$\mathbb{S} = \{MX - XM \mid X \in \mathbb{R}^{n\times n}\}$ gives $\dim \mathbb{S} \leq n^{2} - n$

Question

Fix $M\in\mathbb{R}^{n\times n}$ and define the vector space \begin{align*} \mathbb{S} = \left\{MX - XM \mid X \in \mathbb{R}^{n\times n}\right\}. \end{align*} Show that $\dim \mathbb{S} \leq n^{2} - n$.

Answer 1

Sketch of proof: Define the linear map $T:\Bbb C^{n \times n} \to \Bbb C^{n \times n}$ by $T(X) = MX - XM$. We see that $\ker(T) = \{X \mid MX = XM\}$. By this post or this post, we see that $\dim \ker (T) \geq n$. By the rank-nullity theorem, we have $$ \dim(\Bbb S) = \dim(\operatorname{im(T)}) = n^2 - \dim \ker(T) \leq n^2-n $$ as desired.