How to efficiently re-compute the inverse of positive definite matrix when its i-th row and column are...












1














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







share|cite|improve this question



























    1














    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







    share|cite|improve this question

























      1












      1








      1


      1





      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







      share|cite|improve this question













      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






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      asked Nov 20 '18 at 1:47









      olivia

      766616




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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 $c=v^TA^{-1}v$.






          share|cite|improve this 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 $c=v^TA^{-1}v$.






            share|cite|improve this answer


























              0














              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 $c=v^TA^{-1}v$.






              share|cite|improve this answer
























                0












                0








                0






                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 $c=v^TA^{-1}v$.






                share|cite|improve this answer












                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 $c=v^TA^{-1}v$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 '18 at 7:06









                user1551

                71.3k566125




                71.3k566125






























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