Chapter 5: Complex Inner Product Spaces

Transcription

1 Chapter 5: Complex Inner Product Spaces Let us recall the inner product for the real) n dimensional Euclidean space R n : for vectors x = x 1,x,...,x n ) and y = y 1,y,...,y n ) in R n, the inner product x,y is defined to be x,y = x 1 y 1 +x y + +x n y n, and the norm or the magnitude) x is given by x = x,x = x 1 +x + +x n. For complex vectors, we cannot copy this definition directly. We need to use complex conjugation to this definition in such a way that x,x so that the definition of magnitude x = x,x still makes sense. Denoteby C n thecomplexvectorspaceconsistingofalln tuples z = z 1,z,...,z n ). The addition and the scalar multiplication of vectors in C n are defined in the usual pointwise fashion: for vectors x = x 1,x,...,x n ) and y = y 1,y,...,y n ) in C n, and scalars a C, x+y = x 1 +y 1, x +y,..., x n +y n ) and ax = ax 1, ax,..., ax n ) The inner product or the scalar product) x,y of vectors x and y is defined via x,y = x 1 y 1 +x y + +x n y n 5.1) Notice that x,x = x 1 x 1 +x x + +x n x n = x 1 + x + + x n, which is what we ask for. The norm of x is given by x = x,x 1/ = x 1 + x + + x n. Remark: In 5.1), 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 C n provides us with the typical example of complex inner product spaces, defined as follows: 1

2 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, and x,x = if and only if x =. C) y,x = x,y. C3) The inner product is a sesquelinear map, i.e. a 1 x 1 +a x, y = a 1 x 1,y +a x,y x, b 1 y 1 +b y = b 1 x,y 1 +b x,y. Actually the second identity of C3) above is the consequence of the first, together with C). Besides C n, another example of complex inner product space is given as follows. Consider a space F of well-behaved complex-valued functions over an interval, say [a, b]; here we do not specify the technical meaning of being well-behaved). The inner product f, g of f, g F is given by f,g = 1 b a b a ft)gt)dt, for f,g F. 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 f f,f 1/ 1 = b a b a ft) dt) 1/ for f F. In the future we will take F to be the space of trigonometric polynomials and [a,b] is any interval of length π, such as [, π] and [ π, π]. Now we return to the general discussion of 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 =. From the definition of orthogonality you should recognize that, first, the zero vector is orthogonal to every vector; second, is the only vector orthogonal to itself and hence is the only vector orthogonal to every vector; third, x y implies y x.

3 A set of nonzero vectors S is called an orthogonal system if each vector in S is orthogonal to all other vectors in S. If, furthermore, each vector in S has length 1, then S is called an orthonormal system. Notice the difference of the endings of the words orthogonal and orthonormal.) We have the following generalized Pythagoras theorem: If v 1,v,,v n are an orthogonal system, then v 1 +v + +v n = v 1 + v + + v n. 5.) We prove this by induction on n. When n = 1, 5.) becomes v 1 = v 1 and there is nothing to prove. So let us assume that n and assume that the theorem is true for n 1 vectors. Let w = v +v 3 + +v n. Then, by our induction hypothesis, w = n k= v k. Thus 5.) becomes v 1 +w = v 1 + w which remains to be verified. Notice that v 1,w = w,v 1 = v,v 1 + v 3,v v n,v 1 =. Hence v 1 +w = v 1 +w,v 1 +w = v 1,v 1 + v 1,w + w,v 1 + w,w = v 1,v 1 + w,w = v 1 + w. Given an orthonormal system E = {e 1,e,...,e n } in V, and a vector v, which can be written as a linear combination of vectors in B, say v = v 1 e 1 +v e + +v n e n n v ke k. k= 1 we look for an explicit expression for the coefficients v k in this linear combination. By the linearity of the first slot of inner product, we have n v,e j = v ke k, e j = n v k e k,e j. k= 1 k= 1 Note that e k,e j are zeros except when k = j, which gives 1 in this case; in short, e k,e j = δ jk ). Hence we have v,e j = v j. Thus v = v,e 1 e 1 + v,e e + + v,e n e n, 5.3) Since v,e k e k = v,e k e k = v,e k, the generalized Pythagoras theorem gives v = v,e 1 + v,e + + v,e n, 5.4) if v is in the linear span of the orthonormal system E = {e 1,e,...,e n }. 3

4 Next weconsider aslightly moregeneral problem: given a vectorvinan innerproduct space V and a subspace W, spanned by an orthogonal system S = {w 1,w,...,w r } w k,w j = for k j and w k,w k, where k and j run between 1 and r) find the so-called orthogonal decomposition of v: v = w+h, 5.5) where w W and h W. The vector w here will be called the orthogonal) projection of v onto W. Since w is in W and since W is spanned by w 1,w,...,w r, we can write w = a 1 w 1 +a w + +a r w r. 5.6) We have to find a 1, a,..., a r. Identity 5.5) can be rewritten as v = w+h = r k= 1 a kw k +h. Take any vector from w 1,w,...,w r, say w j, and form the inner product of w j with each side of the above identity. By the linearity of the first slot of inner product, we have v,w j = r k= 1 a k w k,w j + h,w j. Note that w k,w j are zeros except when k = j. Hence r k= 1 a k w k,w j can be reduced to a j w j,w j. On the other hand, h,w j = because h is perpendicular to W and w j is in W. Thus we arrive at v,w j = a j w j,w j, or a j = v,w j / w j,w j. Substitute this expression of a j to D1), switching the index j to k, to obtain: w = r k= 1 v, w k w k,w k w k v, w 1 w 1,w 1 w v, w r w r,w r w r, 5.7) which is the required projection. Nowweconsidertwospecialcases: ThefirstcaseisthatS consistsofasinglenonzero) vector, say u. Write down the orthogonal decomposition v = v,u u+h where h u. u,u The generalized Pythagoras theorem gives v = v,u / u,u u + h. Using u,u = u, we rewrite the first term on the right-hand side as v,u / u. Then we show our generosity by dropping the second term h on the right to obtain the inequality v v,u / u. We can rearrange this into v,u v u, 4

5 which is the celebrated Cauchy Schwarz s inequality. The second special case is that S consists of an orthonormal system, say S = {e 1,e,...,e n }. In this case w = n k= 1 v,e k e k with w = n k= 1 v,e k. The orthogonal decomposition v = w +h tells us that v = w + h. Dropping h, we get v w, or w v. We have arrived at n This is called Bessel s inequality. k= 1 v,e k v. In the rest of this chapter, we give some important examples of orthonormal systems. Example 5.1. In C n, the vectors e 1 = 1,,,...,, ) e =, 1,,...,, ). e n =,,,...,, 1) clearly form an orthonormal basis, usually called the standard basis of C n. Example 5.. As usual, write ω = e πi/n. Consider the following vectors in C n : f k = 1 n 1, ω k 1, ω k 1),..., ω n 1)k 1)) ; 1 k n. We write down the first three of them to see the general pattern: f 1 = 1, 1, 1,..., 1)/ n, f = 1, ω, ω,..., ω n 1 )/ n, f 3 = 1, ω, ω 4,..., ω n 1) )/ n, We claim that f k 1 k n) form an orthonormal basis in C n. First we check that they are uni vectors: f k = 1 n 1 + ω k 1 + ω k 1) + + ω n 1)k 1) ) = 1. 5

6 Next we show that, for k l, f k,f l =. For definiteness, let us assume 1 l < k n. By using ω = ω 1, we get f k,f l = 1+ω k 1 ω l 1 +ω k 1) ω l 1) + +ω k 1)n 1) ω l 1)n 1)) /n = 1+ω k l +ω k l + +ω k l)n 1)) /n where η = ω k l. Now = 1+η +η + +η n 1 )/n, 1 η)1+η +η + +η n 1 ) = 1 η n = 1 ω k l)n = 1 ω n ) k l = 1 1 =. Since < k l < n, η ω k l 1, or 1 η. Hence 1+η+η + +η n 1 =. Now f k,f l = is clear. This example will be referred to in the next chapter when we discuss the finite Fourier transform in Example 6.3. Example 5.3. Consider the space of all periodic functions of period π. The inner product of two such functions f and g is defined to be f,g = 1 π π ft)gt)dt. We claim that the system e int < n < ), where n ranges over all integers, is orthonormal. First we check that each of them is of unit length: Next, for n m, we have e int == 1 π π e int dt = 1 π π 1dt = 1. e int,e imt = 1 π π e int e imt dt = 1 π π e in m)t dt =. The orthogonal decomposition of a function f in this space gives its Fourier series: where Bessel s inequality in this case says ft) = c n = f, e int = 1 π <n< <n< π c n 1 π π 6 c n e int, ft)e int dt. ft) dt,

7 showing that the infinite sum on the left hand side is a finite number. This prompts many mathematicians to say: the sequence of Fourier coefficients is square summable. We will present some basic theory of Fourier series in Chapter 8. Example 5.4. Consider the space of all even functions of period π. The inner product of two such functions f and g is defined to be f,g = 1 π π ft)gt)dt. Then the following functions 1, cost, cost, cos3t,... form an orthonormal system of this space. To show this, we need to check π π cosmt cosnt dt = δ mn = { 1 if m = n if m n 5.8) which is left to the reader as an exercise. Example 5.5. Recall that the Chebyshev s polynomials T n x) is defined via the identity cosnt = T n cost). The orthogonality relation 5.8) above gives π π T m cost) T n cost) dt = δ mn Now apply the change of variable x = cost. Notice that cos = 1, cosπ = 1 and dx = sintdt, which gives dt = dx/sint = dx/ 1 cos = dx/ 1 x ; notice that, for t π, we have sint ) Thus we have π 1 1 T m x)t n x) dx 1 x = δ mn. This shows that if we define the inner product of two polynomial functions f and g by f,g = π 1 1 fx)gx) dx 1 x then the Chebyshev s polynomials T n n = 1,,3,...) form an orthonormal system. 7