Interference Alignment and Degrees of Freedom of the K-User(7)

CADAMBE AND JAFAR: INTERFERENCE ALIGNMENT AND DEGREES OF FREEDOM OF THE-USER INTERFERENCE CHANNEL3431use (13)–(16) to nd follows. Let be the,and. are picked as column vector. . . We now choose and asis a root of the linear equation. 1) are 2) All the coef cients forming the linear equation in equal to , so that the singularity condition is trivially satis ed for all values of . is a random variable drawn from a continuous distriSince bution, the probability of taking a value which is equal to the root of this linear equation is zero. Therefore, the second event happens with probability greater than , and we can writeIt can be easily veri ed that , and satisfy the three equaand satisfy the intertions (10)–(12). Therefore, ference alignment equations in (7), (8) and (9). Now, consider the received signal vectors at Receiver 1. The vectors while desired signal arrives along the and the the interference arrives along the vectors vectors . As enforced by (7) the interference vectors are perfectly aligned. Therefore, in order to prove that there are interference free dimensions it suf ces to show that the -dimensional matrix columns of the square, (17) are linearly independent almost surely. Multiplying by the full and substituting the values of , rank matrix equivalently, we need to show that almost surely (see (18) at the bottom of the page) has linearly independent column vecis a diagonal matrix. In other tors where with probability 1. The words, we need to show proof is obtained by contradiction. If possible, let be singular . Furwith nonzero probability. For instance, and the ther, let the diagonal entries of be be . Then the equation diagonal entries of shown at the bottom of the page is true with nonzero probaindicate the cofactor of the th row and th column bility.Let of . Expanding the determinant along the rst row, we getConsider the equationSince the terms do not depend on , the above equation is a polynomial of degree in . Again, as before, there are two possibilities. The rst possibility is that takes a value equal to one of the roots of the above equation. Since is drawn from a continuous distribution, the probability of this event happening is zero. The second possibility is that all the coef cients of the above polynomial are zero with nonzero probability and we can writeWe have now shown that if the determinant of the matrix is equal to with nonzero probability, then the determinant of following: matrix (obtained by stripis equal to with ping off the rst row and last column of nonzero probability as shown in the equation at the bottom of the following page with probability greater than . Repeating the above argument and eliminating the rst row and last column at each stage, we get . . . . . . . . . .. . . . .with probability greater than . But this is a Vandermonde matrix and its determinantNone of “cofactor” terms in the above expansion depend and . If all values other than are given, then the on above is a linear equation in . Now, implies one of the following two events:is equal to only if for some . Since are drawn independently from a continuous distribution, they are . all distinct almost surely. This implies that(18). . .. . .. . ..... . .. . .. . ..... . .Authorized licensed use limited to: Harbin Institute of Technology. Downloaded on June 01,2010 at 01:21:44 UTC from IEEE Xplore. Restrictions apply.