EE 263 Final: EE 263 Final Exam Solutions

• School: Stanford University
• Department: Electrical Engineering
• Course: EE 263
• Professor: all

Document Summary

Qt q = (i 2uut )t (i 2uut ) = (i 2uut )(i 2uut ) = i 2uut 2uut + 4uut uut. = i 2uut 2uut + 4uut so q is orthogonal. Qv = v 2uutv = v using ut v = 0 using ut u = 1 using ut u = 1 (c) we know det(q) =!n i=1 i. Since q is symmetric, all eigenvalues are real and we can construct an orthonormal eigenvector basis. From parts (a) and (b), u is an eigenvector with associated eigenvalue 1, and any vector v orthogonal to u is an eigenvector with associated eigenvalue 1. The nullspace of ut has dimension n 1, so we can construct an orthogonal eigenbasis with all eigenvalues 1 except for the 1 eigenvalue with eigenvector u. 1 = det(q). (d) since q is orthogonal, qt q = i has all eigenvalues 1, hence all singular values of.