Question

Can i will be possible to find lu decompostion for non invertible matrix

Answer 1

Any square matrix A
A
admits an LUP factorization. If A
A
is invertible, then it admits an LU (or LDU) factorization if and only if all its leading principal minors are non-zero. If A
A
is a singular matrix of rank k
k
, then it admits an LU factorization if the first k
k
leading principal minors are non-zero, although the converse is not true.
This implies that for a square matrix:

LUP always exists (We can use this to quickly figure out the determinant).
If the matrix is invertible (the determinant is not 0), then a pure LU decomposition exists only if the leading principal minors are not 0.
If the matrix is not invertible (the determinant is 0), then we can't know if there is a pure LU decomposition.
The problem is this third statement here. “If A
A
is a singular matrix of rank k
k
, then it admits an LU factorization if the first k
k
leading principal minors are non-zero”, gives us a way to find out if LU decomposition exists for a singular (non-invertible) matrix. However, it then says, “although the converse is not true”, implying that even if a leading principal minor is 0, that we could still have a valid LU decomposition that we can't detect.

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