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A singular matrix is a zero divisor

  1. Oct 29, 2010 #1
    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. jcsd
  3. Oct 29, 2010 #2


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    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.
  4. Oct 29, 2010 #3
    Thanks a lot. Proof by construction are not always so obvious.
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