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I want to compute the centre of $GL(n,\mathbb{Z})$

Now I saw a good answer by @1015 on the following proof by @Ben: The Center of $\operatorname{GL}(n,k)$.

Here is the answer: https://math.stackexchange.com/a/299637/381362

I understand that $(I_n+E_{i,j})A=A(I_n+E_{i,j})$ and therefore all the matrices of the form: $(I_n+E_{i,j}) \in Z(GL(n,k))$

Here are the parts where I lack understanding:

  • What is the purpose of the equivalence: $(I_n+E_{i,j})A=A(I_n+E_{i,j})\quad \Leftrightarrow\quad E_{i,j}A=AE_{i,j}$. I understand why that equivalence is true but $E_{i,j} \notin GL(n,k)$ so I don't understand why that piece of information is necessary.

  • What does he means by: "Now compute the appropriate coefficients in the latter"?

  • How does he know that this represents all the elements of $Z(GL(n,k)$?

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Sylvester Stallone
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    "I understand that $(I_n+E_{i,j})A=A(I_n+E_{i,j})$ And therefore All the matrices of the form: $(I_n+E_{i,j}) \in Z(GL(n,k))$". This is wrong: $I + E_{ij}$ is not generally an element of $Z(GL(n,k))$. What we can say though is that if $A$ is an element of the center, then it must commute with $I + E_{ij}$, which is what 1015 is actually saying – Ben Grossmann Oct 24 '16 at 18:01
  • Wait If A is an element of the center of $GL(n,k)$ doesn't it commute with every element of $GL(n,k)$? – Sylvester Stallone Oct 24 '16 at 18:05
  • Yes, and $(I + E_{ij})$ is such an element – Ben Grossmann Oct 24 '16 at 18:05
  • So what does $I+E_{i,j}$ tell us about A? (which in its generality is $Z(GL(n,k)$ – Sylvester Stallone Oct 24 '16 at 18:07
  • It tells us that $A E_{ij} = E_{ij} A$, which tells us something about the entries ("coefficients") of $A$. – Ben Grossmann Oct 24 '16 at 18:19

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In his proof, the answerer starts by establishing that $$ A \in Z(GL) \implies A(I + E_{ij}) = (I + E_{ij})A \quad \forall i\neq j \implies \\ AE_{ij} = E_{ij}A \quad \forall i\neq j $$ Perhaps it seems strange that we're not "using every element of $Z$", but this will be enough to show that $A$ must be a multiple of the identity, as we show with a "computation of appropriate coefficients".

In particular: note that $AE_{ij} = E_{ij}A$ means that all their entries are equal. However, the $p,q$ entry of $AE_{ij}$ is given by $$ [AE_{ij}]_{p,q} = [Ae_ie_j^T]_{p,q} = e_p^T Ae_i e_j^Te_p = \delta_{jq} A_{pi} $$ where $\delta$ is a Konecker delta, and $$ [E_{ij}A]_{p,q} = [e_ie_j^TA]_{p,q} = e_p^T e_i e_j^TAe_p = \delta_{ip} A_{jp} $$ From there, one must note that these will be equal for all combinations of $i,j,p,q$ only if $A$ is a multiple of the identity.

Alternative explanation: $AE_{ij} = (Ae_i)e_j^T$ is a matrix whose $j$th column is occupied by the $i$th column of $A$, and whose other entries are all $0$. $E_{ij}A = e_i(e_j^TA)$ is a matrix whose $i$th row is occupied by the $j$th row of $A$, and whose other entries are all $0$.

These two products must be equal for all $i \neq j$. This is enough for us to deduce that $A$ is a multiple of the identity.

Ben Grossmann
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