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Unformatted text preview: MATH 110 - SOLUTIONS TO FINAL LECTURE 1, SUMMER 2009 GSI: SANTIAGO CA NEZ 1. (10 points) Give two equivalent definitions (or characterizations) of each of the following. (a) A normal operator on an inner-product space V . (b) A generalized eigenvector of an operator T . (c) A positive operator on an inner-product space V . (d) An isometry on an inner-product space V . Solution. Here are some possible answers. (a) An operator T such that TT * = T * T ; or, an operator T such that k Tv k = k T * v k for all v V . (b) A vector v such that for some eigenvalue of T there exists k 1 such that ( T- I ) k v = 0; or, a vector v such that for some eigenvalue of T , ( T- I ) dim V v = 0. (c) A self-adjoint operator T such that h Tv,v i 0 for all v V ; or, an operator T so that there exists S L ( V ) such that T = S * S . (d) An operator T such that k Tv k = k v k for all v V ; or, an operator T such that T * T = I . 2. (15 points) Give examples, with brief justification, of each of the following. (a) An operator on R 2 which is not self-adjoint with respect to the standard inner product. (b) An isometry on R 4 with no (real) eigenvalues. (c) An operator on C 4 whose characteristic polynomial equals the square of its minimal polynomial. Solution. Here are some possible answers. (a) With respect to the standard inner product, the adjoint of a matrix is just its tranpose, so any non-symmetric matrix would work say 1 1 0 1 . (b) First note that the rotation by 90 operator on R 2 :- 1 1 , is an isometry and has no eigenvalues. So, the following analog of this on R 4 : - 1 0 1- 1 1 is an isometry on R 4 with no eigenvalues. Date : August 13, 2009. 2 GSI: SANTIAGO CA NEZ (c) The following matrix in Jordan form: 1 1 0 0 0 1 0 0 0 0 1 1 0 0 0 1 works. The characteristic polynomial is ( z- 1) 4 , and the minimal polynomial is (...
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This note was uploaded on 08/31/2009 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.
- Spring '08