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Adjointness and Basis Representation

  1. May 17, 2014 #1


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    Hi, Let V be a fin. dim. vector space over Reals or Complexes and let L: V-->V be a linear operator.
    I am just curious about how to use a choice of basis for general V, to decide whether L is self-adjoint. The issue, specifically, is that the relation ## L= L^T ## ( abusing notation ; here L is a matrix representing L in some choice of basis ) which holds for self-adjoint operators in f.dim. space, will most likely not hold under a change of basis. But we may be able to find a special basis in a given V:

    I think for ## \mathbb R^n ## , if we use the standard basis e_i=δi,j , then L is
    self adjoint if , when it is represented as a matrix M in this basis, we have that ## M^T = M ## , i.e., M equals its transpose ( if V is complex, we need the resp. matrix to equal the transpose of the conjugate ) . (Phew !) Now, can we find some specific basis ## B_V ## in a general f.dim vector space V so that we can conclude L : V-->V is self adjoint if/when its representing matrix M satisfies ## M= M^T ## (or equals its conjugate transpose if the base field is C)? I thought we may use an vector space isomorphism between V and ## \mathbb R^n ## to pull back the basis {#e_i#} , and then this "pulled-back" basis would do the job?

    Thanks in Advance.
  2. jcsd
  3. May 18, 2014 #2
    Well, one way is to just find an orthonormal basis, if you want something that's equivalent to the standard basis (apply Graham-Schmidt process to any basis for an explicit construction). That's what you'd need to make your reasoning work because you need an isomorphism as an inner-product space, rather than just a vector space. Self-adjointness is a property that relates to the inner product, so that has to figure in somewhere. Similarly, for the complex case.

    You could also use an arbitrary basis and check <v, Aw> = <Av,w> for all the basis elements, using bilinearity to extend that to everything else.
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