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Chapter VI. Inner Product Spaces

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Chapter VI. Inner Product Spaces Notes ⃝c F.P. Greenleaf 2014-15 LAI-f14-iprods.tex version 2/9/2015 Chapter VI. Inner Product Spaces. VI.1. Basic Definitions and Examples. In Calculus you encountered Euclidean coordinate spaces Rn equipped with additional structure: an inner product B : Rn Rn R. × → n Euclidean Inner Product: B(x, y)= i=1 xiyi which is often abbreviated to B(x, y)=(x, y). Associated! with it we have the Euclidean norm n x = x 2 =(x, x)1/2 ∥ ∥ | i| i=1 " which represents the “length” of a vector, and a distance function d(x, y)= x y ∥ − ∥ which gives the Euclidean distance from x to y.Notethaty = x +(y x). − Figure 6.1. The distance between points x, y in an inner product space is interpreted as the norm (length) ∥y − x∥ of the difference vector ∆x = y − x. This inner product on Rn has the following geometric interpretation (x, y)= x x cos (θ(x, y)) ∥ ∥·∥ ∥· where θ is the angle between x and y,measuredintheplaneM = R-span x, y ,the2- dimensional subspace in Rn spanned by x and y. Orthogonality of two vectors{ } is then interpreted to mean (x, y)=0;thezerovectorisorthogonaltoeverybody,bydefinition. These notions of length, distance,andorthogonality do not exist in unadorned vector spaces. We now generalize the notion of inner product to arbitrary vector spaces, even if they are infinite-dimensional. 1.1. Definition. If V is a vector space over K = R or C,aninner product is a map B : V V K taking ordered pairs of vectors to scalars B(v1,v2) K with the following properties× → ∈ 1. Separate Additivity in each Entry. B is additive in each input if the other input is held fixed: B(v + v ,w)=B(v ,w)+B(v ,w) • 1 2 1 2 B(v, w + w )=B(v, w )+B(v, w ). • 1 2 1 2 106 Figure 6.2. Geometric interpretation of the inner product (x, y)=∥x∥∥y∥·cos(θ(x, y)) in Rn.Theprojectedlengthofavectory onto the line L = Rx is ∥y∥·cos(θ). The angle θ(x, y)ismeasuredwithinthetwo-dimensionalsubspaceM = R-span{x, y}.Vectorsare orthogonal when cos θ =0,so(x, y)=0.Thezerovectorisorthogonaltoeverybody. for v, vi,w,wi in V . rm2. Positive Definite. For all v V , | ∈ B(v, v) 0 and B(v, v)=0if and only if v =0 ≥ 3. Hermitian Symetric. For all v, w V , ∈ B(v, w)=B(w, v) when inputs are interchanged. Conjugation does nothing for x R (x = x for x R),soaninnerproductona real vector space is simply symmetric,∈ with B(w, v)=∈ B(v, w). 1. Hermitian. For λ K, v, w V , ∈ ∈ 4.B(λv, w)=λB(v, w) and, B(v, λw)=λ¯B(v, w). • An inner product on a real vector space is just a bilinear map –onethatisR-linear in each input when the other is held fixed – because conjugation does nothing in R. The Euclidean inner product in Rn is a special case of the standard Euclidean inner product in complex coordinate space V = Cn, n (z, w)= zjwj , j=1 " which is easily seen to have properties (1.)–(4.) The corresponding Euclidean norm and distance functions on Cn are then n n 1/2 1/2 1/2 2 2 z =(z, z) = [ zj ] and d(z, w)= z w = [ zj wj ] ∥ ∥ | | ∥ − ∥ | − | j=1 j=1 " " Again, properties (1.) - (4.) are easily verified. For an arbitrary inner product B we define the corresponding norm and distance functions v = B(v, v)1/2 d (v ,v )= v v ∥ ∥B B 1 2 ∥ 1 − 2∥B which are no longer given by such formulas. 1.2. Example. Here are two important examples of inner product spaces. 107 1. On V = Cn (For Rn)wecandefine“nonstandard”innerproductsbyassigning different positive weights αj > 0toeachcoordinatedirection,taking n n 1/2 2 Bα(z, w)= αj zjwj with norm z α = [ αj zj ] · ∥ ∥ ·| | j=1 j=1 " " This is easily seen to be an inner product. Thus the standard Euclidean inner n n product on R or C ,forwhichα1 = ...= αn =1,ispartofamuchlargerfamily. 2. The space [a, b]ofcontinuouscomplex-valuedfunctionsf :[a, b] C becomes an inner productC space if we define → b (f,h)2 = f(t)h(t) dt (Riemann integral) #a The corresponding “L2-norm”ofafunctionisthen b 1/2 2 f 2 = [ f(t) dt ] ; ∥ ∥ | | #a the inner product axioms follow from simple properties of theRiemannintegral. This infinite-dimensional inner product space arises in manyapplications,particu- larly Fourier analysis. ! 1.3. Exercise. Verify that both inner products in the last example actually satisfy 2 the inner product axioms. In particular, explain why the L -inner product (f,h)2 has f > 0whenf is not the zero function (f(t) 0forallt). ∥ ∥2 ≡ We now take up the basic properties common to all inner productspaces. 1.4. Theorem. On any inner product space V the associated norm has the following properties (a) x 0; ∥ ∥≥ (b) λx = λ x (and in particular, x = x ); ∥ ∥ | |·∥ ∥ ∥− ∥ ∥ ∥ (c) (Triangle Inequality) For x, y V , x y x + y . ∈ ∥ ± ∥≤∥ ∥ ∥ ∥ Proof: The first two are obvious. The third is important because it implies that the distance function d (x, y)= x y satisfies the “geometric triangle inequality” B ∥ − ∥ d (x, y) d (x, z)+d (z,y), for all x, y, z V B ≤ B B ∈ as indicated in Figure 6.3. This follows directlly from (3.) because d (x, y)= x y = (x z)+(z y) x z + z y = d (x, z)+d (z,y) B ∥ − ∥ ∥ − − ∥≤∥ − ∥ ∥ − ∥ B B The version of (3.) involving a ( )signfollowsfromthatfeaturinga(+)because v w = v +( w)and w = w−. −The proof− of (3.) is∥− based∥ on∥ an∥ equally important inequality: 1.5. Lemma (Schwartz Inequality). If B is an inner product on a real or complex vector space then B(x, y) x y | |≤∥ ∥B ·∥ ∥B for all x, y V . ∈ 108 Figure 6.3. The meaning of the Triangle Inequality: direct distance from x to y is always ≤ the sum of distances d(x, z)+d(z, y)toanythirdvectorz ∈ V . 2 Proof: For all real t we have φ(t)= x + ty B 0. By the axioms governing B we can rewrite φ(t)as ∥ ∥ ≥ φ(t)=B(x + ty, x + ty) = B(x, x)+B(ty, x)+B(x, ty)+B(ty, ty) = x 2 + tB(x, y)+t B(x, y)+t2 y 2 ∥ ∥B ∥ ∥B = x 2 +2t Re(B(x, y)) + t2 y 2 ∥ ∥B ∥ ∥B because B(tx, y)=tB(x, y)andB(x, ty)=tB(x, y)(sincet R), and z + z =2Re(z)= 2x for z = x + iy in C.Nowφ : R R is a quadratic function∈ whose minimum value → occurs at t0 where dφ (t )=2t y 2 +Re(B(x, y)) = 0 dt 0 0∥ ∥B or Re(B(x, y)) t = − 0 2 y 2 ∥ ∥B Inserting this into φ we find the actual minimum value of φ: x 2 y 2 2 Re(B(x, y)) 2 + Re(B(x, y)) 2 0 min φ(t):t R = ∥ ∥B ·∥ ∥B − | | | | ≤ { ∈ } y 2 ∥ ∥B Thus 0 x 2 y 2 Re(B(x, y)) 2 ≤∥ ∥B ·∥ ∥B −| | which in turn implies Re B(x, y) x y for all x, y V. | |≤∥ ∥B ·∥ ∥B ∈ If we replace x eiθx this does not change x since eiθ = cos(θ)+i sin(θ) =1for real θ;intheinnerproductontheleftwehave)→ ∥ ∥ B(eiθx,| y)=| |eiθB(x, y). We may| now take θ R so that eiθ B(x, y)= B(x, y) .Forthisparticularchoiceofθ we get ∈ · | | 0 Re(B(eiθx, y)) = Re(eiθB(x, y)) ≤| | | | =Re(B(x, y) )= B(x, y) x y . | | | |≤∥ ∥B ·∥ ∥B That proves the Schwartz inequality. ! Proof (Triangle Inequality): The algebra is easier if we prove the (equivalent) in- equality obtained when we square both sides: 2 0 x + y 2 ( x + y ) ≤∥ ∥ ≤ ∥ ∥ ∥ ∥ = x 2 +2 x y + y 2 ∥ ∥ ∥ ∥·∥ ∥ ∥ ∥ 109 In proving the Schwartz inequality we saw that x + y 2 =(x + y, x+ y)= x 2 +2Re(x, y)+ y 2 ∥ ∥ ∥ ∥ ∥ ∥ so our proof is finished if we can show 2 Re(x, y) 2 x y .But ≤ ∥ ∥·∥ ∥ Re(z) Re(z) z for all z C ≤| |≤| | ∈ and then the Schwartz inequality yields Re(B(x, y)) B(x, y) x y ≤| |≤∥ ∥B ·∥ ∥B as desired. ! 1.6. Example. On V =M(n, K)wedefinetheHilbert-Schmidt inner product and norm for matrices: 2 2 (44) (A, B) =Tr(B∗A)and A = a =Tr(A∗A) HS ∥ ∥HS | ij | i,j=1 " It is easily verified that this is an inner product. First note that the trace map from M(n, K) K → n Tr(A)= aii i=1 " is a complex linear map and Tr( A )=Tr(A); then observe that n A 2 =(A, A) = a 2 is > 0unlessA is the zero matrix. ∥ ∥2 HS | ij | i,j=1 " 2 Alternatively, consider what happens when we identify M(n, C) = Cn as complex vector ∼ 2 spaces. The Hilbert-Schmidt norm becomes the usual Euclidean norm on Cn ,and likewise for the inner products; obviously (A, B)HS is then an inner product on matrix space. The norm A HS and the sup-norm A discussed in Chapter V are different ways to measure the∥ “size”∥ of a matrix; the HS-norm∥ ∥∞ turns out to be particularly well adapted to applications in statistics, starting with “least-squares regression” and moving on into “analysis of variance.” Each of these norms determines a notion of matrix convergence A A as n in M(N,C). n → →∞ 1/2 (n) 2 2-Convergence: An A HS = [ a aij ] 0asn ∥·∥ ∥ − ∥ | ij − | → →∞ i,j " (n) -Convergence: An A =max aij aij 0asn ∥·∥∞ ∥ − ∥∞ i,j {| − |} → →∞ However, despite their differences both norms determine the same notion of matrix con- vergence.
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