# Linear Algebra with Applications (3rd Edition)

• Homework Help
• davidvictor
• 3

This preview shows page 2 - 3 out of 3 pages.

1 0 1 , 2 3 1 2 1 - 1 2 About the normal basis for the orthogonal complement of V, because V is 2- dimensional, so orthogonal complement must be 1-dimensional. Then let v = a b c , so v u 1 , v u 2 , we have a = - c, b = - a Choose a = 1 Then we have v = 1 - 1 - 1 , because we need normal basis, so the answer is u = v | v | = 1 3 1 - 1 - 1 Problem 3 Let { α 1 , α 2 , . . . , α k } be a set of mutually orthogonal vectors in R n . 2
1. Show that for any v R n , the vector v - (proj α 1 ( v ) + proj α 2 ( v ) + · · · + proj α k ( v )) is orthogonal to each of the α 1 , α 2 , . . . , α k . Solution Because { α 1 , α 2 , . . . , α k } be a set of mutually orthogonal vectors in R n , they are actually linear independent, then they are basis of subspace V of R n , with dimension k. By using Gram-Schmidt, we change the basis to orhornomal basis, because they are mutually orthogonal vectors, the orthnomal basis is just α 1 | α 1 | · · · α k | α k | So v - (proj α 1 ( v ) + proj α 2 ( v ) + · · · + proj α k ( v )) is a vector which is in V , orthogonal to each of the α 1 , α 2 , . . . , α k 1. Verify this claim for R 3 with α 1 = e 1 , α 2 = e 2 , and letting v = 1 2 3 . Draw a picture. Solution v - (proj α 1 ( v ) + proj α 2 ( v )) = 0 0 3 . 3