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I cannot find the $LU$ decomposition of anti-diagonal matrix
$$\begin{bmatrix} 0 &0 &0 &1 \\ 0 &0 &2 &0 \\ 0 &3 &0 &0 \\ 4 &0 &0 &0 \end{bmatrix}.$$

When I find solution using naive ways, some equation rule out.

Also, do all matrices have $LU$ decomposition?

io10
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    Possible duplicate http://math.stackexchange.com/questions/218770/when-does-a-square-matrix-have-an-lu-decomposition – uranix Jun 26 '15 at 11:59

2 Answers2

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LU decomposition is possible only for matrices with nonzero leading principal minors. The first is zero for this matrix

uranix
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  • What is leading principle minor ? – io10 Jun 26 '15 at 12:09
  • @io10 it is the determinant of the upper top block. The first block is $1 \times 1$, the second - $2 \times 2$, etc $$ \Delta_0 = \operatorname{det}||0|| = 0 $$ – uranix Jun 26 '15 at 12:11
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Let $A$ be your matrix above, then an LU factorisation exists if the matrix $A_j$ is invertible for all $j = 1,2,...,n$.

$A_j$ is denoted as the $j^{th}$ principle sub-matrix where $A_j \in \mathbb{C}^{j \times j}$. So for instance,

$A_3 = $$\begin{bmatrix} 0 &0 &0 \\ 0 &0 &2 \\ 0 &3 &0 \end{bmatrix}.$$ $

This is not invertible (neither is it invertible for any $ j = 1,2,3,4$ so no such LU factorisation exists.

user65972
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