Adjoint of Linear Operator

ADJOINT OF LINEAR OPERATOR

Definition

Adjoint of linear operator:

Let V be a finite dimensional inner product space and let T be a linear operator on V. Then there exist a unique function T* : V→ V such that < T(x), y > = < x, T*(y) > for all x, y ∈ V. The linear operator T* is called adjoint of operator T.

Theorem 1

Let V be an inner product space and let T and U be a linear operator on V then

(a) (T+U)* = T* + U*

(b) (cT)* = T* for any c ∈ F

(c) (TU)* = U*T*

(d) T**=T

(e) I*=I

Proof:

(a) (T+U)* = T* +U*

< x, (T + U)* (y) > = < (T + U) (x), y >

 = < T(x)+U (x), y >

= < T(x), y> + <U (x), y>

= <x, T*(y) > + < x, U*(y) >

= < x, T*(y) + U*(y) >

= < x, (T* + U*) y >

 (T + U)* (y) = (T* + U*) (y)

 (T + U)* = T* + U*

(b) (cT)* =T*

 < x, (cT)* y > = < cT (x), y >

 = < x, T*(v) >

 =  < x, T*(v) >

 (CT)* = T*

(c) (TU)*=U*T*

 < x, (TU)* (y) > =  < (TU) (x), y >

 = <T(x) U (x), y>

= <T(x), y> < U(x), y >

= <x, T*(y) ) ( x, U*(y) >.

= < x, U*(y) T*(y) >

 (TU)* = U*T*

(d) T**=T

< x,T(y) > = < T*(x), y >

= < x, T**(y) >

T (y) = T**(y)

  T = T**

Theorem 2

Let V be a finite dimensional inner product space over F, and let g: V →F be a linear transformation. Then there exists a unique vector y ∈ V such that g(x)=<x,y> for all x ∈ V.

Proof:

Let β = {v₁, v₂, v₃... v_n} be an orthonormal basis for V₁ and let

Let us define h: V→F by h(x) = <x, y> which is linear.

Further 1 ≤ j ≤ n we have

Since g and h both on B, we have g=h.

Theorem 3

Let V be a finite dimensional inner product space, and let T be a linear operator on V. Then there exists a unique function. T* : V → V such that

 < T (x), y > = < x, T*(y) > for all x, y ∈ V and T* is linear.

Proof:

Let y ∈ V. Define g: V→F by g(x)= <T(x), y> for all x ∈ V. To show g is linear, let x₁, x₂ ∈ V and c ∈ F.

 g(cx₁+x₂) = < T(cx₁ + x₂), y > = < cT (x₁) + T (x₂), y >

  = c<T(x₁), y> + <T(x₂), y>  = cg(x₁) + g(x₂)

Hence g is linear.

To obtain a unique vector y' ∈ V such that

 g(x) = <x, y'>.

(i.e) < T(x), y> = <x, y'> for all x ∈ V.

Define T* : V→ V by T*(y) = y', we have

 < T(x), y > = <x, T* (y)>

To show T* is linear

Let y₁, y₂ ∈ V and c ∈ F. For any x ∈ V we have

<x, T* (cy₁+ y₂) > = < T (x), cy₁ + y₂ >

=  <T(x), y₁ > + < T (x), y₂ >

=  <x, T* (y₁) > + < x, T*(y₂) >

= < x, cT* (y₁) + T*(y₂) >

Since x is arbitrary T* (cy₁ + y₂) = cT*(y₁) + T*(y₂).

Finally we have to show that T* is unique.

Suppose that U: V→V is linear and it satisfies

 < T(x), y > = < x, U (y) > for all x, y ∈ V.

Then < x, T*(y) > = <x, U (y)> for all x, y ∈ V, so T*= U.

Note: The linear operator T* is called the adjoint of the operator T. The symbol T* is read as “T star”.

T* is the unique operator on V satisfying

 < T(x), y > = <x, T*(y) > for all x, y ∈ V.

 <x, T(y)> =  = < T*(x), y>

 So < x, T(y) > = <T*(x), y> for all x, y ∈ V.

Theorem 4

Let V be a finite dimensional inner product space, and let β be an orthonormal basis for V. If T is a linear operator on V, then [T*]_β = [T]*_β

Proof:

Let A = [T]_β, B = [T]^*_β and let β = {v₁, v₂, v₃... v_n }

Then

 B_ij = < T* (v_i), v_j >

= (A*)_ij

Hence B = A*

Theorem 5

Let A be an n×n matrix. Then L_A^* = (L_A)*.

Proof:

If β is the standard ordered basis for Fⁿ, then

 [L_A]_β = A. Hence [(L_A)*_β] = [L_A]*_β = A* = [L*_A]_β =, and so (L_A)* = L_A*

Corollary:

Let A and B be n×n matries. Then

(a) (A+B)* = A*+B*

(b) (cA)* = A* for all c ∈ F

(c) (AB)* = B*A*

(d) A** = A

(e) I* =I

Theorem 6

Let V be a non−zero finite dimensional inner product space. Then V has an orthonormal basis β. Furthermore if β= { v₁, v₂, ... v_n} and x ∈ V then x = _i=1Σⁿ <x,v_i>v_i.

Proof:

Given V is an finite dimensional inner product space.

Let β₀ be an ordered basis for V.

Then by Gram Schmidt orthogonalization process, we can obtain an orthogonal set β' of non zero vectors with span (β') = span (β₀)

Since β₀ is a basis of V, span (β₀) = V = span (β₀').

By normalizing each vector in β₀', we obtain an orthonormal set β that generates V.

Since β has non−zero orthogonal vectors,

⇒ β is linearly independent.

β is an orthonormal basis for V.

  (By corollary) Let β = (v₁, v₂, …. v_n}

Since x ∈ V = span (β) and β is an orthonormal set then x= _i=1Σⁿ <x,v_i>v_i.