The true formula is given for the probability that an n x n matrix over a finite field has determinant zero. This probability does not (as previously asserted) approach 1 as n grows.
A recent note in this journal [4] derived a formula for the probability that an n x n matrix over a finite field with q elements will have determinant equal to zero. In particular, for fixed q this formula yields the conclusion that the probability approaches 1 as n ~ oo. Since this result has been reported without question in Mathematical Reviews [MR 86a: 15012] and with only a hint of doubt in the Zentralblatt [Zbl. 553.15003], it seems advisable to put on record explicitly the fact that the formula is wrong and the conclusion is false.
The correct formula is essentially known. When the entries of a matrix are independent uniformly distributed random variables, as in , then all matrices are equally likely, and we simply have to determine what portion of them are not invertible. Now the first column of an invertible matrix can be any one of the qn_ 1 nonzero vectors (n-tuples). Once it is chosen, column 2 can then be any one of the qn -q vectors not a multiple of it. Proceeding inductively we see that column k+ 1 can be chosen to be any of the q"--qk vectors not in the (k-dimensional) span of the first k columns. Thus we have IGL(n, q)l = (qn _ 1)(q~ _ q)... (q~ _ q,-1), a result known at least since Dickson's book p. 77] (Gerth [2] refers to an 1893 CreUe paper by G~ Landsberg). Dividing by the total number of matrices, qn2, we find our formula:
The probability that an n x n matrix over the field with q elements has