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Rank(A) + nullity(A) = no. of cols of A (WHY?)

  1. Mar 28, 2007 #1
    Hello! Im confused over why rank(A) + nullity(A) = n = no. of columns of A, not no. of rows or something else.

    My lecturer showed me something like a mxn matrix postmultiplied with a x-vector to get R^n, thus n = no. of cols. Makes sense when he was explaining but when i stepped out i realised that i didnt get it. Any help pls? Thanks!
     
  2. jcsd
  3. Mar 28, 2007 #2

    JasonRox

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    You can probably find a proof somewhere online.

    It's pretty quite intuitive after you get further into Linear Algebra and have become more comfortable with Nullspaces and such. I'll find a link with a proof.
     
  4. Mar 28, 2007 #3

    JasonRox

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    I can't one what doesn't use linear transformations!
     
  5. Mar 29, 2007 #4
    My module hasn't reached transformations yet :(. Is there a explaination why its equal to no. of columns without talking about transformation?
     
  6. Mar 29, 2007 #5

    Hurkyl

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    In my humble opinion, the rank-nullity theorem is not something you really want to "explain" -- it should be part of the foundation of your intuition for linear algebra. If you're looking for understanding, your best bet is probably to review previous exercises where you actually solved systems of equations and apply the rank-nullity theorem to describe the system.

    e.g. you may have done an exercise solving Ax=b, where A is 3x3, and got a two-dimensional space of solutions. The rank-nullity theorem says that the rank of A is one -- so confirm that by computing the rank of A!
     
  7. Mar 29, 2007 #6

    JasonRox

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    Exactly. It comes around. If it hasn't yet, continue solving systems. :biggrin:
     
  8. Mar 29, 2007 #7

    radou

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    Once you learn something about linear operators and their matrix representation, it should become formally clear.

    Edit: actually, you can investigate this fact by going into Gaussian elimination.
     
    Last edited: Mar 29, 2007
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