Let $X$ be a separable Hilbert space, and denote by $\mathcal{L}\left(X,X\right)$ the usual space of bounded linear maps on $X$ (that is, taking $X$ into $X$). Let $n$ be some nonnegative integer and consider the operators $A\in\mathcal{L}\left(X,\mathbb{R}^n\right)$ and $B\in\mathcal{L}\left(\mathbb{R}^n,X\right)$.
Now, both operators are compact, and while in the case of $A$ it is clear that the operator has even finite rank (because $\dim A\left(X\right)<\infty$, of course), this is not necessarily true of $B$ (from my understanding, at least).
There are a lot of papers on systems theory stating the seemingly obvious fact that "...since $B\in\mathcal{L}\left(\mathbb{R}^n,X\right)$ it has finite rank.". I suppose there must be some hidden assumptions underlying a statement of this kind so that $B$ is of finite rank (apart from it being only an element of $\mathcal{L}\left(\mathbb{R}^n,X\right)$); what could the underlying assumptions be?
Finite rank means that the dimension of the image of $B$ is finite, and this image is generated by $B(e_i), e_i=1,..,n$ where $(e_i)$ is a basis of $\mathbb{R}^n$.