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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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An $\mathbf L\mathbf D\mathbf L^T$ decomposition is not a "block LU decomposition". – J. M. Nov 24 '10 at 4:59
In any event, $$\begin{pmatrix}0&1\\1&0\end{pmatrix}$$ – J. M. Nov 24 '10 at 5:01
Sorry about that I was mixing up my questions. – Planeman Nov 24 '10 at 5:03
@Planeman: That matrix has no LU decomposition. You can show this with bare hands: Try writing $\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}a&0\\b&c\end{pmatrix}\begin{p‌​matrix}d&e\\0&f\end{pmatrix}$ and see what happens. – Jonas Meyer Nov 24 '10 at 5:11
@Planeman, the $\mathbf L\mathbf D\mathbf L^T$ decomposition usually does not involve pivoting of any sort. – J. M. Nov 24 '10 at 5:11
up vote 5 down vote accepted

So that this does not remain unanswered:


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. – J. M. Nov 24 '10 at 5:26

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