I've been trying to think of this. My teacher said there could be a nonsingular matrix with all nonzero entries that can't be decomposed without a permutation.
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That works but it has zero entries. – Radagast Oct 05 '17 at 07:24
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Why not just take a non-identity permutation matrix? – Kapil Apr 28 '23 at 03:06
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An LU factorization of a nonsingular matrix without permutations exists if and only if all leading principal submatrices are nonsingular.
We can construct a matrix for which the permutations are necessary by picking one which violates this condition. This is impossible for $1\times 1$ and $2\times 2$ matrices (either there would be a zero element or the whole matrix would be singular).
So let's try a $3\times 3$ one. For example, $$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \end{bmatrix}. $$ The leading principal submatrices are $$ \begin{bmatrix} 1 \end{bmatrix}, \quad \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}, \quad \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 2 \\ 1 & 2 & 3 \end{bmatrix}, $$ where the second one is singular. Hence the LU factorization is impossible without permuting. It is easy to see that by performing the first step of the elimination, which leaves a zero entry in the $(2,2)$ position, which cannot be used as a pivot.
Algebraic Pavel
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