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Diagonalization-ordered basis

  1. May 16, 2009 #1
    I am trying to perpare for my final for linear algebra and am attempting to catch up in readings- in particular about diagonlization.

    Theorem 5.1: A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis B for V consisting of eigenvectors of T. Furthermore, if T is diagonalizable, B = {[tex]v_1, v_2, ..., v_n[/tex]} is an ordered basis of eigenvectors of T, and D = [tex][T]_B[/tex], then D is a diagonal matrix and [tex]D_j_j[/tex] is the eigenvalue corresponding to [tex]v_j[/tex] for 1<= j <= n.

    Question: For some reason in my textbook, they omitted the proof. I'm guessing it's fairly straightfoward, but I was wondering if someone could aid me with a proof.

    Idea for a proof: If T is diagonalizable, then for some [tex]v_j[/tex], [tex]T(v_j)[/tex] = [tex]\lambda_j[/tex][tex]v_j[/tex] (by definition- from last post).
    So it follows, for each eigenvalue [tex]\lambda_j[/tex] ( 1 <= j <= n, where n is the number of elements in our particular basis B ), there exists corresponding eigenvectors [tex]v_j[/tex] by definition. Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors, our basis B = { lambda_1, lambda_2, ..., lambda_n } is composed of eigenvectors.


    Questions:
    1. Can someone tell me if my proof above is correct? I feel as if I'm assuming to much when I said "Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors... "
    2. Also can someone justify the remaining portion of the theorem "... Furthermore, if T is diagonalizable... is the eigenvalue corresponding to [tex]v_j[/tex] for 1 <= j <= n."

    THanks again,


    JL
     
    Last edited: May 16, 2009
  2. jcsd
  3. May 17, 2009 #2

    jbunniii

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    "Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors..."

    This isn't "by definition" - in fact, it's one of the things that you are supposed to prove!

    How about this:

    First suppose that T is diagonalizable. That means that there exists a basis

    [tex]B = \{v_1,v_2,\ldots,v_n\}[/tex]

    with respect to which the matrix [tex]D = [T]_B[/tex] is diagonal.

    Call the entries of the matrix [tex]D_{ij}[/tex], for [tex]i,j=1,2,\ldots,n[/tex]. Then for each j,

    [tex]Tv_j = \sum_{i=1}^n D_{ij} v_i[/tex] (this is the definition of what the matrix elements mean)

    Because D is diagonal, [tex]D_{ij} = 0[/tex] for [tex]i \neq j[/tex], so this reduces to

    [tex]Tv_j = D_{jj} v_j[/tex]

    Since each [tex]v_j[/tex] is a member of the basis B, in particular, [tex]v_j \neq 0[/tex], so [tex]D_{jj}[/tex] is an eigenvalue of T and [tex]v_j[/tex] is a corresponding eigenvector.

    The converse is proved the same way. (If there exists an ordered basis B for V consisting of eigenvectors of T, then T is diagonalizable.) All you have to do is show that T has a diagonal matrix with respect to that basis B.
     
  4. May 17, 2009 #3
    Wow, that was very straightfoward. Thanks for the help.
     
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