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Vector Space-Spanning Linear Algebra

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Bonus] Let E = {“ax+by+cz = d” | a; b; c; d ∈ R} be the set of linear equations
    with real coefficients in the variables x, y and z. Equip E with the usual operations
    on equations that you learned in high school: addition of equations, denoted here
    by “⊕” and multiplication by scalars, denoted here by “~”, as follows:
    “ax + by + cz = d” ⊕ “ex + fy + gz = h” = “(a + e)x + (b + f)y + (c + g)z = d + h”
    and
    ∀k ∈ R; k ~ “ax + by + cz = d” = “ka x + kb y + kc z = k d”:
    You may assume without proof that E is a vector space.
    Find a spanning set for E.
    (You must justify your answer.)

    2. Relevant equations



    3. The attempt at a solution

    I have no idea what this question is asking...isnit the span just span{x,y,z}? o.o
     
  2. jcsd
  3. Sep 30, 2012 #2
    It seems that by {x,y,z} you would mean a vector in [itex]R^{3}[/itex]
    However, E seems to be a vector space where the elements (vectors) are equations of the form: ax+by+cz=d (equations of planes). So you want to find a set of planes, [itex]P_{1}[/itex], [itex]P_{2}[/itex], [itex]P_{3}[/itex] such that their linear combinations can represent any element of E (any plane in [itex]R^{3}[/itex]).

    If you think the plane x+y+z=1 spans E, try to represent the plane x+2y+z=1 (which is in E) by a linear combination of x+y+z=1. Hopefully you see you'll need more than 1 plane to span E.

    Hope this helps!
     
  4. Sep 30, 2012 #3
    I dont think you can get x+2y+z=1 from x+y+z=1

    What I did is: z= 1-y-x so (x,y, 1-y-x)
    which is x(1,0,-1) + y(0,1,-1) + (0,0,1)
    which means E= span{(1,0,-1), (0,1,-1), (0,0,1)}
    And does that mean x+y+z=1 is not in E?
    Sorry I am really confused by this, not good with planes
     
  5. Sep 30, 2012 #4
    Linear Algebra can be really hard. Like this problem shows you need to think about 'vectors' abstractly. Every plane in [itex]R^{3}[/itex] is an element of E. And you need to find a set of planes that are a generating set for E. There is a very simple answer to this problem when you get past the abstractness.

    Remember every element of E takes the form of ax+by+cz=d for some constants a,b,c, and d.

    Now think about what the normal spanning vectors for [itex]R^{3}[/itex] are and try to compare this to what you might need for the planes. Remember you want to find some planes such that taking a linear combination of them, you can get ANY plane.

    What does it mean to take a linear combination of 2 planes?
     
  6. Sep 30, 2012 #5
    Damn...I know what linear combination is, I am just really unsure about how this question is asking the question...it is hard to explain...FOr normal spanning vectors of R^3, isn't it just a(1,0,0) + b(0,1,0) + c(0,0,1)? and I think that in order for a plane to be a vector space doesnt ax+by+cz=d, d has to =0? What do I need for the planes? I am confused about this is because my high school teacher decided not to teach about planes so I only learned lines n 2 and 3 space ..
     
  7. Sep 30, 2012 #6
    You're right about linear combinations in [itex]R^{3}[/itex] (when you say a(1,0,0)+b(0,1,0)+c(0,0,1) you are taking linear combinations of the 3 vectors (1,0,0) (0,1,0) (0,0,1)!). Now look at the problem again and try to find the linear combination of the planes (1x=1) ⊕ 2~(1y=1). What plane does this give you.

    The nice thing about linear algebra is you don't have to visualize any planes, or adding/scalar multiplying them. You just have to look at what the equations say. See what you can do.

    Notice that when you take the linear combinations a(1,0,0)+b(0,1,0)+c(0,0,1) of those 3 vectors in [itex]R^{3}[/itex] you get a vector (a,b,c) for a,b,c in [itex]R[/itex]. This is the form of ANY vector in [itex]R^{3}[/itex]. What set of planes (equations of the form ax+by+cz=d) do you need to get ANY equation in that form (any plane).

    Also remember that the planes themselves don't need to be vector spaces (although if they did you'd be right saying d=0!). This is a vector space of planes. The elements themselves don't necessarily need to be vector spaces.
     
  8. Sep 30, 2012 #7
    cz=d-ax-by (ax, by, d-ax-by)

    ax(1,0,-1)+by(0,1,-1)+d(0,0,1)

    does it have anything to do with that^??

    For: (1x=1) ⊕ 2~(1y=1) ...I don't quite get what you mean...x=1 +2y=2? which would be x+2y=3? Sorry I don't quite get that...


    For any would it be a(1,0,0,0)+b(0,1,0,0) + c(0,0,1,0) + d(0,0,0,-1)? I am just guessing sorry this thing is really confusing the hell out of me.
     
  9. Sep 30, 2012 #8
    When they define ⊕ and ~ in the problem they are defining how you add 2 vectors and scalar multiply (remember here vectors are planes). Thats why they say you add like you would add 2 equations in high school.

    And the standard equation for a plane is ax+by+cz=d (in vector analysis i think?). That's why I am leaving them in that form, instead of writing them in (x=, y=, z=) form like you are (although I don't think there's anything really wrong with that).

    I think you're on the right track. But you need to specify what a(1,0,0,0) means. This appears to be a vector in [itex]R^{4}[/itex] as opposed to an equation for a plane.

    If you mean for the entries in the parenthesis to correspond to the coefficients for the plane then the last plane you wrote would be 0x+0y+0z=d ⇔ d=0 which isn't a plane.

    You're very close though! This is quite a tricky problem.
     
    Last edited: Sep 30, 2012
  10. Sep 30, 2012 #9
    a(1,0,0,0) -->I was trying to put everything to one side so ax+by+cz-d=0 so....

    I am trying all sorts of things now but I am now stuck

    I tried this before (I never mentioned):

    you know fromt he question (a+e)x + (b+f)y....? I multiplied (expanded) them out and got them to be ax+by+cz-d + ex+fy+gz-h=0(or d+h if you didn't move it to the other side)

    I kept on thinking this has something to do with the solution, but I have no idea so far.

    I tried as far as setting the coeficients of the + = to the coefficients of the ~ so (a+e)=ka which obvioiusly didnt work but I just want to let you know I am close to giving up on this. No idea what to do with this one.
     
  11. Sep 30, 2012 #10
    This is how you add equations, just like you surely did in high school! You just add each side together. The ~ is scalar multiplying an equation. For instance 2~x+y+z=1 becomes 2x+2y+2z=2

    These 2 things just describe HOW you take linear combinations in this vector space.

    if you had the 3 planes x=0, y=0, z=0 then you can take linear combinations to get any plane that goes through the origin. Now you just need a way control the "d" term, on the right side.

    Thus I believe [itex]P_{1}[/itex] is x=0, [itex]P_{2}[/itex] is y=0, [itex]P_{3}[/itex] is z=0, and [itex]P_{4}[/itex] is x+y+z=1 is a set of planes that spans E.

    If you want to get any plane ax+by+cz=d, take (a-1)~[itex]P_{1}[/itex]+(b-1)~[itex]P_{2}[/itex]+(c-1)~[itex]P_{3}[/itex]+d~[itex]P_{4}[/itex] does it.

    So S={[itex]P_{1}[/itex],[itex]P_{2}[/itex],[itex]P_{3}[/itex],[itex]P_{4}[/itex]} spans E

    Hopefully this will all make sense with the answer in hand.
     
  12. Sep 30, 2012 #11
    Should it be:
    [tex](a-d)\sim P_1+(b-d)\sim P_2+(c-d)\sim P_3+d\sim P_4[/tex]
     
  13. Sep 30, 2012 #12
    what syznkasz says seems to make sense
     
  14. Sep 30, 2012 #13
    I think my statement is correct. Although, the other linear combination works too. Just replace some constants to get it in the form ax+by+cz=d.

    *explicitly, let a'=a-d+1, b'=b-d+1, c'=c-d+1, d'=d and it will take the form a'x+b'y+c'z=d' and its shown*

    Maybe replace the '+' with '⊕'

    But this isn't all that important (IMO). The question wants a set that spans E.
     
    Last edited: Sep 30, 2012
  15. Sep 30, 2012 #14
    so it just p4~d since the others are 0?
     
  16. Sep 30, 2012 #15
  17. Sep 30, 2012 #16
    nvm i kinda get it but why.the a-1 b-1 ?
     
  18. Sep 30, 2012 #17
    so the span is E=span(0,0,0,x+y+z=1)
     
  19. Sep 30, 2012 #18
    Because that gives you (a-1)x+(b-1)y+(c-1)z=0
    Then when you add P4 x+y+z=d

    you get exactly ax+by+cz=d

    Like I mentioned this isn't super important, its just the 'cleanest' way to do it (that I can see atleast).
     
  20. Sep 30, 2012 #19
    so the span is E=span(0,0,0,x+y+z=1)
     
  21. Sep 30, 2012 #20
    no 0 isn't a plane!

    E=span(x=0,y=0,z=0,x+y+z=1)
     
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