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Problem of the Week #292 - Sep 25, 2019

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Euge

MHB Global Moderator
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Jun 20, 2014
1,896
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Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,896
No one answered this problem. You can read my solution below.


The center is the set of all complex scalar $n\times n$ matrices. Since the identity commutes with every square matrix, so does every scalar matrix; therefore all scalar matrices are central. Conversely, if $X$ is belongs to the center of $M_n(\Bbb C)$, then it commutes with all elementary matrices $E_{ij}$ where $E_{ij}$ has a $1$ in the $(i,j)$-entry and zeros everywhere else. So if $X = [X_{ab}]$, then by comparing the $(a,c)$-entry of $X E_{ji}$ and $E_{ji}X$ we find \[\sum_b X_{ab}\delta_{bj}\delta_{ci} = \sum_b \delta_{aj} \delta_{bi} X_{bc}\] or $X_{aj}\delta_{ci} = \delta_{aj}X_{ic}$. Taking $a = j$ and $c = i$ in the equation yields $X_{jj} = X_{ii}$. Further, taking $a = c = i$ in the equation we deduce $X_{ij} = 0$ if $i \neq j$. Since $i$ and $j$ are arbitrary, $X$ is a scalar matrix.
 
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