# A singular matrix is a zero divisor

1. Oct 29, 2010

### BrianM90

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?

Last edited: Oct 29, 2010
2. Oct 29, 2010

### Dick

Right, the first idea doesn't work. But you knew that. The second idea works fine. If you pick nonzero x in null(A), and you've built B so that B(y)=c*x for all y (where c is a scalar depending on y). So, sure, AB=0.

3. Oct 29, 2010

### BrianM90

Thanks a lot. Proof by construction are not always so obvious.