Question

Prove that the root of unitary and an orthogonal matrix have unit modulus

Answer 1

Consider an example:

A=[cosθsinθ−sinθcosθ]A=[cosθ−sinθsinθcosθ]

The eigen value of A are found by solving D=|A−λI2|=(cosθ−λ)2+sin2θ=0D=|A−λI2|=(cosθ−λ)2+sin2θ=0

Giving eigen values λ=cosθ+isinθ=eiθλ∗=cosθ−iIsinθ=e−iθλ=cosθ+isinθ=eiθλ∗=cosθ−iIsinθ=e−iθ

The modules of λλ and λ∗λ∗ is 1.

b) If A is unitary the A∗A=IA∗A=I

Thus if Ax=λxAx=λx then x∗A∗=λ∗x∗x∗A∗=λ∗x∗

Hence x∗x=x∗A∗A∗x=X∗Xx∗xx∗x=x∗A∗A∗x=X∗Xx∗x

Since x∗x≠0x∗x≠0 we obtain x∗λ=1x∗λ=1

Hence |λ|=1

Answer 2

Step-by-step explanation:

know that a unitary matrix can be defined as a square complex matrix A, such that

AA*=A*A=I

where A∗ is the conjugate transpose of A, and I is the identity matrix. Furthermore, for a square matrix A, the eigenvalue equation is expressed by

Av=λv

If I use the relationship uv=v*u and take the conjugate transpose of this equation then

v*A*=λ*v*

But now I got stuck.