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Sunday, September 12, 2010

Center of General Linear Group

Definition: The general linear group $$\text{GL}_n(\mathbb{R})$$ is the group of all invertible $$n\times n$$ matrices under matrix multiplication.

Definition: For a group G we define the "center" Z(G) to be the subgroup $$\text{Z}(G)=\{z\in G| zg = gz\}$$.

It is a rather well-known result that the center of $$\text{GL}_n(\mathbb{R})$$ are the matrices $$kI$$ where I is the identity matrix and k is some non-zero real number. We will derive this result.

For $$N\times N$$ matrices define $$E_{nm}$$ to be the matrix with 0's everywhere except at the n-th row and m-th coloum where it has 1 as its entry. The problem however is that $$E_{nm}$$ are not invertible unless n=m, so we will rather consider the matrix $$I+E_{nm}$$ for $$n\not = m$$.

If A is some matrix in the general linear group then it commutes with all $$I+E_{nm}$$. Let $$a_{ij}$$ be the ij-entry of A and $$e_{ij}$$ be the ij-entry of $$I+E_{nm}$$. We have that $$A(I+E_{nm}) = (I+E_{nm})A$$. Therefore, $$\Sigma_k a_{ik}e_{kj} = \Sigma_k e_{ik}a_{kj}$$.

If $$j\not = m$$ then $$e_{kj}=0$$ except when k=j, so $$\Sigma_k a_{ik}e_{kj} = a_{ij}$$.

If $$i\not = n$$ then $$e_{ik}=0$$ except when k=i, so $$\Sigma_k e_{ik}a_{kj} = a_{ij}$$ .

If $$j=m$$ then $$e_{kj}=0$$ except when k=j and k=n, so $$\Sigma_k a_{ik}e_{kj} = a_{ij}+a_{in}$$.

If $$i=n$$ then $$e_{ik}=0$$ except when k=i and k=m, so $$\Sigma_k e_{ik}a_{kj} = a_{ij}+a_{mj}$$.

These are our four possible cases. In the second two cases, that is when j=m and i=n we have $$a_{nm}+a_{nn}=a_{nm}+a_{mm}$$ we get that $$a_{nn} = a_{mm}$$. But $$n,m$$ range over $$\{1,2,...,N\}$$ so that means all the diagnol terms are all equal to one another. Thus, A is a matrix with equal diagnol terms.

In the second and third case, that is when i!=n and j=m we get $$a_{im}=a_{im}+a_{in}$$ so we have that $$a_{in}=0$$ for all i!=n. This means all terms in the n-th coloumn are zero except possibly on the diagnol. But $$n\in \{1,2,...N\}$$ so if we vary n over this set we get that all coloumns must be zero except possibly at the diagnol terms.

Thus, A must be a matrix with all terms zero off the diagnol, but all diagnol terms equal. If k is the common value of all these diagnol terms it follows that A=kI where I is the $$N\times N$$ identity matrix. But we require for A to be invertible, therefore k is a non-zero real number.

We have shown it is necessary for any matrix in the center of the general linear group to be of the form kI where k!=0. Obviously, it is sufficient also since the identity matrix communutes with all matrices of that size.

We have finally proved that $$\text{Z}[\text{GL}_n(\mathbb{R})] = \{ kI| k \in \mathbb{R}^{\times} \}$$.

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