# LU decomposition of a matrix given LU decomposition of its blocks.

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?

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

## 1 Answer

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.