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ADDED IN RESPONSE:

Okay then, what you need to essentially prove is 1) that all zero-divisors are singular matrices, and 2) all singular matrices are zero-divisors; this proves a one-to-one relationship.

1) is not too hard; if AB = 0 for some generally non-singular B, then A is a zero-divisor, and simply taking determinants

det(AB) = det(A)det(B) = det(0) = 0, then if det(B) ≠ 0 in general, then we know that det(A) = 0, i.e. all left-hand zero-divisors are singular. (I believe if det(B) = 0, and det(A) ≠ 0 generally, then we have proven this for right-hand zero-divisors as well...)

2)... what we basically need for this is to prove that all singular matrices are zero-divisors, in other words, that if a matrix A is singular, we can find some other non-zero matrix B for which AB = 0; so we just have to prove we can find such a B.

- Let A be a singular matrix.

- In that case, we know we can reduce it using row operations to some alternative form A' = RA (for some unitary non-singular transformation matrix R - it will be non-singular as it will have at least one non-zero entry in every row, being a transformation matrix).

- RA has the same determinant as A (det(R) = 1 for a unitary transformation) but RA has at least one row of zeroes in it - this expresses that all singular matrices A' are rank-deficient.

- In that case, Define B as any matrix that only has entries in the column index which is the row index of the zero row(s) in RA; and zeroes elsewhere (in the column indices that RA has non-zero values - e.g. if the kth row of RA is zeroes, let the kth column of B be non-zero and let the rest of the columns be zero.)

- In that case, the matrix product RAB will be zero. R is non-zero and non-singular; hence A is a left-hand zero-divisor on B.

(Essentially what we proved is that R*A must be a zero-divisor, but R*A can be decomposed into R and A and since R is not one, A must be a zero-divisor.)

The right-hand zero divisor case can be proven the same way, if A is arbitrary and B is singular but with column operations producing B' = BC and defining BC such that ABC = 0.

... and now it's 6am here so I'm signing off till tomorrow. See you soon :)

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Scratch all that - it is possible: An example of a zero-divisor in the 2 by 2 matrices is

[1, 1]

[2, 2], since e.g pre-multiplying this by

[-2, 1]

[-2, 1]

or post-multiplying it by

[-1, -1]

[1, 1]

gives the zero matrix.

However you'll note that both of these matrices B are singular. It is impossible to find a B that is non-singular that will fill this role.

"More generally in the ring of n-by-n matrices over some field, the left and right zero divisors coincide; they are precisely the nonzero singular matrices. In the ring of n-by-n matrices over some integral domain, the zero divisors are precisely the nonzero matrices with determinant zero."

So if the left and right divisors co-incide, you see that all zero-divisors must be singular matrices, since if A is a left-hand zero-divisor then AB = 0 for some B, but we can equally consider B a right-hand zero-divisor on A for the same reason. Hence, since B is a right-hand zero-divisor, it must also be singular.

Define Singular Matrix