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Homework Help: Linear Algebra - Basis

  1. Oct 4, 2005 #1
    Hello... im doing this problem with basis. Infact, im having a lot of problems understanding basis, i did every question in the text book and I still get seem to understand the idea of it.

    So i was hoping somebody can help me with the whole idea about it.

    Say, for example, how would i go abouts a question like this:

    --Let F be a field and let V = F^3. Let
    W = {(a1 a2 a3) E F^3 / 2a1 - a2 - a3 = 0 }

    Find a basis for W --

    If a question like that came on a test, id fail it - sad to say. It would also be good if somebody knows a good website or has sameple tests that covers this mataril so that i may get used to it.

    Thanks
     
  2. jcsd
  3. Oct 4, 2005 #2
    so any element in W can be represented like so:

    w = (a1, a2, a3), where a1, a2, and a3 are arbitrary.

    but W has the additional restriction that a1 = 1/2 (a2 + a3).

    so

    w= ( 1/2 (a2+a3), a2, a3).

    w = a2 ( 1/2, 1, 0) + a3 (1/2, 0, 1). (it's easy to see that this is the same as above.)

    so

    w = span{(1/2, 1, 0), (1/2, 0, 1)}.


    and that set {(1/2, 1, 0), (1/2, 0, 1)} is our basis.


    ah, i miss these problems!
     
  4. Oct 4, 2005 #3

    Gokul43201

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    Keep in mind that the above is not the only basis.

    Also, notice that the given vector space is nothing but a (generalization of a) plane through the origin in [itex]\mathbb{R}^3[/itex]. Any pair of vectors in the plane will serve as a basis.
     
  5. Oct 4, 2005 #4
    Hey.. i was just wondering about Brad Barkers post above...

    he said that a1 = 1/2 (a2 + a3).

    well.. shouldn't it be a1 = 1/2 (a2 - a3) ?

    Does it make a difference?
     
  6. Oct 5, 2005 #5

    Gokul43201

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    No, you said "2a1 - a2 -a3 = 0"

    That gives 2a1 = a2 + a3
     
  7. Oct 5, 2005 #6
    oh that was my silly mistake.. but either way.. i still learned something :)
     
  8. Oct 5, 2005 #7
    Hey.... what about the subspace...

    U = (a+b+c=0/ a b c is in the Real Numbers)

    How would you show the span of that.

    Also, the comment Gokul43201 made, about the basis being the plane through the origin, how did he know that? I mean its a plane because of the two vectors, but how did he know that its through the origin?
     
  9. Oct 5, 2005 #8
    The general equation of a plane is [itex]Ax+By+Cz = D[/itex].
    [itex]D = 0 \Longleftrightarrow[/itex] the plane goes through the origin.
     
  10. Oct 5, 2005 #9
    "and that set {(1/2, 1, 0), (1/2, 0, 1)} is our basis." Can that be a Basis for F^3? what is F (Feild)? isnt that the same as R, like R^3
     
  11. Oct 6, 2005 #10

    Gokul43201

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    Also, for a plane to constitute a vector space with the usual vector addition, it must pass through the origin, since this point is the additive identity element.
     
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