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Closed under addition and closed under scalar

  1. Mar 4, 2014 #1
    Could someone explain to me about what closed under addition and closed under scalar multiplication means? I have a patchy idea of what it is but how does it relates to A = {(x,y) | x^2 + y^2 <= 1}?
    What does A stands for? What does the language implies?

    Edit: My interpretation: Let's suppose there exists a field k with R^n where n = 2) and A is a subset of the field k.
    An element is closed under addition iff an element, uA, and, vA such that
    u^2+v^2 = <=1.
    If u^2+v^2 <=1, then, u and v is a subset of A.
     
    Last edited: Mar 4, 2014
  2. jcsd
  3. Mar 4, 2014 #2

    LCKurtz

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    In other words, you are considering ##R^2## as a vector space over the reals. Addition means ##(u,v)+(x,y) = (u+x,v+y)## and scalar multiplication means ##t(x,y) = (tx,ty)##, the usual vector operations in ##R^2##.

    You have a very confused interpretation. You don't talk about whether an element is closed under addition, you talk about whether the subset ##A## is closed under addition. That is, if ##(a,b)## and ##(c,d)## are in ##A##, is ##(a,b)+(c,d)## always in ##A##? If so, ##A## is closed under addition. Similarly, if ##(x,y)\in A## is ##t(x,y)\in A## or not.
     
  4. Mar 4, 2014 #3
    How does x^2 + y^2 relates the the closed addition and scalar multiplication?
     
  5. Mar 4, 2014 #4

    LCKurtz

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    That's how you tell if ##(x,y)\in A##. So if you add two elements in ##A## is the result in ##A##?
     
  6. Mar 4, 2014 #5

    So x^2 + y^2 <=1 is a condition. If vector u and vector v both of which are elements of S where S is a set of vectors and if u+v <= 1 then it is closed under addition.
    And if it fulfils 2 other necessary conditions- namely, closed under scalar multiplication and where 0 vector is an element of S then S is a subspace.
     
  7. Mar 4, 2014 #6

    Mark44

    Staff: Mentor

    Yes, this is the condition by which you can determine whether a vector <x, y> is a member of set A (the name of the set from post #1). S is not mentioned in post #1.
    No, that isn't it. If u and v are members of A, and if u + v is also in A, then A is closed under vector addition.
    The condition, as you call it, is how you can tell whether a given vector is in set A.
    Just to be clear, assuming that u is in A, and t is a scalar, if tu is also in A, then A is closed under scalar multiplication.
     
  8. Mar 4, 2014 #7
    Is it the same thing to say A closed under addition, closed under scalar multiplication and has zero vector and A is the subspace of a vector space?
     
  9. Mar 4, 2014 #8

    Mark44

    Staff: Mentor

    Not quite. You need some ifs and a then.

    For a set A, a subset of a vector space, if
    1. A is closed under vector addition, and
    2. A is closed under scalar multiplication, and
    3. The zero vector is an element of A,

    Then A is a subspace of the vector space. (Whatever vector space we're working with.)
     
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