Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Linear Algebra With Applications
Found in: Page 217
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

Answers without the blur.

Just sign up for free and you're in.


Short Answer

In Exercises 40 through 46, consider vectorsV1,V2,V3 inR4 ; we are told thatVi,Vj is the entry aijof matrix A.


Find projv(v1), expressed as a scalar multiple ofv2 .

The required projection is projv2v1=59v2

See the step by step solution

Step by Step Solution

Step 1: Formula for the orthogonal projection.

If V is a subspace of Rnwith an orthonormal basis u1,.....,umthenlocalid="1659451887578" projvx=xll=(u1.x)u1+.....+(u1.x)um

For all xin localid="1659436140715" Rn.

Let us write the matrix in the Vi.Vinotation.

Consider the terms below.

localid="1659451940364" A=35115920112049=v1.v1v1.v2v1.v3v2.v1v2.v2v2.v3v3.v1v3.v2v3.v3=v12v1.v2v1.v3v2.v1v22v2.v3v3.v1v3.v2v32

Step 2: obtain the orthonormal basis for span(v→2)

To obtain the span. Since it is a singleton vector. So by normalizing it. Therefore, an orthonormal basis will be.


So, according to the above theorem

projv2v1=u.v1u =v23.v1v23 =v2.v13v23 =59v2

Hence, the role="math" localid="1659438206334" projv2v1=59v2.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.