Inverse of identity minus exponential matrix
$begingroup$
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 elementwise 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}
so that I'm trying to find
begin{align}
W = begin{bmatrix}
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 2^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-1)^2 }{ sigma^2 } right) } \
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-2)^2 }{ sigma^2 } right) } \
vdots & vdots & vdots & ddots & vdots \
- alpha exp{ left( frac{ (1-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (2-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (3-N)^2 }{ sigma^2 } right) } &
dots &
1 - alpha exp{ left( frac{ (N-N)^2 }{ sigma^2 } right) } \
end{bmatrix}^{-1}.
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse
edited Dec 23 '18 at 15:29
user1551
72.5k566127
asked Dec 20 '18 at 21:05
KatieKatie
486
$endgroup$
|
show 3 more comments
$begingroup$
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 elementwise 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}
so that I'm trying to find
begin{align}
W = begin{bmatrix}
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 2^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-1)^2 }{ sigma^2 } right) } \
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-2)^2 }{ sigma^2 } right) } \
vdots & vdots & vdots & ddots & vdots \
- alpha exp{ left( frac{ (1-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (2-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (3-N)^2 }{ sigma^2 } right) } &
dots &
1 - alpha exp{ left( frac{ (N-N)^2 }{ sigma^2 } right) } \
end{bmatrix}^{-1}.
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse
edited Dec 23 '18 at 15:29
user1551
72.5k566127
asked Dec 20 '18 at 21:05
KatieKatie
486
$endgroup$
$begingroup$
Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$...
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:08
$begingroup$
Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation?
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:09
1
$begingroup$
How do you define $e^A$ ?
$endgroup$
– Damien
Dec 20 '18 at 21:21
$begingroup$
Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie.
$endgroup$
– Katie
Dec 20 '18 at 22:18
1
$begingroup$
The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– Damien
Dec 21 '18 at 14:38
|
show 3 more comments
$begingroup$
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 elementwise 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}
so that I'm trying to find
begin{align}
W = begin{bmatrix}
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 2^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-1)^2 }{ sigma^2 } right) } \
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-2)^2 }{ sigma^2 } right) } \
vdots & vdots & vdots & ddots & vdots \
- alpha exp{ left( frac{ (1-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (2-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (3-N)^2 }{ sigma^2 } right) } &
dots &
1 - alpha exp{ left( frac{ (N-N)^2 }{ sigma^2 } right) } \
end{bmatrix}^{-1}.
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse
edited Dec 23 '18 at 15:29
user1551
72.5k566127
asked Dec 20 '18 at 21:05
KatieKatie
486
$endgroup$
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 elementwise 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}
so that I'm trying to find
begin{align}
W = begin{bmatrix}
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 2^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-1)^2 }{ sigma^2 } right) } \
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
1 - alpha exp{ left( frac{ 0^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ 1^2 }{ sigma^2 } right) } &
dots &
- alpha exp{ left( frac{ (n-2)^2 }{ sigma^2 } right) } \
vdots & vdots & vdots & ddots & vdots \
- alpha exp{ left( frac{ (1-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (2-N)^2 }{ sigma^2 } right) } &
- alpha exp{ left( frac{ (3-N)^2 }{ sigma^2 } right) } &
dots &
1 - alpha exp{ left( frac{ (N-N)^2 }{ sigma^2 } right) } \
end{bmatrix}^{-1}.
end{align}
Any help would be much appreciated!
Thank you very much,
Katie
matrices inverse
matrices inverse
edited Dec 23 '18 at 15:29
user1551
72.5k566127
asked Dec 20 '18 at 21:05
KatieKatie
486
edited Dec 23 '18 at 15:29
user1551
72.5k566127
asked Dec 20 '18 at 21:05
KatieKatie
486
edited Dec 23 '18 at 15:29
user1551
72.5k566127
edited Dec 23 '18 at 15:29
user1551
72.5k566127
edited Dec 23 '18 at 15:29
user1551
72.5k566127
72.5k566127
asked Dec 20 '18 at 21:05
KatieKatie
486
asked Dec 20 '18 at 21:05
KatieKatie
486
asked Dec 20 '18 at 21:05
KatieKatie
486
486
$begingroup$
Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$...
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:08
$begingroup$
Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation?
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:09
1
$begingroup$
How do you define $e^A$ ?
$endgroup$
– Damien
Dec 20 '18 at 21:21
$begingroup$
Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie.
$endgroup$
– Katie
Dec 20 '18 at 22:18
1
$begingroup$
The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– Damien
Dec 21 '18 at 14:38
|
show 3 more comments
$begingroup$
Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$...
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:08
$begingroup$
Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation?
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:09
1
$begingroup$
How do you define $e^A$ ?
$endgroup$
– Damien
Dec 20 '18 at 21:21
$begingroup$
Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie.
$endgroup$
– Katie
Dec 20 '18 at 22:18
1
$begingroup$
The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– Damien
Dec 21 '18 at 14:38
$begingroup$
Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$...
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:08
$begingroup$
Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$...
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:08
$begingroup$
Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation?
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:09
$begingroup$
Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation?
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:09
1
1
$begingroup$
How do you define $e^A$ ?
$endgroup$
– Damien
Dec 20 '18 at 21:21
$begingroup$
How do you define $e^A$ ?
$endgroup$
– Damien
Dec 20 '18 at 21:21
$begingroup$
Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie.
$endgroup$
– Katie
Dec 20 '18 at 22:18
$begingroup$
Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie.
$endgroup$
– Katie
Dec 20 '18 at 22:18
1
1
$begingroup$
The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– Damien
Dec 21 '18 at 14:38
$begingroup$
The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– Damien
Dec 21 '18 at 14:38
|
show 3 more comments
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$begingroup$
Are you trying to find $W^{-1}$? That would be the last matrix because $(A^{-1})^{-1}$...
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:08
$begingroup$
Your matrix $W$ has only zeros on the diagonal and is symmetric, have you tried diagonalisation?
$endgroup$
– Viktor Glombik
Dec 20 '18 at 21:09
1
$begingroup$
How do you define $e^A$ ?
$endgroup$
– Damien
Dec 20 '18 at 21:21
$begingroup$
Thank you everyone for your interest in my question. I'm trying to find W, as opposed to $W^{-1}$, as that would simply be $I - alpha e^A$, which would be wonderfully easy! W unfortunately does not have 0's on the diagonal, but rather $1-alpha$. I have looked at diagonalisation but haven't made much progress. Regarding the definition of $e^A$, I wondered if there were multiple definitions of a matrix exponential, which is why I included the expanded matrix at the end. I've simply taken the exponential of each matrix element. This may be wrong? Thanks again, Katie.
$endgroup$
– Katie
Dec 20 '18 at 22:18
1
$begingroup$
The classical definition of $e^A$ corresponds to en.wikipedia.org/wiki/Matrix_exponential
$endgroup$
– Damien
Dec 21 '18 at 14:38