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Solving an example, vectors as rows or columns of a matrix = same solution.

  1. Apr 26, 2009 #1
    Ok from other posts, everyone seems to say that it doesn't matter if you stick vectors into a matrix via columns or rows. So i want to prove this to myself using an example from my notes. The only problem is, I get different answers. Can someone correct me?

    from my notes it says "show (0, 0,−5, 3), (4, 7,−1, 6), (0, 1, 2, 0) in R4 are linearly independent"
    So, down columns (and rearranging)
    -5,-1,2
    0,4,0
    0,7,1
    3,6,0

    which I can see is independent
    -5,-1,2
    0,4,0
    0,0,1
    0,0,0
    but that looks a lot like it belongs to R3, in my opinion..

    Anyways,
    same thing rows instead,
    4,7,-1,6
    0,1,2,0
    0,0,-5,3
    Which looks like it's already in echelon form, to me.
    And also looks like its dependent because x3 = (some constant) x4.

    So something is wrong with what I'm doing, any help?
    Thanks.
     
  2. jcsd
  3. Apr 26, 2009 #2

    matt grime

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    If you're still talking columns, then they are elements of R^4 that span a subspace isomorphic to R^3. They are not 'in R^3'.


    What are x3 and x4?
     
  4. Apr 26, 2009 #3
    x3 and x4, i meant collumns of the matrix (-1,2,5) and (6,0,3).
     
  5. Apr 26, 2009 #4

    HallsofIvy

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    So, in what sense is "x3 = (some constant) x4"?
     
  6. Apr 26, 2009 #5

    matt grime

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    Ok, so you think that there is a linear dependence amongst the 4 columns (there has to be as there are only 3 rows). That's fine. You were attempting to show that the 3 vectors were linearly independent. I.e. that row/column rank of the matrix is 3.

    Note that if you were to add the columns in this case then you've destroyed the fact that you'd written the vectors as the rows. Adding rows to rows preserves the fact that these were vectors in some vector space. Adding columns in this case would lose that.

    You can still calculate the rank by column operations, it is just that these operations do not correspond to do things that are meaningful in the original vector space.
     
  7. Apr 26, 2009 #6
    One more thing...

    Ok, So from what I've read, If the vectors are written across a row like

    4,7,-1,6
    0,1,2,0
    0,0,-5,3

    Applying any operations (column wise) to this will not correspond to things that are meaningful in the original vector space, correct? The only useful information we can obtain through column operations is the rank (rank 3, and 3 vectors = independent), correct?

    We would be able to obtain information that is meaningful in the orginal vector space if we applied row operations, correct?
    So it doesnt matter which way vectors are written in a matrix, but if they are stuck in a matrix across rows, then row operations would need to be applied instead of column operations, correct?

    -----
    Ok so lets change the orginal question a little.
    "vectors (0, 0,−5, 3), (4, 7,−1, 6), (0, 1, 2, 0) in R4 find the basis, and find the Null space"
    From what I remember, the null space should have a rank of one (4-3=1)?
    Ax=0
    We write A along rows as
    4,7,-1,6 x1 0
    0,1,2,0 .x2 = 0
    0,0,-5,3 x3 0
    x4

    And do column operations to get x as
    x = 0.75 x4
    -1.2 x4
    0.6 x4
    1 x4

    Correct?

    To find the basis we write the vectors down columns (and do column operations)

    -5,-1,2
    0,4,0
    0,7,1
    3,6,0

    which reduce to

    -5,-1,2
    0,4,0
    0,0,1
    0,0,0

    So a basis is for the span of the orginal vectors is (-5,0,0,0),(-1,4,0,0),(2,0,1,0), correct?


    Now, if you look at my working, I've only used column operations. I don't think I've ever seen row operations in action. So assuming that I would use row operations, I will need to write vectors down a column (ie basis, independence) or along row (null space), Correct?


    Sorry about all the writing!
    fingers crossed that it's all correct!
     
  8. Apr 26, 2009 #7

    matt grime

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    You can't just re-order the elements of vectors like that. You've taken the vector (0,0,-5,3) and re-ordered it as (-5,0,0,3), which is exactly doing what we've agreed not to do: write as columns then do row operations.

    So, no, that isn't a basis (there are infinitely many possible bases), as you've changed the order of the elements in the vectors.
     
  9. Apr 27, 2009 #8
    ok, so i'll try for the basis again.

    0,4,0
    0,7,1
    -5,-1,2
    3,6,0

    switch column 1 and 2 around.

    4,0,0
    7,1,0
    -1,2,-5
    6,0,3

    row reduce

    4,0,0
    0,1,0
    0,0,-5
    0,0,0

    so a basis is {4,0,0,0},{0,1,0,0},{0,0,-5,0) or more simply {1,0,0,0},{0,1,0,0},{0,0,1,0}?

    So was that the only mistake in my previous post? Was everything else correct?
    Thanks again.
     
  10. Apr 27, 2009 #9

    matt grime

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    You did it again: you switched rows and columns. As you've noted, this stops the answer relating to the original vector space.

    Any linear combination of your vectors is always zero in the final location, unlike 2 of the vectors you were given in the first place, for example.


    Write as rows and use row operations OR write as columns and use column operations.
     
  11. May 2, 2009 #10
    ahh, wait. i think I get it.
    solving for basis
    write
    0,0,-5,3
    4,7,-1,6
    0,1,2,0

    rearrange

    4,7,-1,6
    0,1,2,0
    0,0,-5,3

    Becomes
    4,0,0,-33/5
    0,1,0,6/5
    0,0,-5,3

    so a basis is (4,0,0,-33/5),(0,1,0,6/5),(0,0,-5,3)?

    And simplest (aka standard basis?) would look like
    (1,0,0,-33/20),(0,1,0,6/5),(0,0,1,3/-5)?
     
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