Definition 11 An inner product on a vector space V is a way of assigning a

Definition 11 an inner product on a vector space v is

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Definition 11. An inner product on a vector space V is a way of assigning a scalar value x , y to any pair x , y of vectors such that for all vectors x , y , z 1. Positivity: x , x is a non-negative real number, which can be zero only if x = 0 ; 2. Symmetry: x , y = y , x * ( = y , x for real v.sp); 3. Linearity: x , αy + βz = α x , y + β x , z . Consequences: 1. x , 0 = 0; 2. αx , y = y , αx * = ( α y , x ) * = α * x , y which can be called “antilinearity in the first component”. Definition 12. Length or Norm or Magnitude of x : | x | = x , x , which is always 0 . x is called a unit vector or normalized vector if | x | = 1 . Examples 8. 1. Dot product in R n : x , y = x · y = x 1 y 1 + · · · + x n y n = n i =1 x i y i = x T y More generally, one has weighted dot products: given positive real numbers called weights w 1 , . . . , w n , x , y w = n i =1 w i x i y i
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22 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) 2. in C n : x , y = x * 1 y 1 + · · · + x * n y n = x y or the same with positive weights w i . 3. V = C [0, 1] : f , g = 1 0 f * ( x ) g ( x ) dx and for any positive weight function w ( x ) f , g w = 1 0 f * ( x ) g ( x ) w ( x ) dx Definition 13. x , y V are orthogonal if x , y = 0 . Example 5. The Bessel functions J n , Y n satisfy: 0 J n ( x ) Y n ( x ) x dx = J n , Y n x = 0 and so are orthogonal with respect to a weighted inner product on C (0, ) . There is a reason for this: we will find out later when we study Bessel’s equation. Definition 14. 1. A set of non-zero vectors { u 1 , u 2 , . . . , u n } ∈ V is called an orthog- onal set if they are “mutually orthogonal”: u i , u j = 0 for all pairs i = j ; 2. An orthogonal set which is “normalized”, that is u i , u i = 1 for all i , is called an orthonormal set . In summary, u i , u j = δ ij for all i , j . 3. Kronecker delta : δ ij = 1 if i = j and δ ij = 0 if i = j . Examples 9. 1. It is easy to check that e i e j = δ ij for the standard basis { e 1 , . . . , e N } for C N and so this is an orthonormal basis with respect to the standard dot product. 2. The Fourier exponential functions v n ( x ) = e inx for n Z are an orthonormal set in the inner product space C [0, 2 π ] : v m , v n = 2 π 0 e - imx e inx dx = δ mn , for all integers m , n Question: Why are orthogonal sets useful? Answer: First, orthogonal sets are guaranteed to be linearly independent. If i α i u i = 0, one can take the inner product with u j to show that α j = 0. If the set in fact forms a basis for V , then they are really useful because of the following: any x V has v –coordinates α which can be calculated directly α j = u j , x u j , u j , j = 1, . . . , n as shown in Assignment 2. For a general basis, remember that finding the coordinates requires a major exercise in gaussian elimination.
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1.8. INNER PRODUCT SPACES 23 Example 6. This can be made more concrete in the case of C n with the standard dot product. Suppose { v i } n i =1 is an orthogonal set (hence a basis, since there are n of them). Finding v –coordinates of a vector x means finding a = ( α j ) j =1: n such that x = Va where V = [ v 1 , v 2 , . . . , v n ] is the matrix constructed from the basis. The solution is a = V - 1 x , which requires one to calculate the inverse of V . Normally, this requires Gaussian elimination, but for an orthogonal basis it doesn’t. One can check that V - 1
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