Why choose traceless matrices as basis?

[phoenix95]

While writing down the basis for SU(2), physicists often choose traceless hermitian matrices as such, often the Pauli matrices. Why is this? In particular why traceless, and why hermitian?

 

[fresh_42]

Skew-Hermitian is a consequence of unitary, traceless of determinant one:
$$
U^\dagger U = I \Longrightarrow D(U^\dagger U) = U^\dagger \cdot I + I \cdot U = U^\dagger + U = D(I) = 0\\
\det U = 1 \Longrightarrow D(\det U) = \operatorname{tr}U = D(1) = 0
$$
Here's the computation for the determinant in detail:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/
and here are some remarks on ##\operatorname{SU}(2)##:
https://www.physicsforums.com/insights/representations-precision-important/
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/