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Old November 12th, 2008, 08:39 PM
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Please help me with this
Prove that if T is in L ( V ) and is normal, thennull ( Tk ) = null ( T ) and Im ( Tk ) = Im ( T ) for every positive integer k.
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Old November 13th, 2008, 02:25 PM
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Originally Posted by maria_stoeva View Post
Prove that if T is in L ( V ) and is normal, then null ( Tk ) = null ( T ) and Im ( Tk ) = Im ( T ) for every positive integer k.
I think you need the spectral theorem (the fact that a normal operator is diagonalisable) for this. If you take an orthonormal basis consisting of eigenvectors of T, then the null space of T is spanned by the eigenvectors corresponding to the eigenvalue 0, and the image of T is spanned by the eigenvectors corresponding to nonzero eigenvalues. It's fairly easy to see that the same description works for the null space and image of any positive power of T.
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