This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Purdue University MA 353: Linear Algebra II with Applications Homework 12, due Apr. 16, Some solutions Sec. 6.1:#12: Let { v 1 ,v 2 ,...,v k } be an orthogonal set in V , and a 1 ,a 2 ,...,a k be scalars. Prove that k X i =1 a i v i 2 = k X i =1  a i  2 k v i k 2 . Proof. Note that the vectors a 1 v 1 and a 2 v 2 + a 3 v 3 + + a k v k are orthogonal. Hence by #10 of this section, k X i =1 a i v i 2 = a 1 v 1 + k X i =2 a i v i 2 =  a 1  2 k v 1 k 2 + k X i =2 a i v i 2 . Now applying the same argument to k i =2 a i v i 2 , one obtains k X i =1 a i v i 2 =  a 1  2 k v 1 k 2 +  a 2  2 k a 2 k 2 + k X i =3 a i v i 2 . By repeating the same process, one obtains k X i =1 a i v i 2 = k X i =1  a i  2 k v i k 2 . #13: Suppose that h , i 1 and h , i 2 are inner products on a vector space V . Prove that h , i = h , i 1 + h , i 2 is another inner product on V . Proof. We need to verify the definition of inner product, i.e. all of (a), (b), (c) and (d) as in Definition in p. 329330. For example, (a) is verifies as h x + z,y i = h x + z,y i 1 + h x + z,y i 2 = h x,y i 1 + h z,y i 1 + h x,y i 2 + h z,y i 2 = h x,y i 1 + h x,y i 2 + h z,y i 1 + h z,y i 2 = h x,y i + h z,y i ....
View
Full
Document
This note was uploaded on 04/29/2010 for the course MA 353 taught by Professor Staff during the Spring '08 term at Purdue.
 Spring '08
 Staff
 Linear Algebra, Algebra, Scalar

Click to edit the document details