}
}_{T^top}
begin{bmatrix}
A - B^top C^{-1} B & 0 \ 0 & C
end{bmatrix}
underbrace{
begin{bmatrix}
I & 0 \ C^{-1} B & I
end{bmatrix}
}_{T}.
$$



Namely $M succ 0$ means that $x^top M,x > 0 forall, xneq0$. So when using $x=T,y$ one gets $x^top M,x = y^top T^top M,T,y$. It holds that $xneq0$ iff $yneq0$, since $T$ is full rank (lower triangular with ones on the diagonal), thus it also has to hold that $T^top M,T succ 0$.



But remember that this equivalent matrix inequality formulation only works because the matrix $T$ transforms $M$ into a block diagonal matrix with the blocks the original matrix inequalities. So your proposition only works if there exists a transformation $T$ such that



$$
T^top left(
begin{bmatrix}
Q & -A^top \
-A & Q
end{bmatrix}
+ sum_i
begin{bmatrix}
0 & A_i^top \
A_i & Q
end{bmatrix}
right) T
$$



gives back the original matrix inequalities as blocks on the diagonal.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 2:28

























answered Dec 2 '18 at 4:17









Kwin van der VeenKwin van der Veen

5,4952828




5,4952828












  • $begingroup$
    Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:40










  • $begingroup$
    @AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 12:50










  • $begingroup$
    Nevermind, you only have one $B$ and multiple $A$/$A_i$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 13:14


















  • $begingroup$
    Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
    $endgroup$
    – Alex bGoode
    Dec 2 '18 at 12:40










  • $begingroup$
    @AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 12:50










  • $begingroup$
    Nevermind, you only have one $B$ and multiple $A$/$A_i$.
    $endgroup$
    – Kwin van der Veen
    Dec 2 '18 at 13:14
















$begingroup$
Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
$endgroup$
– Alex bGoode
Dec 2 '18 at 12:40




$begingroup$
Thank you, I will need to take some time and understand your answer. Do you have any pointers on how I can transform my original problem into an LMI or should I post another question for this?
$endgroup$
– Alex bGoode
Dec 2 '18 at 12:40












$begingroup$
@AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
$endgroup$
– Kwin van der Veen
Dec 2 '18 at 12:50




$begingroup$
@AlexbGoode what are the sizes of $Y$ and $B$? If $Y$ has a rank equal to the size of $Q$ then you could define $A$ as a variable and later use that to solve for $B$.
$endgroup$
– Kwin van der Veen
Dec 2 '18 at 12:50












$begingroup$
Nevermind, you only have one $B$ and multiple $A$/$A_i$.
$endgroup$
– Kwin van der Veen
Dec 2 '18 at 13:14




$begingroup$
Nevermind, you only have one $B$ and multiple $A$/$A_i$.
$endgroup$
– Kwin van der Veen
Dec 2 '18 at 13:14


















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