Centre of the general linear group

Let $𝑉 \in 𝖵 𝖾 𝖼 𝗍 𝕂$ . The centre $Z (𝑉)$ of the general linear group $G L (𝑉)$ consists of all homotheties, i.e. multiples of identity, and is isomorphic to the multiplicative group $𝕂 \times$ of the underlying field $𝕂$ . #m/thm/group

Proof

It is clear that scalar transformations form a group isomorphic to $𝕂 \times$ and that $𝕂 \times \subseteq Z (𝑉)$ . Now let $𝐴 \in Z (𝑉)$ and assume there exists a nonzero $𝑥 \in 𝑉$ such that $𝑥$ not an eigenvector of $𝐴$ . Then $𝑦 = 𝐴 𝑥$ is linearly independent from $𝑥$ , and there exists some vector basis $𝐵$ with $𝑥, 𝑦 \in 𝐵$ and $s p a n 𝐵 = 𝑉$ . Let $𝑈 = s p a n {𝑥, 𝑦}$ and $𝑈' = s p a n (𝐵 ∖ 𝑉)$ . Define linear maps such that in the $𝑥, 𝑦$ basis

𝑆 ↾ 𝑈 = [1001] 𝑇 ↾ 𝑈 = [1011] s p a n - 1 ↾ 𝑈 = [10 - 1 1]

and $𝑆 ↾ 𝑈' = 𝑇 ↾ 𝑈' = i d 𝑈'$ . Then

𝐴 𝑦 = 𝐴 𝑆 𝑥 = 𝑆 𝐴 𝑥 = 𝑆 𝑦 = 𝑥 𝐴 𝑇 𝑥 = 𝐴 𝑥 + 𝐴 𝑦 = 𝑥 + 𝑦 𝑇 𝐴 𝑥 = 𝑇 𝑦 = 𝑦

implying $𝑥 = 𝑥 + 𝑦$ , a contradiction. Hence every $𝑥$ must be an eigenvector of $𝐴$ , so $Z (𝑉) \subseteq 𝕂 \times$ .

Properties

Scalar transformation criterion

#state/tidy | #lang/en | #SemBr