Question

x belongs to R is a zero divisor iff det x=0

Answer 1

det x suggests that x is a matrix? But R is the set of real numbers right? So... I'm confused by your question. Can you please clarify?

Answer 2

here R is a ring

Answer 3

\(a\) is a left zero divisor if \(ax=0\) for some \(x\). Unless I am confusing something, det(x) refers to the determinant of a matrix x. I don't even understand now...

Answer 4

For some \(x\neq0\) that is.

Answer 5

i wanted to ask the same thing :P Perhaps you are considering some sort of vector space over R?

Answer 6

Maybe he is considering the ring of matrices! That would make sense right?

Answer 7

yup

Answer 8

This can be proved using rank-nullity theorem. I am too lazy so I will leave this to you lol.

Answer 9

XD Thxs

Answer 10

consider M2(R) for example

Answer 11

Lol, i will give just a very very general sketch idea of why. If you want a proper rigorous answer, ask again. :P
So basically, if x is a zero divisor then there exists a matrix A such that xA=0 or Ax=0. Then it follows that the columns/rows (corresponding to whether x is a left or right zero divisor) of A are in the nullspace of x. As the nullspace of x is not {0}, it is singular and therefore det(x)=0.
On the other hand if det(x)=0. Then its nullspace is not equal to {0}. Thus there must be a v in the nullspace. Let a matrix A be the matrix such that it has v in its first column and 0 in all the other entries. Then xA=0 and so x is a zero divisor.

Answer 12

thanks

Answer 13

It is pretty rigorous IMO since it follows directly from rank-nullity theorem.

Answer 14

lol, it didn't start off rigorous, but then the rigor kinda materialised haha