It is said that the off-diagonal elements of density matrix are "coherence". When a system interacts with its environment the off-diagonal elements decay and the final density matrix is the diagonal one, a statistical mixture. This process is called decoherence.
We know that every density matrix can be diagonalized in some basis.
What would decoherence be when the density matrix is diagonal in some basis?
Regarding to the question of basis dependency of decoherence, I can perhaps give some examples to clear things up. For a pure qubit state $|0\rangle$, when complete decoherence happens, the basis where the maximum coherence loss is witnessable is in the maximum coherent state $$ \rho= H | 0 \rangle \langle 0 | H^{\dagger} = \frac{1}{2} \left( \begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right) \rightarrow \frac{1}{2} \left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right) , $$ where $H$ is a unitary transform by a Hadamard matrix, $\rho$ is turned into the maximally mixed state. No matter what unitary operation you perform on it, coherence cannot be recovered.
However, if $|0\rangle$ and the environment interact in a way where partial decoherence happens and a full decoherence is only viewable from a non-maximal coherence basis $$ \rho = U | 0 \rangle \langle 0 | U^{\dagger} = \frac{1}{3} \left( \begin{matrix} 1 & -\sqrt{2} \\ -\sqrt{2} & 2 \\ \end{matrix} \right) \rightarrow \frac{1}{3} \left( \begin{matrix} 1 & 0 \\ 0 & 2 \\ \end{matrix} \right), $$
where $$ U= \left( \begin{matrix} \sqrt{\frac{1}{3}} & \sqrt{\frac{2}{3}} \\ -\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{3}} \\ \end{matrix} \right). $$ The sight from a Hadamard matrix point of view(where maximal coherence is exhibited) of the decohered state would be $$ \frac{1}{6} \left( \begin{matrix} 3 & -1 \\ -1 & 3 \\ \end{matrix} \right), $$ which still has coherence.
