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I Distinctiveness of the set of nxn matrices as a ring

  1. Apr 5, 2017 #1
    So I know that in general, for the ring of ##n \times n## matrices, if ##AB = 0##, then it is not necessarily true that ##A=0## or ##B=0##. However, in other rings, for example the integers ##\mathbb{Z}##, I know that this statement is true. So what property is the ring of matrices lacking such that it is not true in general?
     
  2. jcsd
  3. Apr 5, 2017 #2

    fresh_42

    Staff: Mentor

    A ring is called an integral domain, if ##ab = 0 \Longrightarrow a=0 \vee b=0##. Elements ##a## for which there is an element ##b## with ##a \cdot b = 0## are called zero-divisors. So an integral domain is a ring without zero-divisors or more precisely: with ##0## as only zero-divisor. E.g. ##\mathbb{Z}_6## has also zero-divisors, namely ##2## and ##3##. (I'm not quite sure, whether ##0## is excluded in the definition of a zero-divisor or not. It's not really important, but "has no zero-divisors" is a usual phrase, so ##0## is probably excluded.)
     
  4. Apr 5, 2017 #3
    One more question. Why doesn't a subring necessarily have to have the same multiplicative identity as the bigger ring?
     
  5. Apr 5, 2017 #4

    fresh_42

    Staff: Mentor

    Do you have an example? I can only think of examples like ##n\cdot \mathbb{Z} \subseteq \mathbb{Z}## where the subring has none.
    In general, a multiplicative identity doesn't always exist. The multiplicative structure of a ring doesn't need to define a group structure, but if rings are compared like ring and subring and both have a ##1##, then it's usually required to be the same. At least the ##1## in the ring is also a ##1## in the subring, if it's included. If not, and the subring has another element as a ##1## it's getting a bit messy, since this one will act as a ##1## on certain elements of the ring as well - or you have completely different multiplicative structures, in which case one doesn't speak of a subring.
     
  6. Apr 8, 2017 #5
    How about {0,2,4} as a subring of Z6. The multiplicative identity of the subring is 4.
     
  7. Apr 9, 2017 #6

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    You can see this issue of AB=0 for matrices in terms of the fundamental theorem of linear algebra , re the orthogonality of the column space of A with the row space of B:

    https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra

    Basically , if A has rows ## a_1,a_2 ,..,a_n ## and B has columns ## b_1, b_2,.., b_n ## then you have that
    ## a_ i. b_j =0 ## for all ## 0 \leq i,j \leq n ## so that every row of A is orthogonal to every column of B . This implies, by linearity, that the row space of A is orthogonal to the column space of B and the column space of B is contained in the nullspace of A..
     
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