How to efficiently re-compute the inverse of positive definite matrix when its i-th row and column are...
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Given a positive definite matrix $Ain R^{ntimes n}$ with its inverse $B$ and the Singular Value Decomposition $A=USU^T$. If I replace the $i-$th row and column of $A$ with a vector $x^Tin R^{1times n}$ and $xin R^n$ respectively, so that the new matrix $C$ is generated.
Here, assume that $C$ is still a positive definite matrix, then my question is that how can I efficiently calculate the inverse of $C$ with the results $B$ and $A=USU^T$?
Maybe we should also know some other calculation results about $A$.
calculus linear-algebra matrices
asked Nov 20 at 1:47
olivia
747616
add a comment |
up vote
1
down vote
favorite
Given a positive definite matrix $Ain R^{ntimes n}$ with its inverse $B$ and the Singular Value Decomposition $A=USU^T$. If I replace the $i-$th row and column of $A$ with a vector $x^Tin R^{1times n}$ and $xin R^n$ respectively, so that the new matrix $C$ is generated.
Here, assume that $C$ is still a positive definite matrix, then my question is that how can I efficiently calculate the inverse of $C$ with the results $B$ and $A=USU^T$?
Maybe we should also know some other calculation results about $A$.
calculus linear-algebra matrices
asked Nov 20 at 1:47
olivia
747616
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a positive definite matrix $Ain R^{ntimes n}$ with its inverse $B$ and the Singular Value Decomposition $A=USU^T$. If I replace the $i-$th row and column of $A$ with a vector $x^Tin R^{1times n}$ and $xin R^n$ respectively, so that the new matrix $C$ is generated.
Here, assume that $C$ is still a positive definite matrix, then my question is that how can I efficiently calculate the inverse of $C$ with the results $B$ and $A=USU^T$?
Maybe we should also know some other calculation results about $A$.
calculus linear-algebra matrices
asked Nov 20 at 1:47
olivia
747616
Given a positive definite matrix $Ain R^{ntimes n}$ with its inverse $B$ and the Singular Value Decomposition $A=USU^T$. If I replace the $i-$th row and column of $A$ with a vector $x^Tin R^{1times n}$ and $xin R^n$ respectively, so that the new matrix $C$ is generated.
Here, assume that $C$ is still a positive definite matrix, then my question is that how can I efficiently calculate the inverse of $C$ with the results $B$ and $A=USU^T$?
Maybe we should also know some other calculation results about $A$.
calculus linear-algebra matrices
calculus linear-algebra matrices
asked Nov 20 at 1:47
olivia
747616
asked Nov 20 at 1:47
olivia
747616
asked Nov 20 at 1:47
olivia
747616
asked Nov 20 at 1:47
olivia
747616
asked Nov 20 at 1:47
olivia
747616
747616
add a comment |
add a comment |
1 Answer
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This is basically a rank-2 update. Let $mathbf a_i$ be the $i$-th column of $A$, $u=e_i$ (the $i$-th standard basis vector) and $v=x-mathbf a_i-frac{x_{ii}-a_{ii}}2e_i$. Then $C=A+uv^T+vu^T$. Hence you may apply Sherman-Morrison formula twice to obtain
$$
C^{-1}=A^{-1}
+A^{-1}frac{avv^T-(1+b)(vu^T+uv^T)+cuu^T}{(1+b)^2-ac}A^{-1},
$$
where $a=u^TA^{-1}u, b=u^TA^{-1}u$ and $c=v^TA^{-1}v$.
answered 2 days ago
user1551
70.2k566125
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is basically a rank-2 update. Let $mathbf a_i$ be the $i$-th column of $A$, $u=e_i$ (the $i$-th standard basis vector) and $v=x-mathbf a_i-frac{x_{ii}-a_{ii}}2e_i$. Then $C=A+uv^T+vu^T$. Hence you may apply Sherman-Morrison formula twice to obtain
$$
C^{-1}=A^{-1}
+A^{-1}frac{avv^T-(1+b)(vu^T+uv^T)+cuu^T}{(1+b)^2-ac}A^{-1},
$$
where $a=u^TA^{-1}u, b=u^TA^{-1}u$ and $c=v^TA^{-1}v$.
answered 2 days ago
user1551
70.2k566125
add a comment |
up vote
0
down vote
This is basically a rank-2 update. Let $mathbf a_i$ be the $i$-th column of $A$, $u=e_i$ (the $i$-th standard basis vector) and $v=x-mathbf a_i-frac{x_{ii}-a_{ii}}2e_i$. Then $C=A+uv^T+vu^T$. Hence you may apply Sherman-Morrison formula twice to obtain
$$
C^{-1}=A^{-1}
+A^{-1}frac{avv^T-(1+b)(vu^T+uv^T)+cuu^T}{(1+b)^2-ac}A^{-1},
$$
where $a=u^TA^{-1}u, b=u^TA^{-1}u$ and $c=v^TA^{-1}v$.
answered 2 days ago
user1551
70.2k566125
add a comment |
up vote
0
down vote
up vote
0
down vote
This is basically a rank-2 update. Let $mathbf a_i$ be the $i$-th column of $A$, $u=e_i$ (the $i$-th standard basis vector) and $v=x-mathbf a_i-frac{x_{ii}-a_{ii}}2e_i$. Then $C=A+uv^T+vu^T$. Hence you may apply Sherman-Morrison formula twice to obtain
$$
C^{-1}=A^{-1}
+A^{-1}frac{avv^T-(1+b)(vu^T+uv^T)+cuu^T}{(1+b)^2-ac}A^{-1},
$$
where $a=u^TA^{-1}u, b=u^TA^{-1}u$ and $c=v^TA^{-1}v$.
answered 2 days ago
user1551
70.2k566125
This is basically a rank-2 update. Let $mathbf a_i$ be the $i$-th column of $A$, $u=e_i$ (the $i$-th standard basis vector) and $v=x-mathbf a_i-frac{x_{ii}-a_{ii}}2e_i$. Then $C=A+uv^T+vu^T$. Hence you may apply Sherman-Morrison formula twice to obtain
$$
C^{-1}=A^{-1}
+A^{-1}frac{avv^T-(1+b)(vu^T+uv^T)+cuu^T}{(1+b)^2-ac}A^{-1},
$$
where $a=u^TA^{-1}u, b=u^TA^{-1}u$ and $c=v^TA^{-1}v$.
answered 2 days ago
user1551
70.2k566125
answered 2 days ago
user1551
70.2k566125
answered 2 days ago
user1551
70.2k566125
answered 2 days ago
user1551
70.2k566125
70.2k566125
add a comment |
add a comment |
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