Compute the centralizer of

(0 -1)

(1 0)

in GL_{2}(R).

I know that the centralizer of an element g of a group G is all the elements of G that commute with g, but I am unsure as to how to compute it.

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- Oct 11th 2012, 12:49 PMTheHowlingLungCentralizers
Compute the centralizer of

(0 -1)

(1 0)

in GL_{2}(R).

I know that the centralizer of an element g of a group G is all the elements of G that commute with g, but I am unsure as to how to compute it. - Oct 13th 2012, 09:18 AMGJARe: Centralizers
Hi HowlingLung.

I'm going to call the matrix you gave M. If we want to determine what the elements that commute with M look like in $\displaystyle GL_{2}(R)$, we take an arbitrary matrix

$\displaystyle

A=

\begin{bmatrix}

a & b\\

c &d

\end{bmatrix}$

then compute AM and MA. Since we want A to commute with M we set up the equality AM=MA. This should give a=d and c=-b. Now replace d with a and c with -b in A and we'll have what an element of the centralizer of M looks like. The last thing we need to do is make sure that the matrix A we have just found really does belong to $\displaystyle GL_{2}(R)$. This means we want A to be invertible. Looking at A for a moment, we see that this is the case when $\displaystyle a\neq 0$ or $\displaystyle b\neq 0.$

Does this help? Let me know if anything is unclear.

Good luck! - Oct 13th 2012, 02:01 PMDevenoRe: Centralizers
first let's find all matrices that commute with:

$\displaystyle A = \begin{bmatrix}0&-1\\1&0 \end{bmatrix}$

so if:

$\displaystyle M = \begin{bmatrix}a&b\\c&d \end{bmatrix}$

then:

$\displaystyle MA = \begin{bmatrix}b&-a\\d&-c \end{bmatrix};\ AM = \begin{bmatrix}-c&-d\\a&b \end{bmatrix}$

comparing the two matrices, we obtain:

a = d, b = -c.

so M must be of the form:

$\displaystyle M = \begin{bmatrix}a&b\\-b&a \end{bmatrix} $

and it should be clear that every matrix of this form does indeed commute with A.

so which are these matrices M are invertible?

calculating det(M) = a^{2}+ b^{2}, we see it is precisely the non-zero matrices of the given form.

(this is really the same as GJA's answer, just formatted a little cleaner. note that we can identify the matrix A with the complex number 0+1i = i, and since the complex numbers form a field, it follows that the multiplicative group of C (as matrices of the form for M) forms an abelian subgroup of GL(2,R). so we know that C(A) contains at least all of this subgroup, since of those matrices commute with each other. so all we really need to do is show that the matrices outside of the complex numbers (as matrices of the form M) do not commute with A, which is what the first part of this post shows).