MODULE 3. 0, y = 0 for all y


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1 Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i) x, y = y, x ii) αx, y = α x, y iii) x + y, z = x, z + y, z where z is ny other vector in V iv) x, x 0 nd x, x = 0 if nd only if x = 0. We note from these properties tht: 0, y = 0 for ll y x, αy = αy, x = α y, x = α x, y nd x, y + z = y + z, x = y, x + z, x = x, y + x, z For rel vector spces the complex conjugtion hs no effect so tht the inner product is liner in the first nd second rgument, i.e. x + αy, z = x, z + α y, z x, y + αz = x, y + α x, z. For complex sclrs we hve wht is sometimes clled ntilinerity in the second rgument. x, y + αz = x, y + α x, z. Exmples: 1) Setting: V = R n nd F = R i) x, y = n j=1 x jy j This is the fmilir dot product nd we shll often use the more fmilir nottion x y insted of x, y. ii) x, y = n j=1 x jy j w j where w i > 0 for i = 1,..., n. 12
2 Note tht this inner product cn be written s x, y = W x y where W is the digonl mtrix W = dig{w 1,..., w n }. iii) x, y = Cx y where C is positive definite symmetric mtrix. The proof tht this defines n inner product will be deferred until we hve discussed mtrices nd their eigenvlues. 2) Setting: V = C n nd F = C n x, y = x j y j This is the complex dot product nd likewise is more commonly denoted by x y. Note tht the order of the rguments now mtters becuse the components of y re conjugted. 3) Setting: V = C 0 [], F = C f, g = f(x)g(x)w(x)dx where the soclled weight function w is continuous nd positive except t isolted points of the given (possibly multidimensionl) domin. For rel vlued functions the conjugtion hs no effect nd f, g = g, f. In generl when checking whether function defined on pirs of vectors is n inner product properties ii) nd iii) re esy to estblish. Properties i) nd iv) my require more work. For exmple, let us define on R 2 the function x, y = Ax y j=1 where A is the mtrix Then A = ( ) x, y = x 1 y 1 + x 2 y 1 + x 2 y 2 while y, x = Ayx = y 1 x 1 + y 2 x 1 + x 2 y 2. 13
3 Hence x, y y, x for ll x nd y (tke, for exmple, x = (1, 0) nd y = (1, 1)) nd property i) does not hold. Looking hed, if for vectors x, y R n we require tht x, y = Ax y = y, x = Ay x for given rel n n mtrix A then A must be symmetric mtrix, i.e., A = A T. However, A = A T is not sufficient to mke x, y n inner product. For exmple, if A = dig{1, 1} then A is symmetric but x, x = Ax x = 0 for the nonzero vector x = (1, 1). Thus, property iv) does not hold. It turns out tht this mtrix A is not positive definite. As finl exmple consider the following function defined on C 0 [ 2, 2] f, g = 2 2 f(t)g(t)w(t)dt where w(t) = mx{0, t 2 1}. We see tht properties i) iii) hold but tht for the nonzero function f(t) = mx{0, 1 t 2 } we obtin f, f = 0. Clerly, the weight function w my not be zero over ny intervl. The following theorem shows tht inner products stisfy n importnt inequlity which is known s Schwrz s inequlity. Theorem: Let V be vector spce over C with inner product x, y. Then for ny x, y V we hve x, y 2 x, x y, y. Proof: Let x nd y be rbitrry in V. If y = 0 then the inequlity is trivilly true. Hence let us ssume tht y 0. Next let us choose θ such tht for ˆx = e iθ x the inner product ˆx, y is rel. Then from the properties of the inner product we see tht g(λ) = ˆx λy, ˆx λy = ˆx, ˆx 2λ x, y + λ 2 y, y 0 for ll rel λ. In prticulr, the minimum of g is 0 This minimum is chieved t λ θ = ˆx, y / y, y so tht g(λ θ ) = ˆx, ˆx ˆx, y 2 / y, y 0 from which we obtin ˆx, y ˆx, ˆx y, y. Finlly, we observe tht this inequlity remins unchnged when ˆx is replced by x since the phse fctor e iθ drops out. 14
4 Two illustrtions: i) It is usully shown in first course on vectors in R 2 tht x y = x 2 y 2 cos θ where θ is the ngle between the vectors x nd y. Since cos θ 1 it follows tht x, y x y x 2 y 2. ii) Let be the tringle with vertices (0, 0), (1, 0), (0, 1). Let w be continuous function positive function in C 0 []. For ny f, g C 0 [] define the inner product f, g = f(x, y)g(x, y)w(x, y)dx dy Then for ɛ > 0 we obtin from Schwrz s inequlity f, g 2 = ɛf, g/ ɛ < ɛf, ɛf g/ ɛ, g/ ɛ = ɛ f(x, y) 2 w(x, y)dx dy 1 g(x, y) 2 w(x, y)dx dy ɛ Theorem: Let V be vector spce with inner product,. Then x = x, x 1/2 is norm on V. Proof. Properties i) nd ii) of the norm re direct consequence of properties ii) nd iv) of the inner product. To estblish the tringle inequlity we observe tht x + y 2 = x + y, x + y = x, x + 2 Re x, y + y, y x x, y + y 2. By Schwrz s inequlity x, y x y so tht x+y 2 ( x + y ) 2 which estblishes the tringle inequlity nd hence tht x, x 1/2 = x is norm on V. For exmple, ( f = ) 1/2 f(x, y) 2 w(x, y)dx dy defines norm on the vector spce C 0 [] provided w is positive weight function. 15
5 efinition: Two vectors x, y in vector spce V with inner product x, y re orthogonl if x, y = 0. Exmples: i) In R 2 with inner product x y two vectors re orthogonl if they re perpendiculr to ech other becuse x ( ) y = x y cos θ. 2 1 ii) Let A = nd define x, y = Ax y. Tke for grnted for the time being tht 1 2 x, y is n inner product. Let x = (1, 0), then y = (1, 2) stisfies x, y = 0 nd hence is orthogonl to x. Note tht orthogonlity in this cse sys nothing bout the ngle between the vectors. iii) Let f, g = π π f(t)g(t)dt be the inner product on C0 ( π, π) then the functions {cos nt, sin mt} re ll mutully orthogonl. iv) The functions t 2k nd t 2n+1 for ny nonnegtive integers k nd n re lso mutully orthogonl with respect to the inner product of iii). We hve repetedly introduced norms nd inner products expressed in terms of integrls for continuous functions defined on some set. However, integrls remin defined for much more generl functions, for exmple, functions with certin discontinuities. In fct, even the notion of the integrl cn be extended beyond the concept of the Riemnn integrl fmilir to us from clculus. We shll introduce nd use routinely the following vector (i.e., function) spce. efinition: L 2 () = {ll functions defined on such tht f(x) 2 dx < }. We observe tht if f L 2 () then αf L 2 (). If f, g L 2 () then it follows from f(x) + g(x) 2 dx = ( f(x) f(x)g(x) + g(x) 2 )dx nd the lgebricgeometric men inequlity tht 2 f(x)g(x) f(x) 2 + g(x) 2 ( ) f(x) + g(x) 2 dx 2 f(x) 2 dx + g(x) 2 dx < so tht L 2 () is closed under vector ddition nd sclr multipliction nd hence vector spce. 16
6 We observe tht C k [] for ny k > 0 is subspce of L 2 () provided tht is bounded set. Finlly, we note tht f, g = f(x)g(x)dx nd f = f, f 1/2 define the inner product nd norm usully ssocited with L 2 (). In generl, the functions will be rel nd the conjugtion cn be ignored. On occsion the inner product nd norm re modified by including weight function w(x) > 0 in the integrl. In this course the integrl will remin to be the Riemnn integrl. In more bstrct setting it should be the Lebesgue integrl. 17
7 Module 3  Homework 1) V = C n, F = C, x, y = x y. Let x = (x 1,..., x n ) where x j = j; y = (y 1,..., y n ) where y j = (1 + i)j nd i 2 = 1. Compute x, y, y, x, x, x, y, y. 2) i) Let A be n m n rel mtrix nd A T its trnspose. Show tht Ax y = x A T y for ll x R n nd y R m. ii) Let A be n m n complex mtrix nd A its conjugte trnspose (i.e., A = A T ). Show tht Ax y = x A y for ll x C n nd y C m. 3) Let nd A = ( ) x, y = Ax y for x, y R 2. Show tht, defines n inner product on R 2. 4) Prove or disprove: is n inner product on C 0 [ 1, 1]. f, g = 1 0 f(t)g(t)dt 5) Show tht if w is positive continuous function on [, b] then b nd in prticulr tht [ b f(t)w(t)dt ] 1/2 b w(t)dt f(t) 2 w(t)dt b f(t)dt b b f(t) 2 dt. 18
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