This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Partial Key to the Final Exam, Linear Algebra, Spring 2003 (The questions will be posted separately) 1) Proofs about orthogonal or symmetric etc are usually easy, because these words can be defined by equations. Which means the proof is mainly a calculation. In this problem, do not talk about perpendicular vectors etc. Instead: Answer: If A and B are orthogonal, we know that A T A = I and B T B = I . We must check this for AB . From matrix algebra and the assumptions, we get ( AB ) T AB = B T A T AB = B T IB = I , as desired. 2) Do the usual 6.1 calculations. Get = 1 or 3 and put the evecs into a matrix X . Then normalize the columns to get: U = 2 1 / 2 1 1 1 1 3) We must use the definition of subspace here. 1) [closure of scalarmult] Assume v U V (so, it is in U and in V ) and that R is a scalar. Since v U and U is a subspace, we know v U . Likewise, we can show v V , so v U V ....
View
Full
Document
 Spring '09
 JULIANEDWARDS

Click to edit the document details