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Do I really need to invert this matrix
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I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation
$$ A=B(mathbb1-B)^-1 $$
where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.
Inspired by this post:
https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/
I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:
$N$ in my case is quite large, its order of magnitude can be ten thousand.
I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.
matrix inverse-problem inverse
asked 4 hours ago
zakk
1623
add a comment |Â
up vote
1
down vote
favorite
I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation
$$ A=B(mathbb1-B)^-1 $$
where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.
Inspired by this post:
https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/
I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:
$N$ in my case is quite large, its order of magnitude can be ten thousand.
I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.
matrix inverse-problem inverse
asked 4 hours ago
zakk
1623
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation
$$ A=B(mathbb1-B)^-1 $$
where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.
Inspired by this post:
https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/
I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:
$N$ in my case is quite large, its order of magnitude can be ten thousand.
I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.
matrix inverse-problem inverse
asked 4 hours ago
zakk
1623
I need to calculate a matrix $A$ (at least some elements of it, see below) as defined by the following equation
$$ A=B(mathbb1-B)^-1 $$
where B is a square matrix of dimension $N$ and $mathbb1$ is $N times N$ identity matrix.
Inspired by this post:
https://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/
I was wondering if I really need to invert $mathbb1-B$ in my case, of if there's some easier way. Keep in mind that:
$N$ in my case is quite large, its order of magnitude can be ten thousand.
I don't need to know the full matrix $A$, I just need a few elements in the upper left corner, let's say $A_00$, $A_01$, $A_11$, $A_02$, $A_12$, $A_22$ would be perfect.
matrix inverse-problem inverse
matrix inverse-problem inverse
asked 4 hours ago
zakk
1623
asked 4 hours ago
zakk
1623
asked 4 hours ago
zakk
1623
asked 4 hours ago
zakk
1623
asked 4 hours ago
zakk
1623
1623
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Since
$$
A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
$$
So you want to solve
$$
(I-B)A=B
$$
You seem to need only the first three columns of $A$. Solve the matrix problems
$$
(I-B)a_i = b_i, qquad i=0,1,2
$$
where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.
answered 2 hours ago
Praveen Chandrashekar
894510
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since
$$
A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
$$
So you want to solve
$$
(I-B)A=B
$$
You seem to need only the first three columns of $A$. Solve the matrix problems
$$
(I-B)a_i = b_i, qquad i=0,1,2
$$
where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.
answered 2 hours ago
Praveen Chandrashekar
894510
add a comment |Â
up vote
2
down vote
accepted
Since
$$
A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
$$
So you want to solve
$$
(I-B)A=B
$$
You seem to need only the first three columns of $A$. Solve the matrix problems
$$
(I-B)a_i = b_i, qquad i=0,1,2
$$
where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.
answered 2 hours ago
Praveen Chandrashekar
894510
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since
$$
A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
$$
So you want to solve
$$
(I-B)A=B
$$
You seem to need only the first three columns of $A$. Solve the matrix problems
$$
(I-B)a_i = b_i, qquad i=0,1,2
$$
where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.
answered 2 hours ago
Praveen Chandrashekar
894510
Since
$$
A = B(I-B)^-1 = (I-B)^-1(I-B)B(I-B)^-1 = (I-B)^-1B(I-B)(I-B)^-1 =(I-B)^-1B
$$
So you want to solve
$$
(I-B)A=B
$$
You seem to need only the first three columns of $A$. Solve the matrix problems
$$
(I-B)a_i = b_i, qquad i=0,1,2
$$
where $b_0,b_1,b_2$ are first three columns of $B$. Then $a_0,a_1,a_2$ are the first three columns of $A$.
answered 2 hours ago
Praveen Chandrashekar
894510
answered 2 hours ago
Praveen Chandrashekar
894510
answered 2 hours ago
Praveen Chandrashekar
894510
answered 2 hours ago
Praveen Chandrashekar
894510
894510
add a comment |Â
add a comment |Â
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