Mixing
efficient computation of Cholesky decomposition during tridiagonal matrix inverse
$begingroup$
I have a symmetric, block tridiagonal matrix $A$. I am interested in computing the Cholesky decomposition of $A^{-1}$ (that is, I want to compute $R$, where $A^{-1}=RR^T$). I know how to compute the blocks of the inverse efficiently using an iterative algorithm. However, is there an efficient algorithm for computing the cholesky factors $R$ directly (rather than first computing the inverse, and them performing the Cholesky decomposition)?
numerical-methods numerical-linear-algebra
edited Apr 19 '15 at 19:48
BigM
2,56611530
asked Apr 19 '15 at 19:41
yepyep
50428
$endgroup$
add a comment |
$begingroup$
I have a symmetric, block tridiagonal matrix $A$. I am interested in computing the Cholesky decomposition of $A^{-1}$ (that is, I want to compute $R$, where $A^{-1}=RR^T$). I know how to compute the blocks of the inverse efficiently using an iterative algorithm. However, is there an efficient algorithm for computing the cholesky factors $R$ directly (rather than first computing the inverse, and them performing the Cholesky decomposition)?
numerical-methods numerical-linear-algebra
edited Apr 19 '15 at 19:48
BigM
2,56611530
asked Apr 19 '15 at 19:41
yepyep
50428
$endgroup$
add a comment |
$begingroup$
I have a symmetric, block tridiagonal matrix $A$. I am interested in computing the Cholesky decomposition of $A^{-1}$ (that is, I want to compute $R$, where $A^{-1}=RR^T$). I know how to compute the blocks of the inverse efficiently using an iterative algorithm. However, is there an efficient algorithm for computing the cholesky factors $R$ directly (rather than first computing the inverse, and them performing the Cholesky decomposition)?
numerical-methods numerical-linear-algebra
edited Apr 19 '15 at 19:48
BigM
2,56611530
asked Apr 19 '15 at 19:41
yepyep
50428
$endgroup$
I have a symmetric, block tridiagonal matrix $A$. I am interested in computing the Cholesky decomposition of $A^{-1}$ (that is, I want to compute $R$, where $A^{-1}=RR^T$). I know how to compute the blocks of the inverse efficiently using an iterative algorithm. However, is there an efficient algorithm for computing the cholesky factors $R$ directly (rather than first computing the inverse, and them performing the Cholesky decomposition)?
numerical-methods numerical-linear-algebra
numerical-methods numerical-linear-algebra
edited Apr 19 '15 at 19:48
BigM
2,56611530
asked Apr 19 '15 at 19:41
yepyep
50428
edited Apr 19 '15 at 19:48
BigM
2,56611530
asked Apr 19 '15 at 19:41
yepyep
50428
edited Apr 19 '15 at 19:48
BigM
2,56611530
edited Apr 19 '15 at 19:48
BigM
2,56611530
edited Apr 19 '15 at 19:48
BigM
2,56611530
2,56611530
asked Apr 19 '15 at 19:41
yepyep
50428
asked Apr 19 '15 at 19:41
yepyep
50428
asked Apr 19 '15 at 19:41
yepyep
50428
50428
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is my matlab code
function[L]=MyChol(A)
[n,m]=size(A);
L=eye(n);
for k=1:n-1
L(k,k)=sqrt(A(k,k));
L(k+1:n,k)=(A(k+1:n,k))/L(k,k);
A(k+1:n,k+1:n)=A(k+1:n,k+1:n)-L(k+1:n,k)*L(k+1:n,k)';
end
L(n,n)=sqrt(A(n,n));
end
$A=LL^T$
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
$endgroup$
1
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is my matlab code
function[L]=MyChol(A)
[n,m]=size(A);
L=eye(n);
for k=1:n-1
L(k,k)=sqrt(A(k,k));
L(k+1:n,k)=(A(k+1:n,k))/L(k,k);
A(k+1:n,k+1:n)=A(k+1:n,k+1:n)-L(k+1:n,k)*L(k+1:n,k)';
end
L(n,n)=sqrt(A(n,n));
end
$A=LL^T$
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
$endgroup$
1
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
add a comment |
$begingroup$
Here is my matlab code
function[L]=MyChol(A)
[n,m]=size(A);
L=eye(n);
for k=1:n-1
L(k,k)=sqrt(A(k,k));
L(k+1:n,k)=(A(k+1:n,k))/L(k,k);
A(k+1:n,k+1:n)=A(k+1:n,k+1:n)-L(k+1:n,k)*L(k+1:n,k)';
end
L(n,n)=sqrt(A(n,n));
end
$A=LL^T$
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
$endgroup$
1
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
add a comment |
$begingroup$
Here is my matlab code
function[L]=MyChol(A)
[n,m]=size(A);
L=eye(n);
for k=1:n-1
L(k,k)=sqrt(A(k,k));
L(k+1:n,k)=(A(k+1:n,k))/L(k,k);
A(k+1:n,k+1:n)=A(k+1:n,k+1:n)-L(k+1:n,k)*L(k+1:n,k)';
end
L(n,n)=sqrt(A(n,n));
end
$A=LL^T$
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
$endgroup$
Here is my matlab code
function[L]=MyChol(A)
[n,m]=size(A);
L=eye(n);
for k=1:n-1
L(k,k)=sqrt(A(k,k));
L(k+1:n,k)=(A(k+1:n,k))/L(k,k);
A(k+1:n,k+1:n)=A(k+1:n,k+1:n)-L(k+1:n,k)*L(k+1:n,k)';
end
L(n,n)=sqrt(A(n,n));
end
$A=LL^T$
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
edited Apr 19 '15 at 19:50
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
answered Apr 19 '15 at 19:45
Alonso DelfínAlonso Delfín
3,78411132
3,78411132
1
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
add a comment |
1
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
1
1
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
$begingroup$
Thanks, however I am curious about an algorithm for computing $A^{-1} = L L^T$, especially assuming that $A$ is symmetric & block tridiagonal.
$endgroup$
– yep
Apr 19 '15 at 19:52
add a comment |
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