is the case u 1 u m is said to be linearly dependent 10 CHAPTER 1 LINEAR

# Is the case u 1 u m is said to be linearly dependent

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is the case, { ˛ u 1 , . . . , ˛ u m } is said to be linearly dependent .

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10 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) 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 i , i = 1 : m such that m ÿ n =1 n ˛ u n = ˛ 0 is i = 0, for all i . Definition 5. A set { ˛ u 1 , . . . , ˛ u m } of vectors in W forms a basis for W i ff 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 = q m n =1 n ˛ u n is uniquely solvable for any ˛ x œ 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 = q i i e i is uniquely solved by x i = 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 = ( 1 , 2 , 3 ) T such that X = ( x 1 , x 2 , x 3 ) T = 1 u 1 + 2 u 2 + 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 œ C 3 , its coordinates A œ 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 ] æ [ I | U 1 ] Exercise: Review Algorithm 1 and compute the inverse of U = Q c a 1 1 1 1 1 1 0 0 1 R d b .
1.4. LINEAR OPERATORS 11 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 . 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 . Examples 4. 1. R 3 has dimension 3 . So does C 3 . 2. V = C [0, 1] is infinite dimensional because one can show that { 1, x , x 2 , . . . , x m } is linearly independent for all m . (Homework!) 3. The solution space of (1.1) is two-dimensional. Given a specific basis { ˛ u n } N n =1 of an N -dimensional vector space V , every ˛ x œ V can be uniquely expressed in the form ˛ x = N ÿ n =1 n ˛ u n The components of a = ( 1 , . . . , N ) T form a column N -vector a œ C N is called the coordinate N -vector of ˛ x relative to the basis { ˛ u } , or the u

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