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.