Suppose that $A \in M_n$ is strictly diagonally dominant. Show that $|a_{kk}|$$\gt C_k'$, for at least one value of $k=1,\dots, n$, where $C_k'$ denotes $A$'s deleted absolute column sums ($a_{kk}$ is deleted).
My guess is that since $A$ is strictly diagonally dominant, $A$ must be nonsingular, and so $A^{T}$ is nonsingular as well. However, if we deny the assertion, then $0$ must be in the union of the Gersgorin discs. I think it's possible to lead to the contradiction that $0$ is actually an eigenvalue of $A^{T}$ in this case, but I'm stuck here.
Any solutions, hints, or suggestions would be appreciated.
Hint: note that $$ \sum_{i=1}^n |a_{ii}| > \sum_{i=1}^n \sum_{j \neq i} |a_{ij}| = \sum_{1 \leq i \neq j \leq n} |a_{ij}| = \sum_{j=1}^n C_{j}' $$