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The center of $M_{n} (k)$ consists of the scalar matrices

Statement

Let $k$ be a field and let $n \geq 1$ . Then

$Z (M_{n} (k)) = {λ I_{n} : λ \in k} .$

Facts & Assumptions

Given: A field $k$ and an integer $n \geq 1$ .

[L1]

Matrix multiplication is given by $(A B)_{i k} = \sum_{j < n} a_{i j} b_{j k},$ and $I_{n}$ is the identity matrix (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

Proof

technique · direct

1.1L1givenalgebra

Let $X = (x_{a b}) \in Z (M_{n} (k))$ . For each $i$ , let $E_{i i}$ be the diagonal matrix with a single $1$ in position $(i, i)$ . Because $E_{i i} X = X E_{i i}$ and [L1] computes matrix products entrywise, the $(a, b)$ -entry comparison shows $x_{a b} = 0$ whenever $a \neq b$ . So $X$ is diagonal.

2.1step 1.1L1givenalgebra∎

Write $X = diag (d_{1}, \dots, d_{n})$ . For $i \neq j$ , let $E_{i j}$ be the matrix unit with a single $1$ in position $(i, j)$ . Then [L1] gives $E_{i j} X = d_{j} E_{i j}, X E_{i j} = d_{i} E_{i j} .$ Since $X$ is central, these are equal, so $d_{i} = d_{j}$ for all $i, j$ . Thus $X = λ I_{n}$ for one scalar $λ \in k$ . Conversely every scalar matrix commutes with every matrix because scalar multiplication is entrywise and $I_{n}$ is the identity from [L1]. Hence the center consists exactly of the scalar matrices.

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Webb, A Course in Finite Group Representation Theory, Lemma 3.4.1(1) (standard reference, not scraped)

The center of Mn(k)Mn​(k) consists of the scalar matrices