Find the center of the group $\operatorname{GL}(n,\mathbb R)$ of invertible $n \times n$ matrices. Find the center of the group  $\operatorname{GL}(n,\mathbb R)$ of invertible $n \times n$ matrices.
Please can someone please help me? I know that by definition the center $Z$ of a group $G$ is defined by $Z(G) = \{g \in G\ |\ ag = ga ,\, \forall a \in G\}$.
I know that the identity matrix commutes with any matrix. I also notice by computing several matrix products that if we have a matrix with a main diagonal and all other entries are zero, then the given matrix commutes. In addition, I know that the determinant cannot be zero since zero times another matrix will only be zero.
Please I would really appreciate the help.
Thank you.
 A: Outline: Let $I \in GL(n, \mathbb{R})$ be the identity matrix.  Fix indices $i,j$, and
let $E_{ij}$ be the matrix that is zero everywhere except in the $i$th row and $j$th column, where it has the entry $1$.  Let $A$ be an arbitrary member $GL(n,\mathbb{R})$ with entries $a_{ij}$.  We note that
$$
A(I + E_{ij}) = AI + AE_{ij} = A + 
\pmatrix{
&&&a_{1i}&&&\\
0&\cdots&0& \vdots&0&\cdots&0\\
&&&a_{ni}&&&}
$$
That is, every column of $A E_{ij}$ is all zeros except the $j$th, which is simply the $i$th column of $A$.  Similarly, we compute
$$
(I + E_{ij})A = IA + E_{ij}A = A + 
\pmatrix{
0&\cdots&0\\
&\vdots&\\
a_{j1} & \cdots &a_{jn}\\
&\vdots&\\
0&\cdots&0}
$$
That is, every row of $E_{ij}A$ is all zeros except the $i$th, which is simply the $j$th row of $A$.
Now, for $A$ to be in the center, we must have $(I + E_{ij})A = A(I + E_{ij})$.  It follows that $E_{ij}A = A E_{ij}$.  Based on this, what can we conclude about the entries of $A$?  (In particular, show that the off-diagonal entries must be zero, and that $a_{ii} = a_{jj}$ for each pair of indices $i,j$).
Show that any matrix satisfying the above conditions lies in the center.
