This preview shows pages 1–3. 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: Math 416  Abstract Linear Algebra Fall 2011, section E1 Midterm 3, November 16 Name: Solutions No calculators, electronic devices, books, or notes may be used. Show your work. No credit for answers without justification. Good luck! 1. /10 2. /10 3. /10 4. /10 Total: /40 1 Problem 1. (10 pts) Let A : R 4 R be the linear map defined by A x 1 x 2 x 3 x 4 = x 1 + x 2 + x 3 + x 4 . Find the projection of the vector v = 2 1 3 onto the subspace ker A = { x R 4  Ax = 0 } . Noting ker A = Span { w = 1 1 1 1 } , the projection is Proj ker A ( v ) = Proj Span { w } ( v ) = v Proj w v = v ( v,w ) ( w,w ) w = v 4 4 w = v w = 2 1 3  1 1 1 1 = 1 2 2 1 . Alternate method: GramSchmidt. ker A has basis { 1 1 , 1 1 , 1 1 } . Applying (some variant of) GramSchmidt yields the orthogonal basis { v 1 = 1 1 ,v 2 = 1 1 2 ,v 3 = 1 1 1 3 } ....
View
Full
Document
This note was uploaded on 12/16/2011 for the course MATH 416 taught by Professor Frankland during the Fall '11 term at University of Illinois at Urbana–Champaign.
 Fall '11
 FRANKLAND
 Math, Linear Algebra, Algebra

Click to edit the document details