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

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 $n\geq 2$ 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 $i\neq j$.

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

Suppose $A$ is a matrix with $a_{ji}\neq 0$ for some $i\neq j$. 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 $a_{ii}\neq a_{jj}$. 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 $R$.