Center of general linear group is group of scalar matrices over center
- 1 Statement
- 2 Proof
Let be a (not necessarily commutative) unital ring, and be a natural number. Let denote the group of invertible matrices over . The center of is the subgroup comprising scalar matrices whose scalar entry is a central invertible element of .
In particular, for a field , the center comprises scalar matrices with a nonzero scalar value.
Note that the proof is clear for , so we consider the case here.
First step: any element of the center commutes with off-diagonal matrix units
Suppose are distinct elements of and . Define to be the matrix with in the entry and zeroes elsewhere. is termed the matrix unit.
Define as the sum of the identity matrix and .
- and are two-sided multiplicative inverses for any . Thus, .
- Any matrix that commutes with must also commute with , because of distributivity and the fact that the matrix commutes with the identity. Thus, any matrix in the center of commutes with for .
Second step: Anything that commutes with off-diagonal matrix units is a diagonal matrix
Suppose is a matrix with for some . Consider the matrix . Then, the entry of is nonzero, while the entry of is zero. Thus, any matrix that commutes with all the off-diagonal matrix units cannot have any off-diagonal entries.
Third step: Any diagonal matrix that commutes with all permutation matrices is scalar
Suppose is a diagonal matrix with . Then does not commute with the permutation matrix corresponding to the transposition of and , because conjugation by that matrix switches with .
Combining the facts
Combining the first two steps yields that any matrix in the center of must be diagonal, and the third step then yields that it must be scalar. Looking at when two scalar matrices commute, we see that the matrix must in fact be a scalar matrix with the scalar value itself a central and invertible element of .