1. The problem statement, all variables and given/known data Show a matrix A belonging to the Mn(F) is a zero divisor if and only if A is singular. ( By the way Mn(F) is the n x n matrices with entries from a field.) 2. Relevant equations det(A) = 0 if a is singular definition of a zero divisor 3. The attempt at a solution The forward direction I have. (===>) Assume by contradiction a is a unit, But a is a zero divisor, so there exist a b belong to R b not equal to 0 such that ab = 0 but we assume a is a unit so a^-1(ab) = (a^-1)0 = 0 which implies b = 0 which is a contradiction. The reverse I am having trouble with. I have an attempt though (<===) I am assuming A is a singular matrix. Then det(A) = 0 by definition. Am i allowed to now say let assume we have a non-singular matrix B and let us look at the determinant of the product of A and B det(AB) = det(A)*det(B) = 0 (Since det(A) is 0 and B is not singular so det(B) does not equal zero) Does this work? (To me it seems it doesn't) Another idea I had was that if A is a singular matrix, then the null space of A does not simply consist of just the zero vector. ( So maybe I have to do a proof by construction here?) Assume A an n x n singular matrix with entries from a field. Then null(A) has a non trivial solution. Assume some arbitrary vector (x1,x2,x3,x4,...,xn) belongs to the Null(A). Take any n non-zero multiplies of (x1,x2,x3,x4,...,xn) and form a non-zero n x n matrix B. But A*B = 0, where A and B are not equal to 0, thus A is a zero divisor? (It is interesting though because B will also happen to singular, they are both zero divisors then i suppose, but certainly at least A is a zero divisor) I mean i know if you used a specific example this will obviously work. If this proof okay? Does it need to be cleaned up? Or is it no good at all?