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

Linear Operator and Self Adjoint

  1. Oct 17, 2009 #1
    I would be grateful for some help/tips/with this question.

    Let (V,<,>) be a complex inner product space with an orthonormal basis {v1,v2,.......vn}. Let L:V------>V be a linear operator. Explain what is meant by saying that L is self-adjoint. Let A=a_ij be the matrix representing L with respect to the basis {v1,......v_n}. prove the following.

    i) L is self-adjoint if and only if <Lvi,vj>=<vi,Lvj>.
    ii) a_ij=<Lvj,vi>.

    L is self adjoint means L = L*, but we know L* is the one and unique operator for which <Lv, u> = <v, L*u> for all u,v. How do i prove i) and ii).

    Thanks
     
  2. jcsd
  3. Oct 19, 2009 #2
    Let's begin with the statement i). You gave a definition of a self-adjoint operator. This definition implies (since L=L*) that L is self-adjoint iff <Lu,v>=<u,Lv> for all u,v in V. Thus, it's obvious that if L is self-adjoint then <Lvi,vj>=<vi,Lvj>. Now we'll prove the converse assertion. That is, we should show that <Lu,v>=<u,Lv>. Note first that any vector v is written as v=b_{i}v_{i}, where b_{i},i=1,...,n, are complex numbers. In addition we remember that <Lvi,uj>=<vi,Luj>. We have:
    <Lu,v>=<L(b_{i}v_{i},c_{j}u_{j})>=b_{i}c*_{j}<Lvi,uj>=b_{i}c*_{j}<vi,Luj>=<b_{i}v_{i},c_{j}Lu_{j}>=<v,Lu>

    To prove ii) you should recall the meaning of numbers a_ij. Being more precise, if v_i is a basis vector then Lv_{i}=a_{ij}v_{j}.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear Operator and Self Adjoint
  1. Adjoint operators (Replies: 10)

Loading...