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What would be an example of a matrix that would not have a $LDL^T$ decomposition? This is a trivial case but I was thinking of a zero matrix which would result in L being an identity matrix and D would be a zero matrix. When you compute $LDL^T$ it still gives you the original zero matrix so I don't believe this is correct.

What about a singular matrix?

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    Thank you both, it makes sense now. I was just not clear before on the use of the permutation matrix.2010-11-24

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So that this does not remain unanswered:

$\begin{pmatrix}0&1\\1&0\end{pmatrix}$

is the simplest matrix that does not possess an $\mathbf L\mathbf D\mathbf L^T$ decomposition.

In general, any symmetric matrix whose leading submatrix (the submatrix formed by the first few rows and columns) is singular will not possess such a decomposition.

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    By extension, Cholesky too will fail if given a matrix with a singular leading submatrix.2010-11-24