I have some troubles to show that the operator norm of a normal operator is always equal to its largest eigenvalue, how can I proof this? Does anybody of you have a hint? My problem is, that I do not know where to use that this operator is normal?
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Just a reminder: the statement is not necessarily true if the vector norm that defines the operator norm is not the same norm induced by the inner product that the normal operator is subject to. – user1551 May 29 '13 at 10:34
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I'm assuming we are talking finite dimension here, since you say "matrix norm".
Being normal, your operator is unitary diagonalizable. And the operator norm of a diagonal operator is easily seen to be the largest eigenvalue in absolute value.
Martin Argerami
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Notably, by "diagonalizable" here unitary diagonalization is meant, hence norm-preserving diagonalization. – Jonas Meyer May 29 '13 at 06:00
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