Considering $T$ as a linear transformation from $\mathbb{F}^n$ to $\mathbb{F}^m$ of full rank $m\leq n$, we know that $\dim(\operatorname{ker}(T))=n-\dim(\operatorname{im}(T))=n-m$ by the rank–nullity theorem. This means that we can take an $m\times n$ matrix $T_{\text{im}}$ to represent $\operatorname{im}(T)$ and an $n\times (n-m)$ matrix $T_{\text{ker}}$ to represent $\operatorname{ker}(T)$. The question is that, does the row space of $T_{\text{im}}$ always have no intersection with the column space of $T_{\text{ker}}$?
I'm wondering if there is a concrete counterexample for this, particularly when $\mathbb{F}$ is a finite field. Because if the above vector spaces are all finite-dimensional and equipped with an inner product, we can always take $T_{\text{im}}$ and $T_{\text{ker}}$ consisting of orthonormal row vectors and orthonormal column vectors, respectively. In this case, one can easily verify that the row space of $T_{\text{im}}$ is the orthogonal complement of the column space of $T_{\text{ker}}$ in $\mathbb{F}^n$, so the intersection between them would always be empty.
Best Answer
Take the map $$T: \mathbb F^2 \to \mathbb F^2, (x,y) \mapsto (0,x).$$ Then you have equality $\operatorname{ker}(T) = \operatorname{im}(T)$, and so $\dim (\operatorname{ker}(T) + \operatorname{im}(T)) = \dim \operatorname{ker}(T) = 1$.
This works for any field $\mathbb F$.
Not sure what this has to do with the inner product, but in general this is false, as shown by my example.