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

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SUMMARY

The center of the ring \( M_n(\mathbb{C}) \) of all \( n \times n \) complex matrices consists of scalar matrices, which are matrices of the form \( cI_n \), where \( c \) is a complex number and \( I_n \) is the identity matrix of size \( n \). This conclusion is derived from the property that a matrix commutes with all other matrices in the ring if and only if it is a scalar multiple of the identity matrix. The discussion highlights the importance of understanding matrix commutativity in the context of linear algebra.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly matrix operations.
  • Familiarity with the properties of complex numbers.
  • Knowledge of commutative properties in algebraic structures.
  • Basic understanding of matrix theory, specifically regarding \( n \times n \) matrices.
NEXT STEPS
  • Study the properties of scalar matrices in linear algebra.
  • Learn about the structure of rings in abstract algebra.
  • Explore the concept of commutativity in matrix multiplication.
  • Investigate the implications of the center of a ring in algebraic systems.
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Mathematicians, students of linear algebra, and anyone interested in the properties of matrix rings and their applications in advanced mathematics.

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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.
 

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