# Linear Operator and Self Adjoint

1. Oct 17, 2009

### charikaar

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. Oct 19, 2009

### GoodMax

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}.