Dimension of image of Lie bracket

Question

Is there a method to calculate the dimension of the set of vectors in $\mathfrak{su}(n)$ $\{\ [A,B] \ \text{s.t} \ B \in \mathfrak{su}(n)\}$ for some fixed $A$. Is the dimension the same for all $A$?

Answer 1

In general, the linear operator ${\rm ad}(x)$, defined by ${\rm ad}(x)(y)=[x,y]$ has a non-trivial kernel $ker (ad(x))$, since we always have $[x,x]=0$, and an image ${\rm im}(ad(x))$. As you know, we have $$ \dim \ker(ad(x))+\dim im(ad(x))=\dim (L) $$ for all $x\in L$. The dimensions $\dim im(ad(x))$ are not always the same, e.g., take $x=0$, and take an $x\neq 0$ with non-trivial adjoint operator.