Why is T* the definition of an adjoint operator in functional analysis?

[jy02927731]

if T: H -> H' such <T x , y > = <x , T* y >

i want to ask , why SHOULD T* be the defination http://en.wikipedia.org/wiki/Adjugate_matrix#Definition"


PLZ help , i did think about this Question more than ONE week , that's why i register

 

[Fredrik]

Are you asking why the adjoint T* is equal to the "adjugate" Adj T defined on that page? It isn't. The Wikipedia page says that Adj T=T^-1det T, so So T*=Adj T would imply T*T=(Adj T)T=T^-1T det T=I det T for all invertible matrices. This isn't true. It holds when T is a member of SU(n), but not in general. As a counterexample, consider

$$T=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}$$

We have T*=T=T^-1, but det T=-1. So T*T = I ≠ -I = I det T.

 

[jy02927731]

THX , fredrick
sorry for found a wrong defination, and i just found again , i sure that this is the defination of my book http://en.wikipedia.org/wiki/Conjugate_transpose" .


if <T x , y > = <x , T* y > , why SHOULD T* be the defination http://en.wikipedia.org/wiki/Conjugate_transpose"

 

[Fredrik]

The standard inner product on the space of n×1 matrices is defined by

$$\langle x,y\rangle=x^*y$$

where x* is the complex conjugate of the transpose of x. This is the physicist's convention. Mathematicians usually want their inner products to be linear in the first variable instead of the second. For all matrices M, define M* to be the conjugate transpose of M (the matrix obtained by taking the transpose of M and replacing each component by its complex conjugate). Note that M**=M. If T is an n×n matrix, and x is n×1, then Tx is n×1 and (Tx)* is 1×n. You should be able to use the definition of matrix multiplication to prove that (Tx)*=x*T*. Let us know if you're not.

$$\langle x,Ty\rangle=x^*(Ty)=(x^*T)y=(T^*x)^*y=\langle T^*x,y\rangle$$

This is the explanation of the equalities:

1. Definition of the inner product.
2. Associativity of matrix multiplication.
3. The easy to prove result (T*x)*=x*T**=x*T.
4. Definition of the inner product.

 

[Fredrik]

If T is defined as a linear operator, not a matrix, we have to define its matrix of components using a basis. It's convenient to use an orthonormal basis. Then the component on row i, column j of that matrix is

$$(Te_j)_i=\langle e_i,Te_j\rangle$$

(See this post if you don't know why). Let's call that matrix A. We need to show that if T* is the adjoint, defined using the inner product as in your posts, then its matrix of components, let's call it B, is the conjugate transpose of A.

$$A_{ij}=(Te_j)_i=\langle e_i,Te_j\rangle=\langle T^*e_i,e_j\rangle=\langle e_j,T^*e_i\rangle^*=((T^*e_i)_j)^*=(B_{ji})^*$$

Edit: You said T→H'. My reply is for the case H'=H. If H' is a different inner product space, we have to use two bases, one for each space:

$$A_{ij}=(Te_j)_i=\langle f_i,Te_j\rangle=\cdots$$

 

[jy02927731]

Fredrik,than you very much.
U explain it very clear.
Are u also student .

 

[Fredrik]

You're welcome. I'm not a student, but I am studying. Right now I'm studying functional analysis, trying to understand the mathematics of quantum mechanics.