A singular matrix is a zero divisor

[BrianM90]

Homework Statement

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.)

Homework Equations

det(A) = 0 if a is singular
definition of a zero divisor

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?

 

[Dick]

Homework Statement

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.)

Homework Equations

det(A) = 0 if a is singular
definition of a zero divisor

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?

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.

 

[BrianM90]

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