quiz5-solns - = range S * for any operator S , we see that...

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MATH 110 - SOLUTION TO QUIZ 5 LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA ˜ NEZ Suppose that V is a finite dimensional inner-product space and that T ∈ L ( V ). Prove that λ is an eigenvalue of T if and only if λ is an eigenvalue of T * . Proof. Suppose that λ is an eigenvalue of T . Then T - λI is not injective, so null( T - λI ) 6 = { 0 } , or equivalently null( T - λI ) 6 = V where the second statement follows from the decomposition V = null( T - λI ) null( T - λI ) . Now, since ( T - λI ) * = T * - λI and (null S )
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Unformatted text preview: = range S * for any operator S , we see that range( T *-I ) 6 = V. Thus T *-I is not surjective, and hence not injective since V is nite dimensional. This shows that is an eigenvalue of T * . The converse (i.e. backwards direction) follows by applying the above to the eigen-value of T * , and using the fact that = and ( T * ) * = T . Date : July 30, 2009....
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