Idea The trick of writing m i 1 α i u i Ua 13 for a α 1 α m T is very helpful

# Idea the trick of writing m i 1 α i u i ua 13 for a

• Notes
• 26

This preview shows page 9 - 12 out of 26 pages.

Idea: The trick of writing m i =1 α i u i = Ua (1.3) for a = ( α 1 , . . . , α m ) T is very helpful! 1.3 Basis Given a set { u 1 , . . . , u m } of vectors in V , the span W = span ( u 1 , . . . , u m ) is always a subspace of V . This means every x W can be written as x = m n =1 α n u n (1.4) for some scalars α 1 , α m . However, some vector, say u i , may be redundant in the sense that u i is itself a linear combination of the remaining vectors in { u 1 , . . . , u m } . When this is the case, { u 1 , . . . , u m } is said to be linearly dependent . We want to focus on sets { u 1 , . . . , u m } that have no such redundancy. Such a set of vectors is called linearly independent ; formally, this means that the only set of scalars α i , i = 1 : m such that m n =1 α n u n = 0 is α i = 0, for all i . Subscribe to view the full document.

10 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) Definition 5. A set { u 1 , . . . , u m } of vectors in W forms a basis for W iff these two properties are true: 1. span ( u 1 , . . . , u m ) = W 2. { u 1 , . . . , u m } is linearly independent . In other words, “ { u 1 , . . . , u m } is a basis for W ” means x = m n =1 α n u n is uniquely solvable for any x W . Examples 3. 1. Standard basis for both for R 3 and C 3 : e 1 = (1, 0, 0) T , e 2 = (0, 1, 0) T , e 3 = (0, 0, 1) T . Check this is a basis: For any X = ( x 1 , x 2 , x 3 ) T , X = i α i e i is uniquely solved by x i = α i , i = 1, 2, 3 . Note that in case V = C 3 (respectively R 3 ), the scalars are complex (real). 2. u 1 = (1, 1, 0) T , u 2 = (1, - 1, 0) T , u 3 = (1, 1, 1) T is another basis for R 3 and C 3 . Why? Answer by row reduction: For any X , to find A = ( α 1 , α 2 , α 3 ) T such that X = ( x 1 , x 2 , x 3 ) T = α 1 u 1 + α 2 u 2 + α 3 u 3 we use the same trick as before. We write U = [ u 1 , u 2 , u 3 ] and solve X = UA for the 3 unknown components of A . By row reducing the augmented matrix [ U | X ] , we can show the system is uniquely solvable for any X . More succinctly, for any X C 3 , its coordinates A C 3 are computed by A = U - 1 X where U - 1 is the inverse of U . Algorithm 1. (Computing U - 1 ): use Gauss-Jordan elimination on the fully aug- mented [3, 6] matrix [ U | I ] [ I | U - 1 ] Exercise: Review Algorithm 1 and compute the inverse of U = 1 1 1 1 - 1 1 0 0 1 . 3. V = C [0, 1]: The three functions { cos x , sin x , 1 } are linearly independent, which means they form a basis for their span. On the contrary, the three functions { cos 2 x , sin 2 x , 1 } are linearly dependent because of the identity 1 · cos 2 x + 1 · sin 2 x + ( - 1) · 1 = 0 (1.5) 4. The functions v n ( x ) = e inx for n = - N : N form a basis for the vector space W N . 1.4. LINEAR OPERATORS 11 5. The pair of functions J n , Y n form a basis for the solution space of (1.1). We some- times say they form a fundamental set of solutions . Definition 6. 1. V has dimension N means it has a set of N linearly independent vectors, but no set of N + 1 linearly independent vectors 2. V is infinite dimensional means it has a set of N linearly independent vectors for every N . Theorem 1. (“Dimension theorem”) Any two bases { u 1 , . . . , u m } and { v 1 , . . . , v n } for a vector space V with finite dimension N have the same number of vectors m = n = N . Subscribe to view the full document. • Winter '10
• kovarik

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern