# 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 $n \ge 2$ 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 $R$.