Because the mean is described by a rank-1 matrix.
Let A = rand(m,m), and o = ones(m) (the vector of 1’s). Then A - oo^T/2 has zero mean, and the expected value E[A] = oo^T/2 has 1 eigenvalue of m/2 and m-1 eigenvalues of 0. The actual eigenvalues of A are distributed around these two possibilities. Similarly for the complex case.
(Caveat: I have never properly studied random-matrix theory, so I’m just going off a back-of-the-envelope understanding. @alanedelman is the expert here.)