Eigenvalues of Unitary Matrix proof

• Apr 19th 2010, 06:20 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.

$Uv= \lambda v$

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

$v= \lambda U^* v$

$v/\lambda=U^* v$
so v is also a eigenvector for U* with eigenvalue of $1/\lambda$.
Not sure how to continue?
any help would be appreciated
• Apr 19th 2010, 07:38 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.

$Uv= \lambda v$

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

$v= \lambda U^* v$

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