Center of a matrix group

**sashikanth** · June 25th 2011, 03:04 AM

The question is for me to find the center of a group comprising of non singular square matrices of order n.

So if any member of the center is A, and G is any matrix in the group, we need AG = GA --> A = G-1 * A * G. (Where G-1 = G inverse)

I can intuitively say that only elements of the form A = nI satisfy this equation, but how do I prove it?

Please help

**girdav** · June 25th 2011, 03:26 AM

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

**topspin1617** · June 25th 2011, 03:29 AM

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 $A$ is in the center of the group, then we should have

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

What does this tell us about $A$ ?

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

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