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Linear Algebra proofs

  1. Dec 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Ok so I am stick on three proofs for my linear algebra final adn help on any of all of them would really help with my studying

    For the first 2 assume that A is an nxn matrix

    1.If the collumns of A span Rn then the homogenous system Ax = 0 has only the trivial solution
    2. If the collumns of A are linearly independent, then the columns of A span Rn

    These 2 have to be proved without referencing other parts of the invertible matrix theorem

    And then,

    3. Be able to prove: If A is an nxn matrix then lambda is an eigencalue of A if and only if det(A-lamda*In) = 0


    2. Relevant equations



    3. The attempt at a solution

    the first two i have a better idea at than the third, since its an nxn i know that if it spans Rn then it has a pivot in every row and coincidently in every row and therefore every column. This same fact can be used to explain number 2 with them being linearly independent. My problem with these two is that Im having trouble since I cant reference them being part of the Invertible matrix theorem.

    For the third one I think I am on the right track but not sure
    The determinat of A-IL must equal zero because the determinant of A is simply the product of the eigenvalues. If you replace each eigenvalue into the determinat one at a time and multiply it by I, one of the entries will be replaced by zero and any other vlues multiplied by zero will result in zero. As a result, the determinat must equal zero.

    Any help would be great
     
  2. jcsd
  3. Dec 12, 2009 #2
    A somewhat different take on 1 and 2 than yours (I suspect yours could be formalized as well, but I find the following approach more natural):
    For 1: In other words you are told that the column space of A equals R^n. Then what is the rank and nullity of A? Knowing the nullity what can you tell about the space of solutions?

    For 2: In other words the n-columns are linearly independent in the column space whose dimension is at most n. Can n-linearly independent vectors possibly fail to span a n-dimensional space?

    For 3: For [itex]\lambda[/itex] to be an eigenvalue means that there exists a non-zero vector v such that,
    [tex]Av = \lambda v = \lambda I_n v[/tex]
    which is equivalent to:
    [tex](A-\lambda I_n)v = 0[/tex]
    If [itex]A-\lambda I_n[/itex] has an inverse B, try to left-multiply the above equality by B.

    For the other direction assume [itex]\det(A-\lambda I_n)=0[/itex] which is equivalent to [itex]A-\lambda I_n[/itex] not having an inverse, or having nullity >0. Since the nullity is >0 the null space contains a non-zero vector v which you can show is a eigenvector with eigenvalue [itex]\lambda[/itex].
     
  4. Dec 12, 2009 #3
    Ok I think i can take it from here, thanks a lot you really saved me with this
     
  5. Dec 13, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In (1) you have n vectors that span an n dimensional space.

    In (2) you have n vectors in an n dimensional space.

    Do you recall that a basis for a space has three properties- but that any two are enough to prove the third?
     
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