$Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $
This article is the general case for $GL(n,k)$ where $k$ is a field. Could I prove it only with a basic linear algebra?
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2Well, the answer in that question is basic linear algebra. – ronno May 23 '13 at 15:41
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@robjohn It even links to that question. – ronno May 23 '13 at 16:04
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@ronno: then this is not a duplicate since it is asking for a simpler proof in a restricted case. – robjohn May 23 '13 at 16:09
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1@robjohn Well up to people who can cast close votes. In my opinion, the existing answers there already answer this one as "simply" as it wants. – ronno May 23 '13 at 16:13
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Let $A \in Z(GL_n(\mathbb R))$.
Let $B$ be defined as $b_{ii}=i$ and $b_{ij}=0$ if $i\neq j$ (or any other diagonal matrix with pairwise distinct entries).
Then since $AB=BA$, for all $i \neq j$ we have
$$\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^n b_{ik}a_{kj} \Rightarrow a_{ij}b_{jj}=b_{ii}a_{ij}\,.$$
Since $i \neq j$ we have $b_{ii} \neq b_{jj} \Rightarrow a_{ij}=0$.
This proves that $A$ is a diagonal matrix.
Now, use the fact that $A$ commutes with permutation matrices. Let $\sigma=(i,j)$ be any transposition and $P_\sigma$ the corresponding matrix.
Then $AP_\sigma=P_\sigma A$ implies $a_{ii}=a_{jj}$.
P.S. If you work over a finite field with less elements that the size of the matrix, instead of a matrix $B$ in the first part you need to consider a family $B_{i,j}$ of diagonal matrices so that $b_{ii} \neq b_{jj}$. AND, you need to study separately the case $k=F_2$.
N. S.
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We know that $G=GL_n(\mathbb{R})$ is generated by elementary matrices. Therefore $A$ is in $Z(G)$ if and only it commutes with all elementary matrices. By checking this condition, you will see that $A$ has to be diagonal.
M Turgeon
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@MartinBrandenburg Maybe we don't have the same definition of elementary matrices. In any case, it wasn't stated explicitly. – M Turgeon May 23 '13 at 16:29
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You don't need that $G$ be generated by elementary matrices (not so easy). You simply need that it contain the transvections (trivial). – Julien May 23 '13 at 16:39
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@julien I agree that you need less. But that $G$ is generated by elementary matrices is exactly that any invertible matrices can be brought to the identity matrix by elementary row (or column) operations, which is basic linear algebra. – M Turgeon May 23 '13 at 16:51
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I did not say it was hard, I am just saying that the fact the transvections are invertible is comnpletely trivial. Anyway, which elementary matrices will you use to deduce that $A$ is scalar? If not the transvections. – Julien May 23 '13 at 17:26
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