Adjoint of Linear Operator: Example Solved Problems

ADJOINT OF LINEAR OPERATOR

WORKED EXAMPLE PROBLEMS

Example 1

Define g: R²→R by g (a₁, a₂) = 2α₁+a₂. Prove that g is a linear transformation.

Solution

Let β = {e₁, e₂} and let y=g(e₁)e₁ + g(e₂)е₂

= 2e₁ + e₂ = (2, 1)

Then g(a₁, a₂) = < (a₁, a₂), (2, 1) > = 2a₁ + a₂

Example 2

Let T be the linear operator on C² defined by

 T (a₁, a₂) = (2ia₁ +3a₂, a₁ − a₂).

If β is the standard ordered basis for C², then find the adjoint.

Solution

 Here T*(a₁, a₂) = (−2ia₁ + a₂, 3a₁−a₂)

Example 3

Let V = R². Define T(a, b) = (2a+b, a−3b) and x = (3,5). Determine T* for the given vector V.

Solution:

Let β = {(1, 0), (0, 1)} be the standard basis for V=R²

Then T (1, 0) = [2 (1) +0, (1) − 3(0)] = (2, 1)

T (0, 1) = [2 (0) + 1, 0 −3 (1)] = (0 +1, 0−3) = (1,3)

[T]β = 

Since the innerproduct space V = R²  ⇒ [T]_β = [T ]^*_β

The self adjoint of T* at x = (3, 5) is

  [T*(x)]β = [T*]β  [x]β   = [T]*β [x]β  = [T]β [x]β 

 The self−adjoint of T* at x = (3, 5) is T* (x) = (11, ‒12)

Example 4

Let V=C². Define T(x₁,x₂) = (2x₁+ ix₂, (1−i) x₁) and x = (3 − i, 1+ 2i). Compute T* for the given vector in V−C².

Solution:

Let V=C²

It is given that T (x₁, x₂) = (2x₁ + ix₂, (1 − i) x₁) \\ \

and the given x is that x = (3− i, 1 + 2i).

Given that T(x₁, x₂) = (2x₁ + ix₂, (1 − i) x₁)

= (2x₁+ix₂, (1−i)x₁+ 0x₂)

To find [T]^*_β:

The self adjoint of T* at x = (3−i, 1+2i) is

 [T*(x)]_β = [T*(x)]_β [x]_β = [T]*_β [x]_β

 The adjoint of T* at x=(3−i, 1+2i) is obtained as

[T*(x)]_β = (5 + i, − 1 − 3i).

Example 5

Let V = P₁(R) defined by <f,g> = _-1ʃ¹ f(t)g(t) dt and T[f] = f ' + 3f where f(t) = 4−2t. Determine T* for the given V.

Solution:

It is given that V = P₁(R), and

 <f, g> = _-1ʃ¹ f(t)g(1) dt and T(f)=f '+3f and f(t)=4−2t.

By the definition of adjoint of linear operator, we know that

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

Here < T(g), f > = < g, T*(f) > for all f,g ∈ V.

Let us consider g (t) = a + bt which is an unknown arbitrary function.

Let us consider the L.H.S of adjoint of operator.

 < T(g), f > = < T(a + bt), 4 ‒ 2t >

But T(f) = f '+ 3f

 < (T(g), f > = < (a, bt)' + 3(a+bt), 4−2t >

 = < b+3(a+bt), 4−2t >

 = < 3a+3bt + b, 4 − 2t >

But we know that <f,g> = _‒1ʃ¹ f(t)g(t) dt

= _‒1ʃ¹ (3a + 3bt + b) (4 – 2t) dt

= _‒1ʃ¹ [12a+12bt + 4b − 6at ‒ 6bi² – 2bt] dt

= _‒1ʃ¹ [12a+10bt + 4b − 6at – 6bt²] dt

= [ 12at + 5b₁t² + 4bt ‒ 3at² ‒ 2bt³ ]¹₋₁

= 24a + 4b

  < T(g), f > = 24a+4b

                    …………….(1)

Now let us consider the R.H.S of adjoint of operator.

  (ie) < g, T*(f) >

Now let us assume that T*(f) = c+dt

 < g, T*(f) > = < a+bt, c+dt >

= bd/3 + bd/3 +ac + ac

< g, T*(f) > = 2ac + 2bd/3

We know that  < T(g), f > = < g, T*(f) >

From equations (1) and (2), we have

 24a+4b = 2ac + 2/3 bd

By comparing the coefficients of a and b on bothsides, we get

24=2c

c = 12

4= 2d/3

 d = 4. 3/2 = 6

 T*(f) = c + dt, we have T*(f) = 12 + 6t.

The self−adjoint of T* at f(t)=4−2t is obtained as T*(f) = 12 + 6t.