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find the center of the group of invertible 2 x 2 matrices with real entries.

Attempt: By definition, the center of a group Z(G), is where all the elements are commutative. If G = { invertible 2 x 2 matrices}, then doing several multiplications of matrices, Z(G) is equal to the 2 x 2 matrix where the main diagonal is k, and the rest of the entries are zero, where k is not equal to 0, and k is an element from the real numbers.

Can anyone please help me? I don't know if there are others. I have tries to do several multiplications to see which 2 matrices commute.

Thank you.

Davide Giraudo
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user78424
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1 Answers1

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General proof for the $n\times n$ case:

Let $A\in Z(g)$, $1\le i,j\le n$. Consider the elementary matrix $E_{ij}$ which only non 0 entry is on $(i,j)$.

Then $I + E_{ij}$ is invertible, so$$ A(I + E_{ij}) = (I + E_{ij})A \implies AE_{ij} = E_{ij}A $$

Now, if you know the center of all $2\times 2$ matrices you know that $A$ has the form $xI$.

Otherwise, let us prove it:

$$A=\sum A_{kl}E_{kl}\\ AE_{ij} = E_{ij}A\iff \sum A_{kl}E_{kl}E_{ij} = \sum A_{kl}E_{ij} E_{kl}\\ \iff \sum A_{ki}E_{kj} = \sum A_{kl}E_{il} $$ This gives looking at LHS:$$k\neq i\implies A_{ki}=0\\ k=i\implies A_{ii} = A_{kk} $$that is, $A =A_{11}I$.

Now it remains to check that it is in the center.

mookid
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