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Dimension of a solution space

  1. Dec 12, 2007 #1
    Find the dimension of the solution space of Ax=0, where
    A=1 2 5
    -1 3 1

    is the rank(A)=2
    so the nullity(A)=2?
    is this correct?
     
  2. jcsd
  3. Dec 12, 2007 #2

    HallsofIvy

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    If A is a linear transformation from vector space U to vector space V, then range(A)+ nullity(A)= dimension of U. Here, A is from R3 to R2. Yes, the rank of A is 2. No, the nullity of A is not 2 also.

    You could just as well do this directly: if x+ 2y+ 5z= 0 and -x+ 2y+ z= 0, then, adding the equations, you must also have 4y+ 6z= 0 or z= (2/3)y. You can pick y to be anything you like and then calculate both x and z. What does that tell you about the nullity?
     
  4. Dec 12, 2007 #3
    so the nullity(A)=1?
     
  5. Dec 12, 2007 #4

    Office_Shredder

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    Read HallsofIvy's post again... if rank(A)=2, and rank(A)+nullity(A)=2, what does nullity(A) equal?
     
  6. Dec 12, 2007 #5
    so your saying n=2? i thought it was 3. n is always the number of rows?
     
  7. Dec 12, 2007 #6
    [​IMG]
    //the last blury part is "Furthermore, rank of the matrix is 2,"

    This is an example from my book. When I reduce A i get
    1 0 13/5
    0 1 6/5


    so according to the example in the book
    my rank would be 2
    and nullity would be 1
    with n=3
    Is this wrong?
     
  8. Dec 12, 2007 #7

    HallsofIvy

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    Yes, that is correct.
     
  9. Dec 12, 2007 #8
    just to make sure, what Office_Shredder said was wrong?
     
  10. Dec 12, 2007 #9

    HallsofIvy

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    Yes, I believe he confused the dimension of the domain and range spaces.

    At any rate, as I also pointed out, you could find the nullity directly:
    " if x+ 2y+ 5z= 0 and -x+ 2y+ z= 0, then, adding the equations, you must also have 4y+ 6z= 0 or z= (2/3)y. You can pick y to be anything you like and then calculate both x and z. "

    Since you can pick one number arbitrarily, the kernel has dimension 1.
     
  11. Dec 12, 2007 #10
    great thanks [solved]
     
  12. Dec 12, 2007 #11

    Office_Shredder

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    I apologize, I can't read :/
     
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