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?
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.
