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Can a non-zero matrix multiply itself to become zero matrix?

  1. Nov 18, 2004 #1
    In real number, only zero multiplies itself to become zero.
    Can a non-zero matrix multiply itself to become zero matrix?
    equivalent question: Can the row space of a matrix be orthogonal to the column space?

    I'm sorry if the question looks stupid.
     
  2. jcsd
  3. Nov 18, 2004 #2
    Yeah, an example is:

    Code (Text):

    (1 -1/2)
    (2 -1)
     
    .
     
  4. Nov 18, 2004 #3

    jcsd

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    Yep this is an example of a nilpotent (xn = 0 for some n) element of a ring (iu this case the ring of 2 x 2 matrices) and hence a zero divisor.
     
    Last edited: Nov 18, 2004
  5. Nov 18, 2004 #4
    Thanks
    I also found that if we restrict ourselves to symmetric matrices (Hermitian matrices in complex case).
    No such matrix can be found.
     
  6. Nov 20, 2004 #5
    No, based upon the definition of multiplication, the only way to have a product of zero is if one of the factors are zero. ie. if both factors are non-zero, the product must be non-zero.
     
  7. Nov 21, 2004 #6
    You must've missed the part where kakarukeys said this was about matrices.
     
  8. Nov 21, 2004 #7

    jcsd

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    I think that's true for fields, but it's not true for rings in general, infact a non-zero memenber of a ring that has the property of being (right) multiplied by some other non-zero member to give zero is called a zero divisor.
     
    Last edited: Nov 21, 2004
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