Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Low

Let A be a unitary matrix

$A A^\theta=A^\theta A=I$ (1)

Let $\lambda$ be any eigen value of A

To prove that $\lambda$ is of unit modulus i.e $\mid \lambda \mid=1$

As $\lambda$ is an eigen value of A then there exist $X\neq$0 such that

$(AX-\lambda I)X=0$

$AX-\lambda X=0$

$AX=\lambda X$ (2)

$(AX)^\theta=(\lambda X)^\theta$

$X^\theta A^\theta= \lambda^\theta X^\theta$

$X^\theta A^\theta=(\bar \lambda^\theta) \prime X^\theta$

$X^\theta A^\theta=(\bar \lambda^\theta) X^\theta$

post multiplying by A on both sides, we get,

$X^\theta A^\theta A=(\bar \lambda^\theta) X^\theta A $

$X^\theta I= \bar \lambda X^\theta A$

post multiplying by X on both sides we get,

$X^\theta I X= \bar \lambda X^\theta A X$

$X^\theta X= \bar \lambda X^\theta \lambda X$ from (2)

$X^\theta X= \bar \lambda X^\theta \lambda X$

$-X^\theta X +\bar \lambda X^\theta \lambda X=0$

$(\bar \lambda \lambda -1) X^\theta X=0$

As $X\neq0$ $X^\theta X\neq0$ $\bar \lambda-1=0$

$\bar \lambda \lambda=1$

${\mid \lambda^ \mid}^2=1$

$\mid \lambda \mid=1$

Therefore, eigen values of unitary matrix are of unit modulus