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Basis for subspace

  1. Aug 2, 2005 #1
    Hi everone,
    quick question: if i have 3 vectors and I need to know basis of the subspace they span, so I write each vector as rows of a matrix and when I reduce it, is it the basis of rowspace or column space of matrix that is the basis of the subspace that vectors span?

    Thanks in advance.
     
  2. jcsd
  3. Aug 2, 2005 #2

    matt grime

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    As you start by writing them as rows and then row reduce (hopefully), wanna havae a wild guess?
     
  4. Aug 2, 2005 #3
    basis for the rowspace?

    edit:
    on a test one time I wrote them as colums and found the basis for columnspace -- it was marked wrong, so now I'd want to be sure to know this.
     
  5. Aug 2, 2005 #4

    matt grime

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    as long as you declared you were writing them as columns and did colmun operations, csurely you can see that the answers would be the same? if it wasn't clear that was what you were doing then i can see why they might mark it wrongly. more likely is that you got a different (but equivalent) answer from the mark scheme, the marker looked at the working to see if it was valid and noticed your initial matrix was wrong and gave you no more marks. this is common as unfortunate as it may be since no one actually has patience to accurately mark work as tedious as this in great detail.
     
  6. Aug 2, 2005 #5

    shmoe

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    How did you find the basis for the columnspace? Maybe you did this incorrectly? If you used row operations to reduce it, you find the columns with leading 1's. The corresponding columns in the original matrix make the basis for the comlumn space. The bold part is a common mistake here.
     
  7. Aug 2, 2005 #6
    thanks for the replies :smile:

    What i did was arrange all vectors as colums of a matrix. I did use row operations to reduce it and then went back to the columns of the original matrix, THAT part I did remember. So, was it wrong to do row operations in this case to find a basis of the column space? (edit: that is what I am confused about)
     
    Last edited: Aug 2, 2005
  8. Aug 2, 2005 #7

    matt grime

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    yes. think of a simple example. (1,1,1) and (1,0,0) clearly the span is the 2 d plane spanned by (1,0,0,) and ((0,1,1) try putting them as columns then doing row ops.
     
  9. Aug 2, 2005 #8

    shmoe

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    As long as you go back to the original columns before doing the operations (and didn't make any algebra mistakes), it's good.

    There's the danger something here is getting lost in translation though, if you have your test handy would you like to post the question and your answer? If it's wrong, we can work through what you did in more detail. (Even if it's right we can work through it in more detail if you like too)
     
  10. Aug 2, 2005 #9
    OK, here it is:

    find a basis for the subspace spanned by the vectors
    {(1, -1, 2), (5, -4, 1), (7, -5, -4)}

    so I reduced the matrix which consisted of these vectors arranged as columns and after reduction, I had 2 pivots in first and second column, so my answer was {(1, -1, 2), (5, -4, 1)} and I'm pretty sure I did not make algebra mistakes.
    Oh and the note from grader said: this is the column space.
     
    Last edited: Aug 2, 2005
  11. Aug 2, 2005 #10

    shmoe

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    Assuming no algebra mistakes (ruling out a 'fluke'), what you've done is good.

    No matter how you found your answer, it's correct. The third vector is a linear combination of the first two, and the first two vectors are linearly independant.

    I hate to second guess someone, but TA's sometimes make mistakes. I'd go ask him or her about it, or your professor.
     
  12. Aug 2, 2005 #11
    Thank you, I will take your advice :)
     
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