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A proof

  1. Jun 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Let T: V---->V be a linear operator where dim V=n. Show that V
    has a basis of eigenvectors if and only if V has a basis B such that
    TB is diagonal.


    2. Relevant equations



    3. The attempt at a solution
    Let T=[a1,1......an,1] ai,j=/=0
    [a1,n......an,n]
    Let TB=[a1,1v1......0n,1]
    [01,n.......an,nvn]
    Since this is diagonal, and ai,j=/=0,
    then we have a basis of eigenvectors (these are meant to be vertical)<[v1,...0]......[0.....vn]>
    that, after being mulitplied by T, formed the matrix TB.
    To show that they are eigenvectors, a possible linear combination is
    [v1,...0] and when multiplied by T gives a1,1[v1,...0]+...+a1,n[v1,....0]
    =(a1,1+...+a1,n)[v1,....0]. Since [v1,....0] is an eigenvector, and
    the others follow the same logic, mulitplying the basis of eigenvectors by T
    should produce a diagonal matrix, as shown.
     
    Last edited: Jun 12, 2009
  2. jcsd
  3. Jun 12, 2009 #2

    jbunniii

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    Suppose V has a basis [tex]B=\{b_1,\ldots,b_n\}[/tex] consisting of eigenvectors of T. Then for each j, [tex]T b_j = \lambda_j b_j[/tex]. If I express this equation in terms of the basis B, then
    [tex][T]_B [b_j]_B = \lambda_j [b_j]_B[/tex]
    But [tex][b_j]_B[/tex] is simply a column vector containing all zeros except for a 1 in the j'th position. Carrying out the matrix multiplication then shows that
    [tex][T]_B = \Lambda[/tex]
    where [tex]\Lambda[/tex] is the diagonal matrix whose diagonal entries are [tex]\{\lambda_1,\ldots,\lambda_n\}[/tex]. This gives you one direction of the proof. For the other direction, you can essentially reverse the steps above to verify that the [tex]\lambda_j[/tex]'s must be eigenvalues and the [tex]b_j[/tex]'s must be eigenvectors.
     
  4. Jun 12, 2009 #3
    Yeah, I did the other direction by assuming that TB is diagonal then showing that
    the basis consists of eigenvectors and that TB consists of eigenvalues, though I
    think I messed up when I mentioned a1,1+...+a1,n is an eigenvalue since I multiplied wrong.
    It should be a1,1 for the eigenvalue of [v1.....0]. Made a mistake with multiplication.
    Oh and when I said "then we have a basis of eigenvectors" I shouldn't've called
    the basis of eigenvectors a basis of eigenvectors, since I was proving it in the first place,
    and I should've called the scalar an eigenvalue. It's these small mistakes that make a big
    difference.
     
    Last edited: Jun 12, 2009
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