Eigenvalues of Unitary Matrix proof

**firebio** · April 19th 2010, 05:20 PM

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
Thanks in advance

**aliceinwonderland** · April 19th 2010, 06:38 PM

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
Thanks in advance

The eigenvalues of $U^*$ are the complex conjugates of the eigenvalues of $U$ (link).

You have

$Uv = \lambda v$ ,
$U^*v = (1/\lambda)v$ .

Thus, $\lambda = 1/\bar{\lambda} \Leftrightarrow \lambda \bar{\lambda} =1 \Leftrightarrow |\lambda|^2=1$ .

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