Transformation matrix for Dirac equation The Dirac wavefunction $\psi(x)$, a four component spinor, transforms under Lorentz transformations according to
$$\psi'(x')=S\psi(x)$$
where $S$ is the transformation matrix.
In Ashok Das' QFT book, pg 77, he says that the product $S^\dagger S$ is a non negative matrix, i.e. all the matrix elements are positive. Why is that so?
 A: The question boils down to concepts of pure linear algebra.
According to the derivation of pg.77 of Ashok's book we only need to show in eq. (3.52) that $$Tr(S^\dagger S) > 0$$
The matrix $S$ can be written as a polar decomposition $S=UP$ (where $U$ is an unitarian and $P$ is a positive-semidefinite hermitian matrix) which always exists.
(see https://en.wikipedia.org/wiki/Polar_decomposition ).
EDIT:
The square of the polar factor $P^2$ can be found by the relation ($P^\dagger = P$ since it is hermitian):
$$P^2  =  P^\dagger U^\dagger U P =  S^\dagger S$$
According to wikipedia $P$ is even completely positive-definite if $S$ is invertible (each representation of Lorentz-group is invertible).
Then $P^2$ is also hermitian and positive-definite
and can be diagonalised as it is hermitian into
strictly positive eigenvalues $\lambda_i(P^2)$
So we get:
$$Tr(S^\dagger S)= Tr(P^2) =\sum_i \lambda_i(P^2) >0$$
It has to be recalled that S is a reducible representation $D^{(1/2,0)}\oplus D^{(0,1/2)}$ of the Lorentz-group that is composed of 2 fundamental representations of the SL(2,C)-group that comprises 2x2 complex matrices with determinant 1 that are generally not unitary.
