Centre of the general linear group
Scalar transformation criterion
Let
Proof
Let
be the eigenvalue corresponding to π π . Then π― π π β π = 0 π π + 1 π― π = π π + 1 π― π + 1 = Ξ¦ ( π― π + 1 ) = Ξ¦ ( π β π = 0 π― π ) = π β π = 0 π π π― π’ and since the decomposition of a vector into basis vectors is unique, it follows
for all π π = π π + 1 . The converse is trivial, since every nonzero vector is an eigenvalue of a scalar transformation. π = 0 , β¦ , π + 1