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Set of vectors whose coordinates are integer (is a subspace?)

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data

    For a set of vectors in R3,

    is the set of vectors all of whose coordinates are integers a subspace?


    3. The attempt at a solution

    I do not exactly understand if I should be looking for a violation or a universal proof.

    If x,y, z [itex]\in Z[/itex] then x,y,z can be writted as {(x,y,z)|x,y,z [itex]\in Z[/itex]}

    (1,0,0) has integers as coordiantes but is not in the set because it does not pass through the origin.
     
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  3. May 29, 2014 #2

    HallsofIvy

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    I have no idea what you mean by this. "(1, 0, 0)" certainly is in the set because, as you say, "it has integers as coordinates and that is all that is required. I don't know what you mean by "it does not pass through the origin"- (1, 0, 0) does not "pass" through anything. It is a single point. If you mean that (0, 0, 0) is not in the set, that is not true.

    One of the conditions for a vector space is that it must be "closed under scalar multiplication". You know that (1, 1, 1) is in this set. What can you say about (1/2)(1, 1, 1)?
     
  4. May 29, 2014 #3
    I'm unsure if I understood the question rightly. There would hardly be any substance in my workings if my understanding of the question is misguided.

    I assume the question is asking me to construct a vector whose coordinates are element of integers and subject them to the the condition of 'zero vector','closure by addition' and closure by multiplication'.

    Falling on my initial understanding, (1,0,0) is a vector whose coordinates are element of integers. And the zero vector (0,0,0) does not fall within (1,0,0). But you have verified it to be false.

    You mentioned (1,1,1), and I suppose I am free to equally set a vector (2,1,1)? Just to be sure I am on the right track in my understanding.

    Testing for closure by multiplication:

    k(1,1,1) where k is a constant: we have (1/2)(1,1,1) = (1/2,1/2,1/2)
    We see that (1/2,1/2,1/2) does not close under scalar multiplication because
    x1 = x2 = x3 [itex]\notin Z[/itex]
     
    Last edited: May 29, 2014
  5. May 29, 2014 #4

    LCKurtz

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    @negation: If you don't mind my asking, I am curious to know if English is your first language.
     
  6. May 29, 2014 #5
    Yea it is.

    It's probably shaky understanding of the concept and not employing the appropriate terminology.
     
    Last edited: May 29, 2014
  7. May 29, 2014 #6

    HallsofIvy

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    Why would you assume that? The problem was decide "is the set of vectors all of whose coordinates are integers a subspace?". It is not necessary to construct any vector to answer that. It is only necessary to determine whether the set given is a subspace and to do that you just need to see if the two conditions for a subspace, "closed under vector addition" and "closed under scalar multiplication", are satisfied.

    Again, I have no idea what you mean by this. What definition of "one vector falls within another" did your course give?

     
  8. May 29, 2014 #7
    I admit it was sloppy use of words. The correct word would be 'element'.

    It's fuzzy to me as to why the vector (1,1,1) was chosen. Could we have chosen (2,2,2)?
     
  9. May 29, 2014 #8

    LCKurtz

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    Yes, that's why I asked. I think that part of your problem is that you don't use, or maybe don't understand, the precise definitions you need to know, like the definition of a subspace, or whether a set of vectors is closed under addition or scalar multiplication. I will comment about this in your quote below:

    The coordinates aren't "elements of integers". They are integers. And you don't "subject them to the condition of zero vector". You decide whether the given set is closed under addition and scalar multiplication. You should have the definition of what it means for a set of vectors to be closed under addition and under scalar multiplication memorized word for word. In complete sentences. Do you? It would be appropriate for you to state the definition at this point in your work, because it is what you are trying to prove or disprove.

    Halls has already commented on this.

    "We see that (1/2,1/2,1/2) does not close under scalar multiplication" doesn't make any sense. An individual vector doesn't "close" under scalar multiplication. The problem is that you haven't stated carefully what it means for a set of vectors to be closed under scalar multiplication, so you don't know how to properly phrase that it is not. You need to learn the definitions exactly and use them.
     
  10. May 29, 2014 #9

    Ray Vickson

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    In this case there is not much to remember: just (1) closure under addition; and (2) closure under multiplication by a scalar. That's all there is! The rest comes not from memorization, but from working at it all until you understand it and have a 'feel' for it. A way to accomplish this is to do lots of examples, and to learn from mistakes.
     
  11. Aug 7, 2017 #10
    Let K is a set of elements named "vectors". The latter name, without any special meaning, is used solely to distinguish elements of this set from the elements of the field P, which we name "numbers" also without any particular meaning. However, the "field" has the meaning: it is another from K set with two algebraic operations "addition" (+) and "multiplication" (*). The names only distinguish the two operations. The former is commutative and associative. There is zero element 0 with respect to +: and for any element of the "field" a: a + 0 = a. There is a reverse element for each a named (-a): a + (-a) = 0. The field is the so-called Abel (commutative) group with respect to "addition". For the later, "multiplication", commutative and associative property holds too. But there is also a unit 1 so that 1 * a = a * 1 = a. There is a reverse (with respect to *): a * r = r * a = 1 for any NON-ZERO r. Finally, + and * are connected by the distribution of multiplication relatively to addition: (a + b) * c = a * c + b * c. Some say, the field is a ring with commutative multiplication, unit, and where each NON-ZERO element has a reverse. Now, K is not simply a "linear space" but always "a linear space above the field P", where the following holds (as axioms but without pretending to be unique minimal axiom set): A) K is an Abel group with respect to "addition" of its elements "vectors". This is not the already considered "addition of numbers in P" but a separate "addition of vectors of K", simply, has the same algebraic properties. B) "number" * "vector" is defined and belongs to K. Multiplication of "vectors" by "numbers" is associative with respect to "numbers": a*(b*x)=(a*b)*x, 1*x=x for any x (x is "vector" - element of K). C) Addition and multiplication are connected: a *(x + y)=a * x + a * y AND (a + b) * x = a * x + b * y. Now, the set K is a linear space above the field P, if A, B, C axioms hold. Back to the QUESTION. The "all" 3D vectors with real "coordinates" are the linear space above the field or real numbers. We can select any system of elements of our K. Let us chose only those, which have INTEGER coordinates (we omit a discussion how elements of K relate to "coordinates" and what is the "basis" of K). This subsystem (a system but we say "subsystem" emphasizing that it is not all K "vectors") of "vectors" S has infinite number of "vectors" (unless there is some limit on coordinate values, length or norm of vectors). However, this subsystem CANNOT BE A LINEAR SPACE ABOVE THE SAME FIELD P AS THE ORIGINAL SPACE K because multiplication by "numbers" from the field P is not defined in the sense that it yields the vectors from K but not our selection, our subsystem S. Take any rational or real but not integer and not zero number from P, multiply any vector from our system and the result is in K but not S. It is useful to see that the set of integer numbers with respect to ordinary addition and multiplication of numbers as we know them, is, even, not a field but ring, in contrast, with rational, real, and complex numbers (all three are fields). In other words, the phrase "a linear space K above the field of P of integer numbers ..." would be incorrect, if talking about P we talk about ordinary addition and multiplication of numbers. We can invent "some operations" so that a set of numbers 0, 1, ..., n - 1 is a field, if n is a prime number. It is so with respect to addition and multiplication by modulo of n. But it would be a separate story. Best Regards, Valerii
     
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