Re: Center of a matrix group

Let $\displaystyle A$ a matrix in the center. First we will try to get information about the fact that $\displaystyle AD=DA$ for all diagonal matrix $\displaystyle D$. We must have $\displaystyle (AD)_{ij} = \alpha_ja_{ij}$ and $\displaystyle (DA)_{ij} = a_{ij}\alpha_i$ if $\displaystyle D = \mathrm{diag}(\alpha_1,\ldots,\alpha_n)$. If we fix $\displaystyle i\neq j$ we must have $\displaystyle a_{ij} =0$ since we can choose a matrix $\displaystyle D$ such that $\displaystyle \alpha_i\neq \alpha_j$. Hence $\displaystyle A$ is a diagonal matrix: $\displaystyle A =\mathrm{diag}(a_1,\ldots,a_n)$. Now if you take $\displaystyle G$ with $\displaystyle g_{ij} = i$ you will get that $\displaystyle a_i=a_j$ for all $\displaystyle i$ and $\displaystyle j$.

Re: Center of a matrix group

Any element in the center of the group must commute with an arbitrary matrix. The idea is to use very simple matrices, multiply on both sides, and get conditions that your matrix would have to satisfy so that it would commute with your simple matrix.

These "simple matrices" will be elementary matrices; that is, matrices that correspond to elementary column (when multiplying on the right) or row (when multiplying from the left) operations. For example, if $\displaystyle A$ is in the center of the group, then we should have

$\displaystyle \begin{pmatrix}-1&0&\cdots&\\0&1&0&\cdots\\0&0&1&\cdots\\\cdots&\c dots&&\end{pmatrix}A=A\begin{pmatrix}-1&0&\cdots&\\0&1&0&\cdots\\0&0&1&\cdots\\\cdots&\c dots&&\end{pmatrix}$

What does this tell us about $\displaystyle A$?

(Note: That matrix is supposed to be $\displaystyle I_n-2E_{1,1}$, where $\displaystyle I_n$ represents the $\displaystyle n\times n$ identity matrix, and $\displaystyle E_{i,j}$ the matrix with a 1 in position (i,j) and 0's elsewhere.)