I need to linearize z(i,j)=abs(xixj) for which I write the following code:
where This code is wrong as I am solving in CVX sedumi and sedumi does not accomodate integer variables. How can I handle the linearization? Thanking you. 
Just for clarity then SeDuMi solves linear conic optimization problems. That means you must have something like min c'x st. A x = b x \in K where K is a convex cone. Only some special but important convex cones are allowed. Therefore, SeDuMi cannot deal with integer constrained variables at all. Moreover, it cannot deal with quadratic functions such as x^2 = x because that is a nonconvex set. If your problem is nonconvex e.g. contain integer variables it might be possible to formulate a convex relaxation that can be solved with SeDuMi. answered 22 Aug '13, 09:31 Erling_MOSEK 
I'd try z(i,j)^2 = (xixj)^2, z(i,j) >= 0 Depending on the rest of your model you may get a semi definite positive problem. answered 22 Aug '13, 02:52 jfpuget Thanks for the response. I am more intersted in linearization of abs(x(i)x(j)) where x(i),x(j) are binary. In fact I am intersted in max(x(i),x(j)),where i,j=1,2...n. I have converted max interms of absolute value function. Could you write your model in SDP form?
(22 Aug '13, 04:31)
R K Nayak
I proposed a quadratic form because that's what sedumi handles (from their online documentation) If you want to use a mixed integer model then you should use a solver that handles such models, for instance CPLEX.
(22 Aug '13, 04:36)
jfpuget
Thanks again. But I could not follow your model. Means how to solve in cvx_Sedumi unless the above QP is not converted to SDP. Are you saying to convert your model to SDP first and then solve by CVX_Sedumi?
(22 Aug '13, 04:49)
R K Nayak
sorry, I wanted to say that depending on the rest of the model the quadratic objective function can be semi definite positive. sedumi handles linear and quadratic constraints, which is what I have used. You can express that a variable x is binary with the constraint x = x^2
(22 Aug '13, 05:02)
jfpuget

I'm confused about how \(x_i\) can be binary if the solver cannot handle integer variables. At any rate, if they really are binary then \(\leftx_i  x_j\right = x_i + x_j  2 x_i x_j\). To linearize \(y_{ij} = x_i x_j\), use \(y_{ij} \le x_i\), \(y_{ij} \le x_j\), \(y_{ij} \ge x_i + x_j  1\), \(y_{ij} \ge 0\). answered 22 Aug '13, 17:23 Paul Rubin ♦♦ Thanks for pointing me about x_i as binary. In fact x_i is binary but I am relaxing it by taking as continuous in [0,1] and I am not declaring in my code x as binary variable. But in the above model of me I have to declare y(i,j) as binary so that the linarization of z(i,j)=x_ix_j is possible. If x_i is continous, whether your model works?
(22 Aug '13, 22:59)
R K Nayak
No, my approach does not work for continuous x.
(23 Aug '13, 10:39)
Paul Rubin ♦♦

I suggest you formulate your problem in Yalmip as it has an integer conic solver built in and it is hooked up to sedumi as a backend solver. answered 23 Aug '13, 09:57 Imre Polik Well, if it is mixedinteger SOCP then you can use MOSEK from CVX for instance.
(23 Aug '13, 10:05)
Erling_MOSEK

In the above model, x(i),x(j) are binary.