In pg. 81 of the Physics from Symmetry book, the author deduced that the $(\frac{1}{2}, \frac{1}{2})$ representation of the Lorentz $SO(1,3)$ group acts on $2\times2$ Hermitain matrices using the following arguments:
First assume that the $(\frac{1}{2}, \frac{1}{2})$ representation acts on general $2\times2$ matrices.
Any $2\times2$ matrix can be written as a sum of a Hermitian and an anti-Hermitian matrix.
Transformations in the $(\frac{1}{2}, \frac{1}{2})$ representation always transform a $2\times2$ Hermitian matrix into another $2\times 2 $ Hermitian matrix and equivalently an anti-Hermitian matrix into another anti-Hermitian matrix.
The above point tells us that the if $(\frac{1}{2}, \frac{1}{2})$ representation acts on general $2\times2$ matrices, it is a reducible representation. But this contradicts the fact that the $(\frac{1}{2}, \frac{1}{2})$ representation is irreducible.
Thus the $(\frac{1}{2}, \frac{1}{2})$ representation acts on $2\times 2$ Hermitian matrices.
The leap in logic from the 4th to 5th point confuses me. Why can't the $(\frac{1}{2}, \frac{1}{2})$ representation act on $2\times2$ anti-Hermitian matrices?
