CHAPTER IV OPERATORS ON INNER PRODUCT SPACES 1. Complex Inner Product Spaces § 1.1. Let us recall the inner product (or the dot product) for the real n–dimensional n n Euclidean space R : for vectors x = (x1, x2, . , xn) and y = (y1, y2, . , yn) in R , the inner product x, y (denoted by x y in some books) is defined to be · x, y = x y + x y + + x y , 1 1 2 2 ··· n n and the norm (or the magnitude) x is given by x = x, x = x2 + x2 + + x2 . 1 2 ··· n For complex vectors, we cannot copy this definition directly. We need to use complex conjugation to modify this definition in such a way that x, x 0 so that the definition ≥ of magnitude x = x, x still makes sense. Recall that the conjugate of a complex number z = a + ib, where x and y are real, is given by z = a ib, and − z z = (a ib)(a + ib) = a2 + b2 = z 2. − | | The identity z z = z 2 turns out to be very useful and should be kept in mind. | | Recall that the addition and the scalar multiplication of vectors in Cn are defined n as follows: for x = (x1, x2, . , xn) and y = (y1, y2, . , yn) in C , and a in C, x + y = (x1 + y1, x2 + y2, . , xn + yn) and ax = (ax1, ax2, . , axn) The inner product (or the scalar product) x, y of vectors x and y is defined by x, y = x y + x y + + x y (1.1.1) 1 1 2 2 ······ n n Notice that x, x = x x +x x + +x x = x 2+ x 2+ + x 2 0, which is what 1 1 2 2 ··· n n | 1| | 2| ··· | n| ≥ we ask for. The norm of x is given by x = x, x 1/2 = x 2 + x 2 + + x 2. | 1| | 2| ··· | n| 1 Remark: In (1.1.1), it is not clear why we prefer to take complex conjugates of components of y instead of components of x. Actually this is more or less due to the tradition of mathematics, rather than our preference. (Physicists have a different tradition!) The space Cn provides us with the typical example of complex inner product spaces, defined as follows: Definition. By an inner product on a complex vector space we mean a device of assigning to each pair of vectors x and y a complex number denoted by x, y , such that the following conditions are satisfied: (C1) x, y 0, and x, x = 0 if and only if x = 0. ≥ (C2) y, x = x, y . (C3) The inner product is a “sesquelinear map”, i.e. a x + a x , y = a x , y + a x , y 1 1 2 2 1 1 2 2 x, b y + b y = b x, y + b x, y . 1 1 2 2 1 1 2 2 (Actually the second identity of (C3) above is the consequence of the first, to- gether with (C2). Inner products for real vector spaces can be defined in the similar fashion. It is slightly simpler because there is no need to take complex conjugation. This is simply because the conjugate of a real number is just itself. Besides Cn, another example of complex inner product space is given as follows. Consider a space of well-behaved complex-valued functions over an interval, say [a, b]; F (here we do not specify the technical meaning of being well-behaved). The inner product f, g of f, g is given by ∈ F 1 b f, g = f(t) g(t) dt, for f, g . b a ∈ F − a (On the right hand side, 1/(b a) is a normalization factor added for convenience in the − future.) The norm induced by this inner product is 1/2 1 b f f, f 1/2 = f(t) 2 dt for f . ≡ b a | | ∈ F − a In the future we will take to be the space of trigonometric polynomials and [a, b] is F any interval of length 2π, such as [0, 2π] and [ π, π]. − 2 1.2. Let V be a complex vector space V with an inner product , . We say that · · two vectors x and y in V are orthogonal or perpendicular if their inner product is zero and we write x y in this case. Thus, by our definition here, ⊥ x y x, y = 0. ⊥ ⇐⇒ From the definition of orthogonality you should recognize that, first, the zero vector 0 is orthogonal to every vector (indeed, for each vector x in V , 0, x = 0 + 0, x = 0, x + 0, x by (C3) and hence 0, x = 0); second, 0 is the only vector orthogonal to itself (this follows from (C1)) and hence 0 is the only vector orthogonal to every vector; third, x y implies y x (indeed, if x, y = 0, then y, x = x, y = 0 = 0). ⊥ ⊥ A set of nonzero vectors is called an orthogonal system if each vector in is S S orthogonal to all other vectors in . If, furthermore, each vector in has length 1, S S then is called an orthonormal system. (Notice the difference of the endings of the S words “orthogonal” and “orthonormal”.) We have the following generalized Pythagoras theorem: If v , v , , v are an orthogonal system, then 1 2 ··· n v + v + + v 2 = v 2 + v 2 + + v 2. (1.2.1) 1 2 ··· n 1 2 ··· n We prove this by induction on n. When n = 1, (5.2) becomes v 2 = v 2 and there is 1 1 nothing to prove. So let n 2 and assume that the theorem is true for n 1 vectors. Let ≥ − w = v + v + + v . Then, by our induction hypothesis, w 2 = n v 2. Thus 2 3 ··· n k= 2 k (1.2.1) becomes v + w 2 = v 2 + w 2 which remains to be verified. Notice that 1 1 v , w = v , v + v , v + + v , v = 0. 1 1 2 1 3 ······ 1 n Hence v + w 2 = v + w, v + w 1 1 1 = v , v + v , w + w, v + w, w 1 1 1 1 = v , v + v , w + v , w + w, w 1 1 1 1 = v , v + w, w = v 2 + w 2. 1 1 1 Hence (1.2.1) is valid. Given an orthogonal system = e , e ,..., e in V , and a vector v which can E { 1 2 n} be written as a linear combination of vectors in , say B n v = v1e1 + v2e2 + + vnen vkek, ··· ≡ k= 1 we look for an explicit expression for the coefficients vk in this linear combination. By the linearity in the “first slot” of the inner product, we have n n v, ej = vkek, ej = vk ek, ej . k= 1 k= 1 3 Note that e , e are zeros except when k = j, which gives 1 in this case; (in short, k j e , e = δ ). So the above identity becomes v, e = v . Thus k j jk j j n v = v, ek ek = v, e1 e1 + v, e2 e2 + + v, en en, (1.2.2) k= 1 ··· Since v, e e = v, e e = v, e , the generalized Pythagoras theorem gives k k | k| k | k| v 2 = v, e 2 + v, e 2 + + v, e 2, (1.2.3) | 1| | 2| ······ | n| if v is in the linear span of the orthonormal system = e , e ,..., e . The last E { 1 2 n} identity is a general fact about orthonormal system that should be kept in mind. 1.3. Next we consider a slightly more general problem: given a vector v in an inner product space V and a subspace W of V , spanned by a given orthogonal system S = w , w ,..., w of nonzero vectors ( w , w = 0 for k = j and w , w = 0, where k { 1 2 r} k j k k and j run between 1 and r), find the so-called orthogonal decomposition of v: v = w + h, (1.3.1) where w W and h W (that is, h is perpendicular to all vectors in W ). The vector w ∈ ⊥ here will be called the (orthogonal) projection of v onto W . Since w is in W and W is spanned by w1, w2,..., wr, we can write w = a w + a w + + a w . (1.3.2) 1 1 2 2 ··· r r We have to find a1, a2,..., ar. Identity (1.3.1) can be rewritten as r v = w + h = akwk + h. k= 1 Take any vector from w1, w2,..., wr, say wj, and form the inner product of wj with each side of the above identity. By the linearity of the “first slot” of inner product, we have v, w = r a w , w + h, w . Note that w , w are zeros except when k = j. j k= 1 k k j j k j Hence r a w , w can be reduced to a w , w . On the other hand, h, w = 0 k= 1 k k j j j j j because h is perpendicular to W and w is in W . Thus we arrive at v, w = a w , w , j j j j j or a = v, w / w , w . Substitute this expression of a to (1.3.2), switching the index j j j j j j to k, to obtain: r v, wk v, w1 v, w2 v, wr w = wk w1 + w2 + + wr, (1.3.3) k= 1 wk, wk ≡ w1, w1 w2, w2 ··· wr, wr which is the required projection.