Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Adjointness and Basis Representation

  1. May 17, 2014 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook