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Suppose $A, B, C$ are $n\times n$ matrices. Let $A = L_1U_1$ and $D = L_2U_2$. Then what is the LU decomposition of $$\begin{bmatrix} A&B\\ 0&D\end{bmatrix}$$ How to find this? I am able to find $$\begin{bmatrix} A&B\\ 0&D\end{bmatrix} = \begin{bmatrix} L_1&0\\ 0&L_2\end{bmatrix}\times \begin{bmatrix} U_1&X\\ 0&U_2\end{bmatrix}$$but this means $L_1X = B$ which might not be the case. How to look for this?

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  • $\begingroup$ Why might that not be the case? In the first sentence, do you mean $A,B,D$ are $n \times n$ matrices? $\endgroup$ – Viktor Glombik Nov 24 '18 at 11:39
  • $\begingroup$ Because B can have a different LU decomposition. $\endgroup$ – Mittal G Nov 24 '18 at 11:41
  • $\begingroup$ If $L_1$ is invertible then such a matrix $X$ exists. $\endgroup$ – Mittal G Nov 24 '18 at 11:42
  • $\begingroup$ @Viktor Glombik Yes all are $n\times n$ matrices. $\endgroup$ – Mittal G Nov 24 '18 at 11:51
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If $rank(L_1)=rank([L_1,B])$, then $L_1L_1^+B=B$ and one has the solution

$diag(L_1,L_2)\begin{pmatrix}U_1&L_1^+B\\0&U_2\end{pmatrix}$.

Otherwise, I'm not sure there is a solution in closed form.

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