Find diagonal matrix $D$ such that $A D$ is Hurwitz
$begingroup$
Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...
Obviously $A, D$ must have full rank.
If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).
An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?
Note: This matrix analysis question is motivated by a problem in systems and control theory.
matrices matrix-equations matrix-decomposition semidefinite-programming lmis
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
asked Dec 25 '18 at 15:20
JohnJohn
1238
$endgroup$
add a comment |
$begingroup$
Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...
Obviously $A, D$ must have full rank.
If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).
An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?
Note: This matrix analysis question is motivated by a problem in systems and control theory.
matrices matrix-equations matrix-decomposition semidefinite-programming lmis
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
asked Dec 25 '18 at 15:20
JohnJohn
1238
$endgroup$
add a comment |
$begingroup$
Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...
Obviously $A, D$ must have full rank.
If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).
An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?
Note: This matrix analysis question is motivated by a problem in systems and control theory.
matrices matrix-equations matrix-decomposition semidefinite-programming lmis
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
asked Dec 25 '18 at 15:20
JohnJohn
1238
$endgroup$
Let $A in mathbb{R}^{m times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D in mathbb{R}^{m times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...
Obviously $A, D$ must have full rank.
If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).
An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P succ 0$ such that $P(AD)^{sf T} + (AD)P prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{sf T}A + AK prec 0$ in the variable $K$. If this is feasible for some $K^{star}$, you can then probably argue about the factorization $K^{star} = DP$ with $P succ 0$ and recover an appropriate choice of $D$?
Note: This matrix analysis question is motivated by a problem in systems and control theory.
matrices matrix-equations matrix-decomposition semidefinite-programming lmis
matrices matrix-equations matrix-decomposition semidefinite-programming lmis
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
asked Dec 25 '18 at 15:20
JohnJohn
1238
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
asked Dec 25 '18 at 15:20
JohnJohn
1238
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
edited Dec 25 '18 at 16:52
Rodrigo de Azevedo
13k41958
13k41958
asked Dec 25 '18 at 15:20
JohnJohn
1238
asked Dec 25 '18 at 15:20
JohnJohn
1238
asked Dec 25 '18 at 15:20
JohnJohn
1238
1238
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Your problem is equivalent to the state feedback problem for LTI systems, so
$$
dot{x} = mathcal{A},x + mathcal{B},u
$$
with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your problem is equivalent to the state feedback problem for LTI systems, so
$$
dot{x} = mathcal{A},x + mathcal{B},u
$$
with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
add a comment |
$begingroup$
Your problem is equivalent to the state feedback problem for LTI systems, so
$$
dot{x} = mathcal{A},x + mathcal{B},u
$$
with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
add a comment |
$begingroup$
Your problem is equivalent to the state feedback problem for LTI systems, so
$$
dot{x} = mathcal{A},x + mathcal{B},u
$$
with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
$endgroup$
Your problem is equivalent to the state feedback problem for LTI systems, so
$$
dot{x} = mathcal{A},x + mathcal{B},u
$$
with $u = -K,x$ such that $mathcal{A} - mathcal{B},K$ is Hurwitz. In your case $mathcal{A}=0$ and $mathcal{B} = -A$. When $mathcal{A}=0$ then the pair $(mathcal{A},mathcal{B})$ is stabilizable if and only if $mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
answered Dec 29 '18 at 15:15
Kwin van der VeenKwin van der Veen
5,4952828
5,4952828
add a comment |
add a comment |
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