Cholesky decomposition when deleting one row and one and column.

Given Cholesky decomposition of a symmetric positive semidefinite matrix $A = LL^T$. Now, suppose that we delete the $i$-th row and the $i$-th column of $A$ to obtain $A'$ ($A'$ is also a symmetric positive semidefinite matrix), and the Cholesky decomposition of this new matrix is $A' = L'(L')^T$.

Is there any efficient way to obtain $L'$ from $L$?

Note: In case we delete the last row and the last column of $A$, the problem becomes simple, we just delete the last row and last column of $L$ to obtain $L'$. Other cases, for me, are not trivial.

Thank you so much.

• Have you tried just deleting the $i$th row and $i$th column of $L$ to see if that works? Honestly, not sure but it seems like it's worth a numerical experiment. – Omnomnomnom Aug 18 '16 at 19:55
• I tried. That didn't work. – Tiep Vu Aug 18 '16 at 21:02

Let $$C\equiv \left[\begin{array}{ccc}C_{11} & C_{12} & C_{13} \\ C_{12}^T & C_{22} & C_{23}\\C_{13}^T & C_{23}^T & C_{33} \end{array}\right] = \left[\begin{array}{ccc}L_{11} & 0 & 0 \\ L_{21} & l_{22} & 0\\L_{31} & L_{32} & L_{33} \end{array}\right]\left[\begin{array}{ccc}L_{11} & 0 & 0 \\ L_{21} & l_{22} & 0\\L_{31} & L_{32} & L_{33} \end{array}\right]^T$$ be the Cholesky factorization of the matrix $C$ and let $$\bar{C}\equiv \left[\begin{array}{ccc}C_{11} & 0 & C_{13} \\ 0 & 0 & 0\\C_{13}^T & 0 & C_{33} \end{array}\right] = \left[\begin{array}{ccc}\bar{L}_{11} & 0 & 0 \\ \bar{L}_{21} & \bar{l}_{22} & 0\\\bar{L}_{31} & \bar{L}_{32} & \bar{L}_{33} \end{array}\right]\left[\begin{array}{ccc}\bar{L}_{11} & 0 & 0 \\ \bar{L}_{21} & \bar{l}_{22} & 0\\\bar{L}_{31} & \bar{L}_{32} & \bar{L}_{33} \end{array}\right]^T$$ be the Cholesky factorization of the matrix $\bar{C}$ obtained from the matrix $C$ by zeroing $i$-th row and column. Then $\bar{L}_{11} = L_{11}$, $\bar{L}_{21} = 0$, $\bar{L}_{31} = L_{31}$, $\bar{l}_{22} = 0$, $\bar{L}_{32} = 0$ and \begin{align}C_{33} &= L_{31}L_{31}^T + L_{32}L_{32}^T + L_{33}L_{33}^T \\ C_{33} &= L_{31}L_{31}^T + \bar{L}_{33}\bar{L}_{33}^T\end{align} Thus $$\bar{L}_{33}\bar{L}_{33}^T = L_{33}L_{33}^T + L_{32}L_{32}^T\tag{1}$$ since $L_{32}$ is a vector, (1) gives a rank-1 update of the Cholesky factorization $L_{33}L_{33}^T$. Now removing $i$-th row and column of $\bar{C}$ is trivial.
Rank-1 update of the Cholesky factorization is a standard procedure and can be done efficiently in $O(k^2)$ operations, where $k$ is size of the matrix $L_{33}$.