Inverse of identity minus matrix exponential
$begingroup$
I am trying to analytically find the inverse of a matrix given by:
begin{align}
W = left( I - alpha e^A right)^{-1},
end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and
begin{align}
A = begin{bmatrix}
frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
vdots & vdots & vdots & ddots & vdots \
frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
end{bmatrix}
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse matrix-exponential
asked Jan 14 at 11:58
KatieKatie
636
$endgroup$
add a comment |
$begingroup$
I am trying to analytically find the inverse of a matrix given by:
begin{align}
W = left( I - alpha e^A right)^{-1},
end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and
begin{align}
A = begin{bmatrix}
frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
vdots & vdots & vdots & ddots & vdots \
frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
end{bmatrix}
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse matrix-exponential
asked Jan 14 at 11:58
KatieKatie
636
$endgroup$
add a comment |
$begingroup$
I am trying to analytically find the inverse of a matrix given by:
begin{align}
W = left( I - alpha e^A right)^{-1},
end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and
begin{align}
A = begin{bmatrix}
frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
vdots & vdots & vdots & ddots & vdots \
frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
end{bmatrix}
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse matrix-exponential
asked Jan 14 at 11:58
KatieKatie
636
$endgroup$
I am trying to analytically find the inverse of a matrix given by:
begin{align}
W = left( I - alpha e^A right)^{-1},
end{align}
where $I$ is the identity matrix of appropriate size, $e^A$ denotes matrix exponential of $A$ and
begin{align}
A = begin{bmatrix}
frac{ (1-1)^2 }{ sigma^2 } & frac{ (2-1)^2 }{ sigma^2 } &
frac{ (3-1)^2 }{ sigma^2 } & dots & frac{ (n-1)^2 }{ sigma^2 } \
frac{ (1-2)^2 }{ sigma^2 } & frac{ (2-2)^2 }{ sigma^2 } &
frac{ (3-2)^2 }{ sigma^2 } & dots & frac{ (n-2)^2 }{ sigma^2 } \
vdots & vdots & vdots & ddots & vdots \
frac{ (1-N)^2 }{ sigma^2 } & frac{ (2-N)^2 }{ sigma^2 } &
frac{ (3-N)^2 }{ sigma^2 } & dots & frac{ (N-N)^2 }{ sigma^2 } \
end{bmatrix}
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse matrix-exponential
matrices inverse matrix-exponential
asked Jan 14 at 11:58
KatieKatie
636
asked Jan 14 at 11:58
KatieKatie
636
asked Jan 14 at 11:58
KatieKatie
636
asked Jan 14 at 11:58
KatieKatie
636
asked Jan 14 at 11:58
KatieKatie
636
636
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Define the variables
$$eqalign{
lambda &= log(alpha) &implies alpha = e^lambda cr
X &= -tfrac{1}{2}(A+lambda I) cr
}$$
Then
$$eqalign{
W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
}$$
answered Jan 16 at 1:44
greggreg
8,9551824
$endgroup$
add a comment |
$begingroup$
You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
You can make it upper triangular by a few transformations (after which it's easy to transform to identity).
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Define the variables
$$eqalign{
lambda &= log(alpha) &implies alpha = e^lambda cr
X &= -tfrac{1}{2}(A+lambda I) cr
}$$
Then
$$eqalign{
W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
}$$
answered Jan 16 at 1:44
greggreg
8,9551824
$endgroup$
add a comment |
$begingroup$
Define the variables
$$eqalign{
lambda &= log(alpha) &implies alpha = e^lambda cr
X &= -tfrac{1}{2}(A+lambda I) cr
}$$
Then
$$eqalign{
W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
}$$
answered Jan 16 at 1:44
greggreg
8,9551824
$endgroup$
add a comment |
$begingroup$
Define the variables
$$eqalign{
lambda &= log(alpha) &implies alpha = e^lambda cr
X &= -tfrac{1}{2}(A+lambda I) cr
}$$
Then
$$eqalign{
W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
}$$
answered Jan 16 at 1:44
greggreg
8,9551824
$endgroup$
Define the variables
$$eqalign{
lambda &= log(alpha) &implies alpha = e^lambda cr
X &= -tfrac{1}{2}(A+lambda I) cr
}$$
Then
$$eqalign{
W^{-1} &= Big(I-e^{-2X}Big) &= 2e^{-X}sinh(X) cr
W &= Big(I-e^{-2X}Big)^{-1} &= tfrac{1}{2}e^{X}{,rm csch}(X) cr
}$$
answered Jan 16 at 1:44
greggreg
8,9551824
edited Jan 16 at 2:01
answered Jan 16 at 1:44
greggreg
8,9551824
answered Jan 16 at 1:44
greggreg
8,9551824
answered Jan 16 at 1:44
greggreg
8,9551824
8,9551824
add a comment |
add a comment |
$begingroup$
You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
You can make it upper triangular by a few transformations (after which it's easy to transform to identity).
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
$endgroup$
add a comment |
$begingroup$
You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
You can make it upper triangular by a few transformations (after which it's easy to transform to identity).
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
$endgroup$
add a comment |
$begingroup$
You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
You can make it upper triangular by a few transformations (after which it's easy to transform to identity).
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
$endgroup$
You can take out $frac{1}{sigma^2}$. Notice also that your matrix is symmetric, and of the form: Find rank of the matrix $a_{ij}=(i-j)^2$, $i,j=1,dots, n$
You can make it upper triangular by a few transformations (after which it's easy to transform to identity).
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
answered Jan 14 at 12:26
lightxbulblightxbulb
1,140311
1,140311
add a comment |
add a comment |
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