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Vector Subspaces

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img857.imageshack.us/img857/548/screenshot20120112at853.png [Broken]

    3. The attempt at a solutionI reasoned that if U is a vector subspace, then the zero vector must certainly be an element of U. That is, [itex](0,0,0) \in U[/itex]. If this is true, then we can write for [itex]x_1 + x_2 + x_3 = a[/itex], [itex]0 + 0 + 0 = a = 0[/itex]. If a is a fixed value as is evident in the problem, then a cannot equal anything but zero.

    Does that sound about right?
     
    Last edited by a moderator: May 5, 2017
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  3. Jan 12, 2012 #2

    micromass

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    Correct. You have now proven that if U is a vector subspace, then a must be 0.

    However, you must still prove the converse. That if a=0, then U is a subspace. This requires you to check a couple of properties.
     
  4. Jan 12, 2012 #3
    I only know of two properties of vector subspaces, which are as follows: http://img832.imageshack.us/img832/1343/screenshot20120112at219.png [Broken]
    So, let [itex]A = \{\ (a_1,b_1,c_1) | a_1 + b_1 + c_1 = 0\}\ [/itex] and [itex]B = \{\ (a_2,b_2,c_2) | a_2 + b_2 + c_2 = 0\}\ [/itex]. Let [itex]A,B \in U[/itex]. Then we can write,

    [itex]A + B = (a_1,b_1,c_1) + (a_2,b_2,c_2) \Rightarrow (a_1+a_2 + b_1 + b_2 +c_1 + c_2) = a_1 + b_1 + c_1 + a_2 + b_2 + c_2 = 0 + 0 =0 [/itex]

    So [itex]A +B \in U[/itex]

    Let r be any scalar. Then we can write,

    [itex]rA = r(a_1,b_1,c_1) = (ra_1,rb_1,rc_1) \Rightarrow ra_1 + rb_1 + rc_1 = r(a_1 + b_1 + c_1) = r(0) = 0.[/itex]

    So [itex]rA \in U[/itex].

    This shows that if a = 0, then U is a subspace, right?
     
    Last edited by a moderator: May 5, 2017
  5. Jan 12, 2012 #4

    Fredrik

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    In that image, A and B are vectors in the set U, which is a subset of the vector space V.

    You need to make sure that you understand the notation ##\{x\in X|P(x)\}##. It means "the set of all x in X such that the statement P(x) is true". So what you're saying in this quote defines A and B as sets, not vectors. To be more precise, it defines them both to be the same set, the set of all triples (x,y,z) such that x+y+z=0.

    You defined A and B as sets, so A+B doesn't make sense. If we forget the A and B that you have defined, and say that now A and B are vectors, not sets, what you're saying still doesn't make sense. What does the number ##a_1+a_2 + b_1 + b_2 +c_1 + c_2## have to do with anything?

    What does the number ##ra_1 + rb_1 + rc_1## have to do with anything?

    Also, you aren't using the implication arrow correctly. You are only supposed to write ##P\Rightarrow Q## when the statement "if P is true, then Q is true" is true. What you wrote on the right doesn't seem to have anything to do with what you wrote on the left.
     
  6. Jan 12, 2012 #5
    Okay. I think I should go one step at a time here.

    What notation would I use if I wanted A,B to be two arbitrary vectors in U such that the entries of A and B, not necessarily identical, sum to zero? That's what I wanted when I accidentally defined A,B to be sets instead of vectors.

    When I wrote [itex]a_1 + a_2 + b_1 + b_2 +c_1 +c_2[/itex] I was thinking that since the vectors A,B in U have the property that their entries sum to zero, then the vector A + B if it is in U should also have this property.

    Similarly, for [itex]ra_1 + rb_1 + rc_1[/itex]I wanted to show that any arbitrary vector A in U stays in U when it is multiplied by a scalar multiple.

    As for the arrows, I was looking for something that meant "this leads logically to this." For example, if we have [itex]rA = (ra_1,rb_1,rc_1)[/itex], then it seems to logically follow from the problem that [itex]ra_1 +rb_1 +rc_1 = 0[/itex].
     
  7. Jan 12, 2012 #6

    Fredrik

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    Just say something like this:

    Let U be the subset ##\{(x_1,x_2,x_3)\in\mathbb R^3|x_1+x_2+x_3=0\}##, and let ##A=(a_1,a_2,a_3)## and ##B=(b_1,b_2,b_3)## be arbitrary members of U.

    OK, this is the right approach, but you have to make it clear that this is what you're doing. For the second part, you need to also make the statement "let r be an arbitrary real number", in addition to what I said above. Then you just need to show that A+B and rA is in U.

    OK, I guess it does, but it looks really weird when you write down an equality that really just says that the definition of the scalar multiplication operation can be applied to A, and say that this equality implies the other one. I would rather say that the second equality follows from the assumption that A is in U, which tells us that ##ra_1+rb_2+rb_3=r(a_1+b_1+c_1)=r0=0##. Of course, that isn't 100% true either, because the equalities on the right also rely on the definition of the real numbers.

    I think that the fact that we always use something other than the statement P (a definition, an assumption, or a previously proved result) to prove that Q is true makes it hard to write ##P\Rightarrow Q## without confusing the people you're trying to convince. Feel free to use the implication arrow when you're just proving something to yourself, but always ask yourself if the statement will be understood when you're thinking about using it in a proof that will be read by others.
     
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