MHB What Is the Center of the Ring of All \( n \times n \) Complex Matrices?

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The discussion focuses on computing the center of the ring of all \( n \times n \) complex matrices, denoted as \( M_n(\mathbb{C}) \). The center consists of matrices that commute with all other matrices in this ring. It is established that the center is formed by scalar matrices, which are multiples of the identity matrix. Despite the problem being posed, no participants provided answers, and the original poster shared their solution. The topic emphasizes the unique properties of matrix commutativity within complex matrices.
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Here is this week's POTW:

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Compute the center of the ring $M_n(\Bbb C)$ of all $n\times n$ complex matrices.

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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.