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If BC = 0 where B is invertible, show C = 0

  1. Dec 16, 2009 #1
    Isn’t it true that if B is invertible then B DNE 0, and if B DNE 0 then the only way to have BC = 0 is to have C = 0?

    I'm not seeing (perhaps because I don't want to ;)) how the above is not always true. Can anyone show me an example proving me wrong?

  2. jcsd
  3. Dec 16, 2009 #2


    Staff: Mentor

    If B is invertible, then B-1 exists.

    If you are given that BC = 0, multiply on the left of both sides by B-1. What do you get?
  4. Dec 16, 2009 #3
    Yes, that I understand (and wish I would've put down on the final!), but what I don't understand is why my initial statement is untrue. That is, why is that not a valid proof that C = 0?
  5. Dec 16, 2009 #4
    haha, I've seen thru the cloud of trying to convince myself I was correct. My error is now obvious!
  6. Dec 16, 2009 #5


    Staff: Mentor

    Because you didn't use the given information that B is invertible.
    For example, consider these matrices B and C.
    [tex]B~=~\left[\begin{array}{c c}

    [tex]C~=~\left[\begin{array}{c c}

    For these matrices BC = 0, yet neither B nor C is the zero matrix.
  7. Dec 17, 2009 #6
    Rings of matrices are not in general integral domains (even if the elements are drawn from a field), as Mark44 illustrated with an example. Thus, if you know that BC = 0 and B != 0, you can't conclude that C must be 0 as you would in an integral domain.

    This is fairly common, so you should try to keep it in mind. It pops up in a lot of places; matrices are just one example. For instance, the ring of integers modulo 6 is not an integral domain since 2 and 3 are both nonzero but their product is 0 modulo 6. One can also find positive powers of nonzero elements that are zero. Once again using modular arithmetic for an example, the square of two is congruent to 0 modulo 4.
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