{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture_7_2 - Lecture Note 7-2 Oct 24 Dr Jeff Chak-Fu WONG...

Info icon This preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture Note 7-2: Oct 24 - Oct 27, 2006 Dr. Jeff Chak-Fu WONG Department of Mathematics Chinese University of Hong Kong [email protected] MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff Chak-Fu WONG 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
R EAL V ECTOR S PACES 1. Vector Spaces 2. Subspaces 3. Linear Independence 4. Basis and Dimension 5. Homogeneous Systems 6. The Rank of a Matrix and Applications 7. Coordinates and Change of Basis 8. Orthonormal Bases in R n 9. Orthogonal Complements R EAL V ECTOR S PACES 2
Image of page 2
O RTHONORMAL B ASES IN R n O RTHONORMAL B ASES IN R n 3
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
From our work with the natural bases for R 2 , R 3 , and, in general, for R n , we know that when these bases are present, the computations are kept to a minimum. A subspace W of R n need not contain any of the natural basis, but in this lecture, we want to show that it has a basis with the same properties. That is, we want to show that W contains a basis S such that every vector in S is of unit length and every two vectors in S are orthogonal. The method used to obtain such a basis is the Gram-Schmidt process , which is presented in this lecture. O RTHONORMAL B ASES IN R n 4
Image of page 4
DEFINITION-A set S = { u 1 , u 2 , . . . , u k } in R n is called orthogonal if every pair of distinct vectors in S are orthogonal, that is, if u i · u j = 0 for i 6 = j . An orthonormal set of vectors is an orthogonal set of unit vectors . That is, S = { u 1 , u 2 , . . . , u k } is orthonormal if u i · u j = 0 for i 6 = j and u i · u i = 1 for i = 1 , 2 , . . . , k. Normalizing refers to the process of dividing each vector in an orthogonal set S by its length so S is transformed into an orthonormal set . O RTHONORMAL B ASES IN R n 5
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Example 1 If x 1 = (1 , 0 , 2) , x 2 = ( - 2 , 0 , 1) and x 3 = (0 , 1 , 0) , then { x 1 , x 2 , x 3 } is an orthogonal set in R 3 . The vectors u 1 = ( 1 5 , 0 , 2 5 ) and u 2 = ( - 2 5 , 0 , 1 5 ) are unit vectors in the directions of x 1 and x 2 , respectively. Since x 3 is also of unit length, it follows that { u 1 , u 2 , u 3 } is an orthonormal set. Also, span { x 1 , x 2 , x 3 } is the same as span { u 1 , u 2 , u 3 } . O RTHONORMAL B ASES IN R n 6
Image of page 6
Example 2 The natural basis { (1 , 0 , 0) , (0 , 1 , 0) , (0 , 0 , 1) } is an orthonormal set in R 3 . More generally, the natural basis in R n is an orthonormal set. O RTHONORMAL B ASES IN R n 7
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Theorem 0.1 Let S = { u 1 , u 2 , . . . , u k } be an orthogonal set of nonzero vectors in R n . Then S is linearly independent. Proof Consider the equation c 1 u 1 + c 2 u 2 + · · · c k u k = 0 . (1) Taking the dot product of both sides of (1) with u i , 1 i k , we have ( c 1 u 1 + c 2 u 2 + · · · c k u k ) · u i = 0 · u i . (2) By properties (c) and (d) of Theorem 0.3, Properties of dot Product, the left side of (2) is c 1 ( u 1 · u i ) + c 2 ( u 2 · u i ) + · · · + c k ( u k · u i ) , and the right side is 0 . Since u j · u i = 0 if i 6 = j , (2) becomes 0 = c i ( u i · u i ) = c i k u i k 2 . (3) By (a) of Theorem 0.3, Properties of dot Product, k u i k 6 = 0 , since u i 6 = 0 .
Image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern