M 311 Spring 2004 HOMEWORK SUMMARY REPORT Chapter number: 8 Section number: 1 Exercise number: 24 Number of papers received: 10 Reviewing committee: Gamma List all participating committee members: Patrick Barrett, Mike Shelby Author(s) of paper(s) chosen for publication: James Macfarlane [and Nicole Clark] Comments: We received 10 papers in total and about half were correct. Multiple authors set the vectors up as a linear combination of the two eigenvectors and said that this result was the new eigenvectors. The paper that was published added the interesting fact at the end of "What if a=0" and described that there would be another eigenvectore and thus an eigenbasis would exist. ["a" really means "alpha".] GRADER'S COMMENT: Normally, linear combinations of the eigenvectors are not also eigenvectors, but that is true when the corresponding eigenvalues are equal. Some people seem confused about this point. See also the comments on the related problem Ex. 8.2.24. INSTRUCTOR'S COMMENT: I want to emphasize the point made by the reviewers. In this problem one finds only two linearly independent eigenvectors for a 3-dimensional space. As the grader says, since those two eigenvectors belong to the same eigenvalue, nontrivial linear combinations of them are also eigenvectors; however, those eigenvectors are not "new" in the sense of spanning the unreached part of the space. No eigenbasis exists in this case (except when a = 0). In addition to James (the only one who noted the special case a = 0) I'm also publishing Nicole, who gave the most accurate and complete description of the eigenspace.