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Prove if set A is a subspace of R4

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove if set A is a subspace of R4, A = {[x, 0, y, -5x], x,y E ℝ}
    2. Relevant equations

    3. The attempt at a solution
    Now I know for it to be in subspace it needs to satisfy 3 conditions which are:
    1) zero vector is in A
    2) for each vector u in A and each vector v in A, u+v is also in A
    3) for each vector u in A and each scalar a in ℝ, a*vector u is also in A

    Now the problem I have is proving it...

    Do I need to prove the 3 conditions above by substituting values for x and y?


    1) letting x and y = 0, we get a zero vector, therefore vector 0 is in A (check)
    2) let vector u have x = 1, y = 2 and vector v have x = 3, y =4
    u + v = [1, 0, 2, -5] + [3, 0, 4, -15] = [4, 0, 6, -20] (which is also in A; check)
    3) let a = 2, and vector u having x = 1, y = 3, so 2[1, 0, 3, -5] = [2, 0, 6, -10] (which is also in A; check)

    The reason I am confused is because I have seen methods where the values are multiplied by the entire vector within the set, like...

    1) 0*vector u = vector 0
    2) u = b[itex]_{1}[/itex]u[itex]_{1}+...+[/itex]b[itex]_{k}[/itex]*u[itex]_{k}[/itex] and
    and then you would have v = c[itex]_{1}[/itex]v[itex]_{1}+...+[/itex]c[itex]_{k}[/itex]*v[itex]_{k}[/itex], then u + v = (b[itex]_{1}[/itex]+c[itex]_{1}[/itex])u[itex]_{1}[/itex]+....+(b[itex]_{k}[/itex]+c[itex]_{k}[/itex])u[itex]_{k}[/itex]
    3) and etc.

    But the second method only applies to spans? Which can only be used if you know set A is a subspace of R4?

    Anyways, any clarification would be appreciated, thanks
    I Like Pi
    Last edited: Mar 14, 2012
  2. jcsd
  3. Mar 14, 2012 #2
    I'm confused by this part: x,y E|R}

    Do you mean to say [itex]x,y \in \mathbb{R}[/itex]? I assume so because we are talking about [itex]\mathbb{R}^{4}[/itex], but I just wanted to make sure.

    The first condition is an existence statement; you must be able to find x and y in the reals such that the 0 vector exists. You did that by finding values for x and y.

    The next two conditions are universal statements. They mean for ALL vectors u,v in A, _______ ; for all vectors u in A and scalars c in the reals, ____________.

    So what you do for 2 is take two arbitrary vectors u and v, and add them together to show they are still in the set. When you take arbitrary vectors, you don't assign specific values to the components, you assign variables (different variables for each vector, the vectors can possibly be different from each other!).

    For 3 you take an arbitrary vector u and multiply it by an arbitrary constant c and argue that the new vector is still in your set.

    Can you try it now?
  4. Mar 14, 2012 #3


    Staff: Mentor

    I don't understand what you're asking.
    You can't prove parts 2 and 3 like this. These conditions have to hold for any arbitrary vectors u and v in A.

    Two such arbitrary vectors are u = <x1, 0, y1, -5x1> and v = <x2, 0, y2, -5x2>.

    Now show that u + v is also in A.

    For the 3rd condition the proof is similar.
  5. Mar 14, 2012 #4
    I edited the above post to reflect your concerns and thanks guys! I understand what you mean, so, how about this for the first one:

    let A1 = [x1, 0, y1, -5x1] and A2 = [x2, 0, y2, -5x2]
    for b, c, E R
    bA1+cA2 = b[x1, 0, y1, -5x1]+c[x2, 0, y2, -5x2] = [bx1+cx2,0,by1+cy2,-5bx1-5cx2] = [bx1+cx2,0,by1+cy2,-5(bx1+cx2)], where x = bx1+cx2 and y = by1+cy2

    For the third condition, I'm still stuck on...

    Last edited: Mar 14, 2012
  6. Mar 14, 2012 #5
    You need to state why exactly this is (a little bit more precise than what you've stated). Hint: What were the conditions for x and y in the original vector?
  7. Mar 14, 2012 #6
    Hmm. Not quite.For part b) follow Mark44's advice:

    (you don't multiply those arbitrary vectors by a constant)
  8. Mar 14, 2012 #7
    ahhh, rightt, I don't know what I had in mind... it'd be <x1+x2, 0, y1+y2, -5(x1+x2)> where x = x1+x2 and y = y1+y2!!

    Now for part 3, would it be...
    For b, E R

    b<x1,0,y1,-5x1> = <(bx1),0,(by1), -5(bx1)>, where x = bx1 and y = by1??

    Thanks guys!!!
  9. Mar 14, 2012 #8
    Almost finished, but not quite. You need to explain why <x1+x2, 0, y1+y2, -5(x1+x2)> is a vector in [itex]\mathbb{R}^{4}[/itex]. What specific conditions must be met for [itex]x_1 + x_2 \ \text{and} \ y_1 + y_2[/itex]?

    Similarly you need to explain part c as well.
  10. Mar 14, 2012 #9
    I don't get what you mean by specific conditions? x1+x2 make up x, for example? Since x can be an value of R, doesn't that apply to x1 and x2? I'm sorry, I haven't gone into much detail with subspaces but the very basics...

    Thanks scurty
  11. Mar 14, 2012 #10
    You're right on the edge. I think the reason why it's oblivious to you is because it seems almost common sense! Here is what your set A is:

    [itex]A = \{[x, 0, y, -5x],\ x,y \in \mathbb{R}\}[/itex]

    So, what conditions must x and y satisfy to be in the set A? Likewise, [itex]x_1 + x_2 \ \text{and} \ y_1 + y_2[/itex] must also satisfy these conditions (all you have to do is show why they satisfy these conditions and you are done). And the same for part 3.
  12. Mar 14, 2012 #11
    Wait, x1+x2, x1,x2 E R? haha
  13. Mar 14, 2012 #12
    Yep! Now you just need to write it up nicely and you're good!
  14. Mar 14, 2012 #13
    Haha, thanks so much! It's funny how when you do something complicated, you become oblivious to the obvious!
  15. Mar 14, 2012 #14
    No problem! Yeah, I miss the obvious a lot too. :/
  16. Mar 14, 2012 #15


    Staff: Mentor

    No, he needs to explain why this vector is in A. The vector is obviously in R4.
  17. Mar 14, 2012 #16


    Staff: Mentor

    Hold on here a minute. The only restriction on x and y is that each needs to be real.

    If you mean that x and y are vectors in set A, that causes confusion with how the set is defined. That's why my arbitrary vectors were u and v.

    In any case a vector v in R4 is set A if and only if 1) its 2nd coordinate is 0, and 2) its 4th coordinate is -5 times its first coordinate. The 1st and 3rd coordinates are entirely arbitrary.
  18. Mar 14, 2012 #17
    Maybe I was just describing what I meant wrong. He found adding u and v together to be <x1+x2, 0, y1+y2, -5(x1+x2)>. I was just suggesting that he show [itex]x_1 + x_2, \ y_1 + y_2 \in \mathbb{R}[/itex] which would then satisfy the new vector being in the set A. Was I suggesting for him to show superfluous information?

    In your post above I meant showing the vector was in A, not [itex]\mathbb{R}^4[/itex], sorry about that.
  19. Mar 15, 2012 #18


    Staff: Mentor

    Yes, IMO. x1, x2, y1, and y2 are reals, so of course if you add any two of them you get a real. That's not germane to the discussion, I don't believe.

    What is important is the realization that the coordinates of u + v follow the same pattern as every other vector in A; namely, that the 2nd coordinate is 0, and the 4th coordinate is -5 times the first coordinate.
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