Does anybody has idea how to solve this problem ?
"Show that in Phase I of the simplex method, if an articial variable becomes nonbasic, it need never again become basic. Thus, when an articial variable becomes nonbasic, its column can be eliminated from the tableau"
As mentioned above, this is from the Bertsimas and Tsitsiklis, and the Phase I approach they are referring to is in Section 3.5. The standard form LP they use is \begin{array}{ll} \text{minimize} & c^T x \\ & A x = b \\ & x \geq 0 \end{array} They assume that $b\geq 0$; if this is not the case, negate the corresponding rows to make it so. (And for simplicity, let's assume $b$ has at least one nonzero value.) The corresponding Phase I problem looks like this: \begin{array}{ll} \text{minimize} & \textstyle\sum_i y_i \\ & A x + y = b \\ & x \geq 0, y \geq 0 \end{array} Now you see why $b\geq 0$ is important: $(x,y)=(0,b)$ constitutes a trivial feasible solution, so that's your starting point for the Phase I method, with an initial objective of $\sum_i b_i>0$. If the optimal value of this Phase I model is zero, then original model is feasible; otherwise, the original model is infeasible.
It is important to read the statement carefully. It is not claiming that an artificial variable will never re-enter the basis if you leave it in the tableau. In fact, it can. If you have the book, look at Example 3.8. In one of the steps, one of the nonbasic artificial variables has a negative reduced cost, making it an entirely valid candidate for selection, depending on the pivoting rule you use. (The example chose a different pivot, though.)
The statement, therefore, is not that it will never re-enter the basis. Rather, the statement is saying that it does not need to re-enter the basis. What we wish to show is that eliminating one of these variables from the tableau does not prevent the simplex method from terminating properly.
To see why this is the case, consider this "restarting" approach to solving the Phase I model:
The key to the success of this approach is noting that, after removing a nonbasic artificial variable in Step 5, you're still left with a valid pseudo-Phase I model and a feasible point for that model. The model still has an optimal value of $0$ if the model is feasible. It's also important to note that objective value reached in Step 4 is going to be the initial objective value at Step 2. So we do not lose progress with our restarts.
In fact, one way to look at it is this: we're not really restarting the simplex algorithm per se. Rather, we are just choosing to recompute the inverse basis matrix $B^{-1}$ from scratch every time an artificial variable exits the basis. Still, by expressing the approach in this way, it is hopefully easier to see why the elimination of nonbasic artificial variables does not prevent the algorithm from terminating.
Once you are satisfied with this, you don't need to actually restart at all. You will get exactly the same sequence of simplex pivots if you simply include in your pivot rule the clause: never choose a nonbasic artificial variable.
I am not sure whether I answer your question or not:
In Phase I the objective function is $$\min s_1+s_2+\ldots+s_m$$ where $s_i$ for $1\le i \le m$ are the artificial slack variables of the original problem. But one constraint (of the Phase I LP) is that $$s_i\ge 0, \qquad \text{ for } i=1,2, \ldots, m$$ and therefore if one of these variables becomes $0$ it is "as good as it gets" (for that variable and the objective of the LP of Phase I). This gives the reason that a variable which leaves the basis does not enter the basis again:
A good book which covers (more than) the basics of linear optimization and provides intuitive explanation for most subjects, is that of Bertsimas and Tsitsiklis (1997), the biggest part of which (including the description of Phase I, p. 111) can be found online here.
