Center of general linear group is group of scalar matrices over center

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Statement

Let R be a (not necessarily commutative) unital ring, and n be a natural number. Let GL(n,R) denote the group of n \times n invertible matrices over R. The center of GL(n,R) is the subgroup comprising scalar matrices whose scalar entry is a central invertible element of R.

In particular, for a field k, the center comprises scalar matrices with a nonzero scalar value.

Proof

Note that the proof is clear for n = 1, so we consider the case n \ge 2 here.

First step: any element of the center commutes with off-diagonal matrix units

Suppose i,j are distinct elements of \{ 1,2,3, \dots, n \} and \lambda \in R. Define e_{ij}(\lambda) to be the matrix with \lambda in the (ij)^{th} entry and zeroes elsewhere. e_{ij}(1) is termed the (ij)^{th} matrix unit.

Define E_{ij}(\lambda) as the sum of the identity matrix and e_{ij}(\lambda).

Note that:

  • E_{ij}(\lambda) and E_{ij}(-\lambda) are two-sided multiplicative inverses for any \lambda \in R. Thus, E_{ij}(\lambda) \in GL(n,R).
  • Any matrix that commutes with E_{ij}(1) must also commute with e_{ij}(1), because of distributivity and the fact that the matrix commutes with the identity. Thus, any matrix in the center of GL(n,R) commutes with e_{ij}(1) for i \ne j.

Second step: Anything that commutes with off-diagonal matrix units is a diagonal matrix

Suppose A is a matrix with a_{ji} \ne 0 for some i \ne j. Consider the matrix B = e_{ij}(1). Then, the (jj)^{th} entry of AB is nonzero, while the (jj)^{th} entry of BA is zero. Thus, any matrix that commutes with all the off-diagonal matrix units e_{ij}(1) cannot have any off-diagonal entries.

Third step: Any diagonal matrix that commutes with all permutation matrices is scalar

Suppose A is a diagonal matrix with a_{ii} \ne a_{jj}. Then A does not commute with the permutation matrix corresponding to the transposition of i and j, because conjugation by that matrix switches a_{ii} with a_{jj}.

Combining the facts

Combining the first two steps yields that any matrix in the center of GL(n,R) 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 R.