This is impossible because T v T v v 3 Since T \u03bbI is not injective it follows

# This is impossible because t v t v v 3 since t λi is

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1). This is impossible, because T * ( v ) = - T ( v ) = v . 3. Since T - λI is not injective, it follows that ( T - λI ) 0 = T 0 - λI is not surjective; since V is finite-dimensional, this means that T 0 - λI is not injective, so λ is an eigenvalue of T 0 . 4. Since v and w are eigenvectors of a normal operator for distinct eigenvalues, they are orthogonal; by Pythagoras’ theorem, k Tv + Tw k 2 = k 3 v + 4 w k 2 = k 3 v k 2 + k 4 w k 2 = 9 k v k 2 + 16 k w k 2 = 100 , so k Tv + Tw k = 10 . 5. (i) There are many examples; for example, on C 2 , we can take S = 0 1 - 1 0 and T = 1 0 0 - 1 . (ii) Since S and T are normal, it follows that range( S * ) = range( S ) and range( T * ) = range( T ); in particular, range( S * ) and range( T * ) are also orthogonal. By the Pythagorean theorem, since Sv and Tv , and also S * v and T * v , are orthogonal for any v V, k ( S + T ) v k 2 = k Sv k 2 + k Tv k 2 = k S * v k 2 + k T * v k 2 = k ( S + T ) * v k 2 for all v V . Therefore, S + T is normal. 6. (i) Assume that P = P U . Then P is self-adjoint, because: for any v, w V , since v - Pv, w - Pw range( P ) , it follows that h Pv, w i - h v, Pw i = h Pv, w - Pw i + h Pv - v, Pw i = 0 + 0 = 0 . (ii) Assume that P is self-adjoint, and define U := range( P ). Let v V. For any w U , h v - Pv, w i = P ( v, w ) - P ( Tv, w i ) = P ( v, w ) - P ( v, Tw ) = P ( v, w ) - P ( v, w ) = 0 , so v - Pv U . This implies that Pv = #### You've reached the end of your free preview.

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• Fall '08
• GUREVITCH
• • •  