basic vector being hermitian 
If the space has a mixed metric signature, not all the basis vectors are Hermitian.
  Nevertheless, they are defined to be self-adjoint under reversion. The vector transpose
  conjugate is, therefore, not necessarily the same as the Hermitian conjugate
  of its matrix representation. This distinction becomes important when considering
  Lorentz transformations in Minkowski Space.
  (Classical Mechanics - J. Michael Finn)

What does "basis vectors being Hermitian" mean?
And how can vector transpose conjugate differ from hermitian conjugate?
Thanks.
 A: What a terrible way to write a book. That passage makes no sense without stuff that's very briefly explained a lot later in the book. Take a look at Section 11.3 (p. 463 in the online version I linked to in a comment), which discusses the Clifford algebra of Minkowski space. The term "reversion" is also taken from the Clifford algebra context, but that's never mentioned when it's introduced. See this Wikipedia section for some background on this.
Basically, if you want to understand this without going into the Clifford algebra stuff that's apparently not supposed to be introduced at this point, you can imagine vectors in Minkowski space to have imaginary time components. Then if you want to form their norm, which contains the negative square of the time component, and you want to write it as a product of a row vector with a column vector, you get the wrong sign if you take the row vector to be the conjugate transpose of the column vector; you need to just take the transpose without conjugating it.
