Is there any situation where the LDU decomposition is the same as the eigenvalue decomposition?

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I was just wondering if there are any situation where the LDU decomposition is the same as eigenvalue decomposition (diagonalization)? The only way this can be possible if L and U are inverse so symmetric matrices seems like a good candidate, but do we need any more requirements.

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The inverse of an upper triangular matrix is always upper triangular. So, $L$ and $U$ must be diagonal. Hence, the original matrix must be diagonal.