Do the primal and dual have the same number of basic variables For the primal problem with constraints $Ax\leq b$ and the dual problem with constraints $A^Ty\geq c$. Since $rank(A) = rank(A^T)$, does this mean that the primal and dual will always have the same number of basic variables?
 A: The confusion here stems from the fact that you must first convert the primal and dual forms to one with equality constraints (because this is required when talking about basic feasible solutions.) 
Suppose we are dealing with a maximization primal.
The inequality constraint $Ax \leq b$ can be projected into a higher space by adding slack variables $u$ to get an equality constraint, $$Ax + u = b.$$ Note that $x \geq 0, u \geq 0$. Thus the actual matrix we're looking at is $[A I]$, of dimension $m \times (m + n)$. The rank of this 'primal' matrix is the rank of $A$, so $m$. So, the primal has $m$ basic variables.
The dual of this has the feasible region $A^T y \geq c$ where $y \geq 0$.  However, we need to convert this to an equality constraint too! Add slack variables $s$ to get $$ A^T y - s = c $$ where $y \geq 0, s \geq 0$. 
Note that the matrix we're dealing with in the dual is now $[A^T -I]$, of dimension $n \times (m + n)$. This is not the transpose (or the negative of the transpose, even if $-I$ were $I$.) of the 'primal' matrix, because the dimensions don't match! You can see that the rank of this matrix is $n$, because of the rows of the identity matrix. So, the dual has $n$ basic variables. 
Thus, if $m \neq n$, the dual and primal have a different number of basic variables.
