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# normal - Diagonalization of Normal Operators Here is a...

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Diagonalization of Normal Operators April 18, 2007 Here is a proof of the diagonalizability of normal operators which does not use Schur’s theorem. Lemma Let T : V W be a linear operator. Assume that its adjoint T * exists. Then Ker ( T ) = Im ( T * ) . If V is ﬁnite dimensional, then Ker ( T ) = Im ( T * ). Proof: Recall that an element w of W is zero if and only if ( w | w 0 ) = 0 for all w 0 W . Thus x Ker ( T ) if and only if ( Tx | y ) = 0 for all y W , i.e. , if and only if ( x | T * y ) = 0 for all y W , i.e. , if and only if x is orthogonal to the image of T * . This proves the ﬁrst statement. It follows that ( Ker ( T )) = ( Im ( T * ) ) , and the latter is Im ( T * ) when V is ﬁnite dimensional. Lemma Let A and B be linear operators on V . If A and B commute, then Ker ( A ) and Im ( A ) are B -invariant. Proof: If x Ker ( A ), then AB ( x ) = BA ( x ) = 0, so B ( x ) Ker ( A ). Thus

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normal - Diagonalization of Normal Operators Here is a...

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