$[R(T)]^o$ $?!?!$

Question

I was studying linear algebra(Linear Transformations) a day back and came across this notation and couldn't understand what it meant. Is it the $interior$ of the $Range$ of linear transformation $T$? What does that mean?

Answer 1

If this is indeed linear algebra, then in my opinion $$ R(T) = \text{ image of } T = \{w: \ \exists v:\ w=Tv\} $$ is the image of $T$. And ${}^0$ denotes the annihilator: If $T$ is a linear mapping from $V$ to $W$, then $R(T)=R(V)$ is a subspace of $W$, and its annihilator is $$ [R(T)]^0=\{ f\in W^*: \ f(w) = 0 \ \forall w\in R(T)\}, $$ which is the set of all linear mappings $f\in W^*=L(W,K)$ from the dual space, that are zero on all elements of $R(T)$.

See e.g. German wiki page - Annihilator with the same notation $L^0$, the English wiki page - Annihilator with notation $L^\perp$.

hth