[Math] Cholesky decomposition of the inverse of a matrix

cholesky decompositioninversematricesmatrix decompositionnumerical linear algebra

I have the Cholesky decomposition of a matrix $M$. However, I need the Cholesky decomposition of the inverse of the matrix, $M^{-1}$. Is there a fast way to do this, without first computing $M^{-1}$? In other words, is there a relationship between the Cholesky decompositions of a matrix and of its inverse?

My matrix is a covariance matrix and, hence, positive-definite.

Best Answer

If $L^T L = R$ is the available Cholesky decomposition, then inverting both sides of the equation you get,

$$L^{-1}(L^{T})^{-1} = R^{-1} $$

And since transposition and inverse are interchangeable:

$$L^{-1}(L^{-1})^{T} = R^{-1} $$

So if you define $P = (L^{-1})^T$ this is your desired answer. In other words,

$$P^{T}P=R^{-1}$$

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The Cholesky decomposition allows you to write (for real symmetric positive definite matrix $\Sigma$) $$\Sigma = U^T U$$ with $U$ an upper triangular matrix. Therefore, $$A^T \Sigma A = A^T U^T U A = B^T B$$ with $B= UA$. Note however that $B$ is (in general) not upper triangular anymore such that $B^T B$ is not the standard Cholesky decomposition (but of course very similar to a Cholesky decomposition).

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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

$A=LL^T$

Best Answer

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