Limit of matrix function
$begingroup$
Let $Ainmathbb{R}^{ntimes n}$ be a matrix whose real eigenvalues have negative real part, and $X=X^topinmathbb{R}^{ntimes n}$ be a positive semidefinite matrix, i.e., $Xsucceq 0$. Consider the following matrix-valued function
$$tag{1}label{1}
F(X) = left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2} X left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2}
$$
which maps positive (semi)definite matrices into a positive (semi)definite matrices. (Here $cdot^{1/2}$ denotes the symmetric square root of a positive semidefinite matrix and $e^{cdot}$ is the matrix exponential).
Note that eqref{1} is a continuous function which is not defined for any $Xsucceq 0$ such that $int_0^{infty} e^{At} X e^{A^top t} mathrm{d} t$ is singular. However, I wonder whether it is possible to define a continuous extension of $F(cdot)$ in the set of positive semidefinite matrices.
In more formal terms, let $bar{X}succeq 0$ be such that $int_0^{infty} e^{At} bar{X} e^{A^top t} mathrm{d} t$ is singular, and let ${X_n}_{nge 0}$, $X_nsucc 0$, be any sequence such that $lim_{nto infty} X_n = bar{X}$. Does $
lim_{nto infty} F(X_n)$ exist and is finite?
My question is motivated by the special case of scalar matrices $A=alpha I$, $alpha<0$, for which it is easy to see that the above limit exists and is finite. For general $A$'s, however, it is not clear to me whether this limit still exists. Numerical simulations suggest that the answer is again in the affirmative, but I was not able to prove it.
linear-algebra matrices limits convergence matrix-calculus
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
$endgroup$
|
show 2 more comments
$begingroup$
Let $Ainmathbb{R}^{ntimes n}$ be a matrix whose real eigenvalues have negative real part, and $X=X^topinmathbb{R}^{ntimes n}$ be a positive semidefinite matrix, i.e., $Xsucceq 0$. Consider the following matrix-valued function
$$tag{1}label{1}
F(X) = left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2} X left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2}
$$
which maps positive (semi)definite matrices into a positive (semi)definite matrices. (Here $cdot^{1/2}$ denotes the symmetric square root of a positive semidefinite matrix and $e^{cdot}$ is the matrix exponential).
Note that eqref{1} is a continuous function which is not defined for any $Xsucceq 0$ such that $int_0^{infty} e^{At} X e^{A^top t} mathrm{d} t$ is singular. However, I wonder whether it is possible to define a continuous extension of $F(cdot)$ in the set of positive semidefinite matrices.
In more formal terms, let $bar{X}succeq 0$ be such that $int_0^{infty} e^{At} bar{X} e^{A^top t} mathrm{d} t$ is singular, and let ${X_n}_{nge 0}$, $X_nsucc 0$, be any sequence such that $lim_{nto infty} X_n = bar{X}$. Does $
lim_{nto infty} F(X_n)$ exist and is finite?
My question is motivated by the special case of scalar matrices $A=alpha I$, $alpha<0$, for which it is easy to see that the above limit exists and is finite. For general $A$'s, however, it is not clear to me whether this limit still exists. Numerical simulations suggest that the answer is again in the affirmative, but I was not able to prove it.
linear-algebra matrices limits convergence matrix-calculus
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
$endgroup$
1
$begingroup$
Whenever it's defined, the function $g(Q) = int_0^infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:46
$begingroup$
@Omnomnomnom: Right, thanks! However, I do not see how this can be helpful.
$endgroup$
– Ludwig
Dec 20 '18 at 1:48
$begingroup$
Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:49
$begingroup$
I'm not quite sure if that helps either
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:55
$begingroup$
I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2, bar{X}=0$ and $X_n=diag(frac1n,0)$.
$endgroup$
– user1551
Dec 20 '18 at 4:14
|
show 2 more comments
$begingroup$
Let $Ainmathbb{R}^{ntimes n}$ be a matrix whose real eigenvalues have negative real part, and $X=X^topinmathbb{R}^{ntimes n}$ be a positive semidefinite matrix, i.e., $Xsucceq 0$. Consider the following matrix-valued function
$$tag{1}label{1}
F(X) = left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2} X left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2}
$$
which maps positive (semi)definite matrices into a positive (semi)definite matrices. (Here $cdot^{1/2}$ denotes the symmetric square root of a positive semidefinite matrix and $e^{cdot}$ is the matrix exponential).
Note that eqref{1} is a continuous function which is not defined for any $Xsucceq 0$ such that $int_0^{infty} e^{At} X e^{A^top t} mathrm{d} t$ is singular. However, I wonder whether it is possible to define a continuous extension of $F(cdot)$ in the set of positive semidefinite matrices.
In more formal terms, let $bar{X}succeq 0$ be such that $int_0^{infty} e^{At} bar{X} e^{A^top t} mathrm{d} t$ is singular, and let ${X_n}_{nge 0}$, $X_nsucc 0$, be any sequence such that $lim_{nto infty} X_n = bar{X}$. Does $
lim_{nto infty} F(X_n)$ exist and is finite?
My question is motivated by the special case of scalar matrices $A=alpha I$, $alpha<0$, for which it is easy to see that the above limit exists and is finite. For general $A$'s, however, it is not clear to me whether this limit still exists. Numerical simulations suggest that the answer is again in the affirmative, but I was not able to prove it.
linear-algebra matrices limits convergence matrix-calculus
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
$endgroup$
Let $Ainmathbb{R}^{ntimes n}$ be a matrix whose real eigenvalues have negative real part, and $X=X^topinmathbb{R}^{ntimes n}$ be a positive semidefinite matrix, i.e., $Xsucceq 0$. Consider the following matrix-valued function
$$tag{1}label{1}
F(X) = left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2} X left(int_0^{infty} e^{At} X e^{A^top t} mathrm{d} tright)^{-1/2}
$$
which maps positive (semi)definite matrices into a positive (semi)definite matrices. (Here $cdot^{1/2}$ denotes the symmetric square root of a positive semidefinite matrix and $e^{cdot}$ is the matrix exponential).
Note that eqref{1} is a continuous function which is not defined for any $Xsucceq 0$ such that $int_0^{infty} e^{At} X e^{A^top t} mathrm{d} t$ is singular. However, I wonder whether it is possible to define a continuous extension of $F(cdot)$ in the set of positive semidefinite matrices.
In more formal terms, let $bar{X}succeq 0$ be such that $int_0^{infty} e^{At} bar{X} e^{A^top t} mathrm{d} t$ is singular, and let ${X_n}_{nge 0}$, $X_nsucc 0$, be any sequence such that $lim_{nto infty} X_n = bar{X}$. Does $
lim_{nto infty} F(X_n)$ exist and is finite?
My question is motivated by the special case of scalar matrices $A=alpha I$, $alpha<0$, for which it is easy to see that the above limit exists and is finite. For general $A$'s, however, it is not clear to me whether this limit still exists. Numerical simulations suggest that the answer is again in the affirmative, but I was not able to prove it.
linear-algebra matrices limits convergence matrix-calculus
linear-algebra matrices limits convergence matrix-calculus
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
edited Dec 27 '18 at 19:24
Ludwig
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
asked Dec 20 '18 at 1:29
LudwigLudwig
813715
813715
1
$begingroup$
Whenever it's defined, the function $g(Q) = int_0^infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:46
$begingroup$
@Omnomnomnom: Right, thanks! However, I do not see how this can be helpful.
$endgroup$
– Ludwig
Dec 20 '18 at 1:48
$begingroup$
Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:49
$begingroup$
I'm not quite sure if that helps either
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:55
$begingroup$
I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2, bar{X}=0$ and $X_n=diag(frac1n,0)$.
$endgroup$
– user1551
Dec 20 '18 at 4:14
|
show 2 more comments
1
$begingroup$
Whenever it's defined, the function $g(Q) = int_0^infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:46
$begingroup$
@Omnomnomnom: Right, thanks! However, I do not see how this can be helpful.
$endgroup$
– Ludwig
Dec 20 '18 at 1:48
$begingroup$
Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:49
$begingroup$
I'm not quite sure if that helps either
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:55
$begingroup$
I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2, bar{X}=0$ and $X_n=diag(frac1n,0)$.
$endgroup$
– user1551
Dec 20 '18 at 4:14
1
1
$begingroup$
Whenever it's defined, the function $g(Q) = int_0^infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:46
$begingroup$
Whenever it's defined, the function $g(Q) = int_0^infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:46
$begingroup$
@Omnomnomnom: Right, thanks! However, I do not see how this can be helpful.
$endgroup$
– Ludwig
Dec 20 '18 at 1:48
$begingroup$
@Omnomnomnom: Right, thanks! However, I do not see how this can be helpful.
$endgroup$
– Ludwig
Dec 20 '18 at 1:48
$begingroup$
Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:49
$begingroup$
Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:49
$begingroup$
I'm not quite sure if that helps either
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:55
$begingroup$
I'm not quite sure if that helps either
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:55
$begingroup$
I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2, bar{X}=0$ and $X_n=diag(frac1n,0)$.
$endgroup$
– user1551
Dec 20 '18 at 4:14
$begingroup$
I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2, bar{X}=0$ and $X_n=diag(frac1n,0)$.
$endgroup$
– user1551
Dec 20 '18 at 4:14
|
show 2 more comments
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1
$begingroup$
Whenever it's defined, the function $g(Q) = int_0^infty e^{At}Qe^{A^T}t$ takes input $Q$ and produces a solution $X = g(Q)$ to the continuous Lyapunov equation $$ AX + XA^T + Q = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:46
$begingroup$
@Omnomnomnom: Right, thanks! However, I do not see how this can be helpful.
$endgroup$
– Ludwig
Dec 20 '18 at 1:48
$begingroup$
Whenever the inverse of this $X$ is defined, $Y = X^{-1}$ satisfies the equation $$ YA + A^TY + YQY = 0 $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:49
$begingroup$
I'm not quite sure if that helps either
$endgroup$
– Omnomnomnom
Dec 20 '18 at 1:55
$begingroup$
I don't follow. Isn't the boxed statement obviously false in some cases? E.g. consider the case where $A=-I_2, bar{X}=0$ and $X_n=diag(frac1n,0)$.
$endgroup$
– user1551
Dec 20 '18 at 4:14