Characterize the units and zero divisors in a multiplicative semigroup of square matrices over a field

Let F be a field, let n \geq 1, and let S = \mathsf{Mat}_n(F) denote the multiplicative semigroup of n \times n matrices over F. Characterize the units and zero divisors in S. Does S have a kernel?


In this previous exercise, we showed that a matrix is invertible if and only if it is row equivalent to the identity matrix. In this previous exercise, we showed that a matrix is row equivalent to the identity if and only if it has nonzero determinant (remember, this is over a field). So a matrix is invertible if and only if it has nonzero determinant.

Now suppose a matrix A has determinant 0. Then as a linear transformation, A is not injective. So there exists a nonzero column matrix x such that Ax = 0. Then A[x|\cdots|x] = 0, and in particular A is a zero divisor.

That is, every matrix is either a unit or a zero divisor, and to distinguish the two cases, we need only look at the determinant.

Note that S has a zero, namely the zero matrix. Now every ideal in S contains zero, so that the kernel (i.e. the intersection of all two-sided ideals) is nonempty.

Suppose J \subseteq S is an ideal. Note that if J contains a unit A, then for all B \in S we have BA^{-1}A = B \in J, so that J = S. So every nontrivial ideal of S is contained in the set Z of zero divisors.

Let J \subseteq S be an ideal, and suppose there exists a nonzero matrix A \in J. Suppose the (h,k) entry of A is nonzero. By left- and right- multiplying by appropriate permutation matrices, we can construct matrices A_{i,j} whose (i,j) entry is nonzero for all 1 \leq i,j \leq n. By this previous exercise, left- and right-multiplying by appropriate matrices having only one nonzero entry, we see that J contains matrices B_{i,j} having a nonzero entry in (i,j) and 0 elsewhere for any choice of (i,j). Finally, multiplying by an appropriate scalar matrix \alpha I, in fact J contains every matrix of the form E_{h,k} having a 1 in entry (h,k) and 0 elsewhere for all 1 \leq h,k \leq n. In particular, J contains the ideal generated by such matrices.

So the kernel of S is precisely the ideal generated by the matrices E_{h,k}. In fact, since each E_{h,k} is row- and column- equivalent to E_{1,1}, the kernel of S is the principal ideal generated by the matrix E_1 with a 1 in entry (1,1) and 0 elsewhere.

Recall that every square matrix over F is row- and column-equivalent to a diagonal matrix, and that two matrices A and B are row- and column-equivalent precisely when we have B = PAQ for some invertible matrices P and Q. In particular, A \in J(E_1) if and only if a diagonal matrix to which A is row- and column equivalent is in J(E_1). Moreover, we can assume that these diagonal matrices have only 1 or 0 on the main diagonal.

We claim that a diagonal matrix which has more than one nonzero entry on the main diagonal is not in J(E_1). To see this, recall that J(E_1) = E_1 \cup SE_1 \cup E_1S \cup SE_1S. If D is a diagonal matrix with more than one nonzero entry on the main diagonal, then certainly D \neq E_1. As we saw in this previous exercise, every element of SE_1 and E_1S has at most one nonzero entry on the main diagonal. So it remains to be seen that D \notin SE_1S. Suppose A = [a_{i,j}] and B = [b_{i,j}] and consider AE_1B = [\sum_{k,\ell} a_{i,k}e_{k,\ell}b_{\ell,j}] = [a_{i,1}b_{1,j}]. If this matrix is diagonal with nonzero diagonal entry a_{i,1}b_{1,j}, then a_{t,1} = 0 for all t \neq 1, so that we have at most one nonzero diagonal entry.

So A is in the kernel of S if and only if it is row- and column- equivalent to a diagonal matrix with only one nonzero entry.

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