Question 1204107: We may define a complex conjugation operator K such that Kz =z^ * . Show that K is not a linear operator
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **To show that the complex conjugation operator K is not linear, we need to demonstrate that it violates one of the two defining properties of linearity:**
1. **Homogeneity:** K(αz) = αK(z) for any scalar α and any complex number z.
2. **Additivity:** K(z1 + z2) = K(z1) + K(z2) for any complex numbers z1 and z2.
**Let's check homogeneity:**
* K(αz) = (αz)* = α*z*
* αK(z) = α(z*)
For homogeneity to hold, α*z* must always equal α(z*). However, this is not true in general.
**Counterexample:**
Let α = i (the imaginary unit) and z = 1 + i.
* K(αz) = K(i(1 + i)) = K(-1 + i) = -1 - i
* αK(z) = i * K(1 + i) = i * (1 - i) = i + 1
Since K(αz) ≠ αK(z) in this case, the complex conjugation operator K does not satisfy the homogeneity property of linearity.
**Therefore, the complex conjugation operator K is not a linear operator.**
**Key takeaway:**
Complex conjugation is an example of an **antilinear** operator. It satisfies a modified version of homogeneity: K(αz) = α*K(z).
|
|
|
