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PROOF: Independent vectors and spanning vectors

  1. Mar 13, 2008 #1
    Proof:
    1. why you need at least m vectors to span a space of dimension m.
    2. If m vectors span an m-dimensional space, then they form a basis of the space.
     
  2. jcsd
  3. Mar 13, 2008 #2
    do you know how to do proofs?
     
  4. Mar 14, 2008 #3

    HallsofIvy

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    What does it mean to say that a vector space has finite dimension? (NOT just that "its dimension is finite. You have to have 'finite' dimension before you can define dimension.) What is the definition of "dimension n".
     
  5. Mar 14, 2008 #4

    JasonRox

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    Try reading some definitions in your textbook.

    And reading some proofs of some easy theorems. You'll see how to apply definitions to prove statements like you have.
     
  6. Mar 16, 2008 #5

    mathwonk

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    these are proved in my free linear algebra notes on my web page. the full details are not given, but i believe the proofs can be filled in without huge difficulty, if you try and understand the definitions.
     
  7. Mar 18, 2008 #6
    lets start from the definition that a basis consists only from
    independant vectors
    now in order to span some space in "n" dimention you need to have "n" independant vectors
    which are a basis for this dimention.
     
  8. Mar 19, 2008 #7
    Find the independance

    Find if these vectors are lineary independant :..

    U=(1 2 9) . v=(2 3 5) .


    I know the condition of the lineary independant is

    au+bv=0 but how I can use this condition here .,.,.,

    Please If you know the answer u can send it at >> ahmedtomyus@yahoo.com
     
    Last edited: Mar 19, 2008
  9. Mar 19, 2008 #8
    noooooooooooooo
    dont use that formula
    stack them one upon the other as matrix and make a row reduction
    if you dond have a line of zeros in the resolt then they are independant
     
  10. Mar 20, 2008 #9

    HallsofIvy

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    Normally I would not respond to a question that has nothing to do with the original question but since transgalactic responded to it...

    Why not? au+ bv= a(1, 2, 9)+ b(2, 3, 5)= (a+2b, 2a+ 3b, 9a+ 5b)= (0, 0, 0) seems easy enough. We must a+ 2b= 0, so a= -2b. Then 2a+ 3b= -6a+ 3b= -3b= 0. b must equal 0, so a= -2b= 0 also. Since a and b must both be 0 the two vectors are, by definition, independent.

    Yes, if we have 10 vectors or 10000 then "row reduction" would be simpler but I believe it is better practice to use the basic definitions.
     
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