Is there a linear optimization problem where there is an alternate optimal solution(i.e $z_j-c_j=0$) but all the $y_{ij}$ in the simplex table is negative, i.e $y_{ij}<0$?
In the linear problem of Maximization, we say an optimal solution has been achieved if $z_j-c_j>0$, but if for some $j$ we have $z_j-c_j=0$ then we have an alternate optimal solution.
They shouldn’t be all negative, otherwise we wouldn’t have any basic variables left in the tableau as they are required to have a column of the form:
$$\begin{bmatrix} 0\\\vdots\\1\\\vdots\\0 \end{bmatrix}$$
Thus, in this type of negative-basis tableau, we would be forced to solve for basic variables and that would eventually make all of the $y_{ij}$ non-negative.
Not only that, but if we were to be in such a hypothetical situation, then the other solution we would find after pivoting would be infeasible because it would violates the non-negativity, $x_j\ge0$, constraint for the basic variables.
Thus, this situation would never arise unless some grave miscalculations took place