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Quantum mechanics- eigenvectots of a linear transformation

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data

    My quantum mechanics text (in an appendix on linear algebra) states, "f the eigenvectors span the space... we are free to use them as a basis..." and then states:

    T|f1> = λ1f1
    .
    .
    .
    T|fn> = λnfn


    My question is: is it not true that fewer than n vectors might constitute this "new" basis? In other words, if the eigenvectors span the space, they may not (all together, at least) form a basis.




    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 29, 2013 #2

    vela

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    No, this is a basic result from linear algebra for vector spaces of finite dimension. In an n-dimensional space, every basis has to have n vectors.
     
  4. Apr 29, 2013 #3
    Yes, I see. But when, then, all the talk about "An eigenbasis for a linear operator that operates on a vector space is a basis for that consists entirely of eigenvectors of (possibly with different eigenvalues). Such a basis may not exist."

    See http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors under the heading 'eigenbasis'
     
  5. Apr 30, 2013 #4

    vela

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    It's possible that an operator will not have n linearly independent eigenvectors, so its eigenvectors can not span the vector space and therefore will not form a basis.

    Your text, however, is saying if the eigenvectors span the space, they can be used as a basis. The assumption is that the linear operator has n linearly independent eigenvectors.
     
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