# Eigenvalues of Unitary Matrix proof

• Apr 19th 2010, 05:20 PM
firebio
Eigenvalues of Unitary Matrix proof
Unitary is UU*=I
U* is transpose conjugate

Prove that if a matrix U is unitary, then all eigenvalues of U have absolute value of 1.

$\displaystyle Uv= \lambda v$

$\displaystyle U^* Uv=\lambda U^*v$

$\displaystyle v= \lambda U^* v$

$\displaystyle v/\lambda=U^* v$
so v is also a eigenvector for U* with eigenvalue of $\displaystyle 1/\lambda$.
Not sure how to continue?
any help would be appreciated
• Apr 19th 2010, 06:38 PM
aliceinwonderland
Quote:

Originally Posted by firebio
Unitary is UU*=I
U* is transpose conjugate

Prove that if a matrix U is unitary, then all eigenvalues of U have absolute value of 1.

$\displaystyle Uv= \lambda v$

$\displaystyle U^* Uv=\lambda U^*v$

$\displaystyle v= \lambda U^* v$

$\displaystyle v/\lambda=U^* v$
so v is also a eigenvector for U* with eigenvalue of $\displaystyle 1/\lambda$.
Not sure how to continue?
any help would be appreciated
The eigenvalues of $\displaystyle U^*$ are the complex conjugates of the eigenvalues of $\displaystyle U$ (link).
$\displaystyle Uv = \lambda v$,
$\displaystyle U^*v = (1/\lambda)v$.
Thus, $\displaystyle \lambda = 1/\bar{\lambda} \Leftrightarrow \lambda \bar{\lambda} =1 \Leftrightarrow |\lambda|^2=1$.