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when the difference of two matrices is Hurwitz stable, what does it imply if one of the matrices is Hurwitz stable?
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Consider two constant matrices $A$ and $B$. If the difference matrix
$(A-B)$ and the matrix $A$ are both Hurwitz stable, what does it imply
for the matrix $B$ please?
This is my actual reasoning. Denote the set of $ntimes n$ matrices by $M_n$. Let
begineqnarray*
Sigma & = & Pin M_n:P=P^Tsucc0
endeqnarray*
denote the set of definite positive symmetric matrices and for $Ain M_n,xin R^n$,
let
begineqnarray*
L(A,x) & = & P=P^Tsucc0:x^TPAx<0.
endeqnarray*
Assume that $A$ and $-B$ are Hurwitz stable. Then the matrix $A-B$
is Hurwitz stable if and only if $L(A,x)cap L(-B,x)neqoslash$
for all $0neq xin R^n$ (this is a result of Theorem 2 of the
paper "A Local Lyapunov Theorem and the Stability of Sums" of
Charles R. Johnson https://www.sciencedirect.com/science/article/pii/0024379576900409)
matrices matrix-equations stability-theory
asked Aug 1 at 16:45
G. Trav
1409
add a comment |Â
up vote
-1
down vote
favorite
Consider two constant matrices $A$ and $B$. If the difference matrix
$(A-B)$ and the matrix $A$ are both Hurwitz stable, what does it imply
for the matrix $B$ please?
This is my actual reasoning. Denote the set of $ntimes n$ matrices by $M_n$. Let
begineqnarray*
Sigma & = & Pin M_n:P=P^Tsucc0
endeqnarray*
denote the set of definite positive symmetric matrices and for $Ain M_n,xin R^n$,
let
begineqnarray*
L(A,x) & = & P=P^Tsucc0:x^TPAx<0.
endeqnarray*
Assume that $A$ and $-B$ are Hurwitz stable. Then the matrix $A-B$
is Hurwitz stable if and only if $L(A,x)cap L(-B,x)neqoslash$
for all $0neq xin R^n$ (this is a result of Theorem 2 of the
paper "A Local Lyapunov Theorem and the Stability of Sums" of
Charles R. Johnson https://www.sciencedirect.com/science/article/pii/0024379576900409)
matrices matrix-equations stability-theory
asked Aug 1 at 16:45
G. Trav
1409
1
I don't fully understand what you are asking, but consider $A=mathrmdiag(-1,-1)$ and $B=mathrmdiag(varepsilon,varepsilon)$, $varepsilon ll 1$. Then $A$ is Hurwitz, $A-B$ is Hurwitz, but $B$ is not (Hurwitz from a continuous-time point of view). From a differential equation perspective you can also consider beginalign dotx(t) =& Ax(t) implies x(t)=e^Atx_0,\ dotx(t) =& (A-B)x(t) implies x(t)=e^(A-B)tx_0=e^Ate^-Btx_0. endalign For $e^Ate^-Btx_0to 0$ you can formulate conditions on the spectra of $A$ and $B$. Again, no need for $B$ to be Hurwitz.
– WalterJ
Aug 1 at 18:39
Thank you for your answer @WalterJ. You are right, I was wondering if the Hurwitz stability of the matrices (A-B) and A this implies something for the matrix B.
– G. Trav
Aug 1 at 19:12
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Consider two constant matrices $A$ and $B$. If the difference matrix
$(A-B)$ and the matrix $A$ are both Hurwitz stable, what does it imply
for the matrix $B$ please?
This is my actual reasoning. Denote the set of $ntimes n$ matrices by $M_n$. Let
begineqnarray*
Sigma & = & Pin M_n:P=P^Tsucc0
endeqnarray*
denote the set of definite positive symmetric matrices and for $Ain M_n,xin R^n$,
let
begineqnarray*
L(A,x) & = & P=P^Tsucc0:x^TPAx<0.
endeqnarray*
Assume that $A$ and $-B$ are Hurwitz stable. Then the matrix $A-B$
is Hurwitz stable if and only if $L(A,x)cap L(-B,x)neqoslash$
for all $0neq xin R^n$ (this is a result of Theorem 2 of the
paper "A Local Lyapunov Theorem and the Stability of Sums" of
Charles R. Johnson https://www.sciencedirect.com/science/article/pii/0024379576900409)
matrices matrix-equations stability-theory
asked Aug 1 at 16:45
G. Trav
1409
Consider two constant matrices $A$ and $B$. If the difference matrix
$(A-B)$ and the matrix $A$ are both Hurwitz stable, what does it imply
for the matrix $B$ please?
This is my actual reasoning. Denote the set of $ntimes n$ matrices by $M_n$. Let
begineqnarray*
Sigma & = & Pin M_n:P=P^Tsucc0
endeqnarray*
denote the set of definite positive symmetric matrices and for $Ain M_n,xin R^n$,
let
begineqnarray*
L(A,x) & = & P=P^Tsucc0:x^TPAx<0.
endeqnarray*
Assume that $A$ and $-B$ are Hurwitz stable. Then the matrix $A-B$
is Hurwitz stable if and only if $L(A,x)cap L(-B,x)neqoslash$
for all $0neq xin R^n$ (this is a result of Theorem 2 of the
paper "A Local Lyapunov Theorem and the Stability of Sums" of
Charles R. Johnson https://www.sciencedirect.com/science/article/pii/0024379576900409)
matrices matrix-equations stability-theory
asked Aug 1 at 16:45
G. Trav
1409
edited Aug 1 at 18:17
asked Aug 1 at 16:45
G. Trav
1409
asked Aug 1 at 16:45
G. Trav
1409
asked Aug 1 at 16:45
G. Trav
1409
1409
1
I don't fully understand what you are asking, but consider $A=mathrmdiag(-1,-1)$ and $B=mathrmdiag(varepsilon,varepsilon)$, $varepsilon ll 1$. Then $A$ is Hurwitz, $A-B$ is Hurwitz, but $B$ is not (Hurwitz from a continuous-time point of view). From a differential equation perspective you can also consider beginalign dotx(t) =& Ax(t) implies x(t)=e^Atx_0,\ dotx(t) =& (A-B)x(t) implies x(t)=e^(A-B)tx_0=e^Ate^-Btx_0. endalign For $e^Ate^-Btx_0to 0$ you can formulate conditions on the spectra of $A$ and $B$. Again, no need for $B$ to be Hurwitz.
– WalterJ
Aug 1 at 18:39
Thank you for your answer @WalterJ. You are right, I was wondering if the Hurwitz stability of the matrices (A-B) and A this implies something for the matrix B.
– G. Trav
Aug 1 at 19:12
add a comment |Â
1
I don't fully understand what you are asking, but consider $A=mathrmdiag(-1,-1)$ and $B=mathrmdiag(varepsilon,varepsilon)$, $varepsilon ll 1$. Then $A$ is Hurwitz, $A-B$ is Hurwitz, but $B$ is not (Hurwitz from a continuous-time point of view). From a differential equation perspective you can also consider beginalign dotx(t) =& Ax(t) implies x(t)=e^Atx_0,\ dotx(t) =& (A-B)x(t) implies x(t)=e^(A-B)tx_0=e^Ate^-Btx_0. endalign For $e^Ate^-Btx_0to 0$ you can formulate conditions on the spectra of $A$ and $B$. Again, no need for $B$ to be Hurwitz.
– WalterJ
Aug 1 at 18:39
Thank you for your answer @WalterJ. You are right, I was wondering if the Hurwitz stability of the matrices (A-B) and A this implies something for the matrix B.
– G. Trav
Aug 1 at 19:12
1
1
I don't fully understand what you are asking, but consider $A=mathrmdiag(-1,-1)$ and $B=mathrmdiag(varepsilon,varepsilon)$, $varepsilon ll 1$. Then $A$ is Hurwitz, $A-B$ is Hurwitz, but $B$ is not (Hurwitz from a continuous-time point of view). From a differential equation perspective you can also consider beginalign dotx(t) =& Ax(t) implies x(t)=e^Atx_0,\ dotx(t) =& (A-B)x(t) implies x(t)=e^(A-B)tx_0=e^Ate^-Btx_0. endalign For $e^Ate^-Btx_0to 0$ you can formulate conditions on the spectra of $A$ and $B$. Again, no need for $B$ to be Hurwitz.
– WalterJ
Aug 1 at 18:39
I don't fully understand what you are asking, but consider $A=mathrmdiag(-1,-1)$ and $B=mathrmdiag(varepsilon,varepsilon)$, $varepsilon ll 1$. Then $A$ is Hurwitz, $A-B$ is Hurwitz, but $B$ is not (Hurwitz from a continuous-time point of view). From a differential equation perspective you can also consider beginalign dotx(t) =& Ax(t) implies x(t)=e^Atx_0,\ dotx(t) =& (A-B)x(t) implies x(t)=e^(A-B)tx_0=e^Ate^-Btx_0. endalign For $e^Ate^-Btx_0to 0$ you can formulate conditions on the spectra of $A$ and $B$. Again, no need for $B$ to be Hurwitz.
– WalterJ
Aug 1 at 18:39
Thank you for your answer @WalterJ. You are right, I was wondering if the Hurwitz stability of the matrices (A-B) and A this implies something for the matrix B.
– G. Trav
Aug 1 at 19:12
Thank you for your answer @WalterJ. You are right, I was wondering if the Hurwitz stability of the matrices (A-B) and A this implies something for the matrix B.
– G. Trav
Aug 1 at 19:12
add a comment |Â
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I don't fully understand what you are asking, but consider $A=mathrmdiag(-1,-1)$ and $B=mathrmdiag(varepsilon,varepsilon)$, $varepsilon ll 1$. Then $A$ is Hurwitz, $A-B$ is Hurwitz, but $B$ is not (Hurwitz from a continuous-time point of view). From a differential equation perspective you can also consider beginalign dotx(t) =& Ax(t) implies x(t)=e^Atx_0,\ dotx(t) =& (A-B)x(t) implies x(t)=e^(A-B)tx_0=e^Ate^-Btx_0. endalign For $e^Ate^-Btx_0to 0$ you can formulate conditions on the spectra of $A$ and $B$. Again, no need for $B$ to be Hurwitz.
– WalterJ
Aug 1 at 18:39
Thank you for your answer @WalterJ. You are right, I was wondering if the Hurwitz stability of the matrices (A-B) and A this implies something for the matrix B.
– G. Trav
Aug 1 at 19:12