CHAPTER 5. INNER PRODUCT SPACES AND LINEAR OPERATORS

MODULE 1. INNER PRODUCT SPACES

INDRANATH SENGUPTA

Contents 1. Inner products and Norms 1 2. Cauchy-Schwarz inequality and Triangle inequality 3 3. Metric 4

We assume that the field F is either R or C. Let V be a vector space over F.

1. Inner products and Norms Definition 1. An inner product on V is a map h , i : V × V → F, such that the following properties are satisfied for every x, y, z ∈ V and c ∈ F: (i) hx + y, zi = hx, zi + hy, zi; (ii) hcx, zi = chx, zi; (iii) hx, zi = hz, xi; (iv) hx, xi > 0, if x 6= 0.

Remarks. (1) If F = R, then hx, zi = hx, zi = hz, xi. (2) If h , i is an inner product on V , then r · h , i is also an inner product for every r > 0. (3) For a fixed z ∈ V , the function ϕ = h−, zi : V → F defined as ϕ(x) = hx, zi is a linear map over F. If F = R, then it is also linear with respect to the other variable. In other words, if F = R, then h−, −i : V × V → F is bilinear over F.

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n Example 1.1. Given x = (x1, . . . , xn) and y = (y1, . . . , yn) in F , we define hx, yi = x1y1 + x2y2 + ··· + xnyn. This is called the usual inner product or the dot product of vectors in Rn or Cn.

Example 1.2. Let V = C[0, 1], the vector space of all real-valued R 1 continuous functions on [0, 1]. Define hf, gi = 0 f(t)g(t)dt.

Example 1.3. Let V = C[0, 2π]C denote the vector space of all complex- 1 R 2π valued continuous functions on [0, 2π]. Define hf, gi = 2π 0 f(t)g(t)dt.

Definition 2. Let A = (aij) be an n × n matrix over F. The adjoint ∗ t of A is defined to be the matrix A = (aij) . This is nothing but the conjugate transpose of A.

 i 0 −i 1 − i Example 1.4. if A = , then A∗ = . 1 + i 1 0 1

Example 1.5. V = Mn(F) is an inner product space with respect to the inner product hA, Bi = trace(B∗A). We observe the following identity which is useful: n n n n n ∗ X ∗ X X ∗ X X 2 hA, Ai = trace(A A) = (A A)ii = (A )ikAki = |Aki| . i=1 i=1 k=1 i=1 k=1

Definition 3. The pair (V, h , i) is called an inner product space over the field F if h , i is an inner product on V . We often hide h , i and call V an inner product space to mean that there is an inner product h , i defined on V . The vector space V is called a real inner product space if F = R and a complex inner product space if F = C.

It is easy to check that h , i is an inner product in all the examples discussed above.

Theorem 1.6. Let V be an inner product space over F. The following statements are true for every x, y, z ∈ V and c ∈ F. (i) hx, y + zi = hx, yi + hx, zi; Chapter 5. Inner Product Spaces and Linear Operators 3

(ii) hx, cyi = chx, yi; (iii) hx, 0i = h0, xi = 0; (iv) hx, xi = 0 if and only if x = 0; (v) if hx, yi = hx, zi for all x ∈ V , then y = z. Proof. Follows from the definition of h , i.  Definition 4. Let V be an inner product space over F. For every x in V , we define the norm of x to be ||x|| = phx, xi.

n n Example 1.7. If V = R and x = (x1, . . . , xn) ∈ R , then 1 p 2 2
2 ||x|| = hx, xi = |x1| + ··· + |xn| ; which is the usual length of the vector x or the Euclidean distance of the vector x from the origin (0,..., 0). The following proposition is easy to verify using the definition of norm.

Proposition 1.8. For every x ∈ V and c ∈ F, (i) ||c · x|| = |c| · ||x||; (ii) ||x|| = 0 if and only if x = 0

2. Cauchy-Schwarz inequality and Triangle inequality Theorem 2.1. Let V be an inner product space over F. (i) (Cauchy-Schwarz Inequality) |hx, yi| ≤ ||x|| · ||y||, for all x, y ∈ V . (ii) (Triangle Inequality) ||x+y|| ≤ ||x||+||y||, for all x, y ∈ V .

(iii) |||x|| − ||y||| ≤ ||x − y||, for all x, y ∈ V , for all x, y ∈ V . Proof. (i) If y = 0 then the statement is true. We therefore assume that y 6= 0. Hence, hy, yi 6= 0. We have hx, yi hx, yi hx, yi 0 ≤ ||x − y||2 = hx − y, x − yi hy, yi hy, yi hy, yi hx, yi hx, yi hx, yi hx, yi = hx, xi − hx, yi − hy, xi + hy, yi hy, yi hy, yi hy, yi hy, yi |hx, yi|2 = ||x||2 − . ||y||2 4 Module 1

(ii) ||x + y||2 ≤ hx + y, x + yi = hx, xi + hy, xi + hx, yi + hy, yi = ||x||2 + 2Rehx, yi + ||y||2 ≤ ||x||2 + 2|hx, yi| + ||y||2 ≤ ||x||2 + 2||x|| · ||y|| + ||y||2 = (||x|| + ||y||)2. (iii) ||x|| = ||(x − y) + y|| ≤ ||x − y|| + ||y|| by the triangle inequality. Therefore, ||x||−||y|| ≤ ||x−y||. Interchanging x and y in the inequality we get ||y||−||x|| ≤ ||y−x||. The proof follows from these observations. 

Remark 2.2. It can be proved easily that |hx, yi| = ||x|| · ||y|| if and only y = λx for some λ ∈ F. We also not that in case of R2, the Cauchy-Schwarz inequality is easy to prove because |hx, yi| = | ||x|| ||y|| cos θ | = ||x|| ||y|| | cos θ| ≤ ||x|| ||y||.

3. Metric Definition 5. Let X be a non-empty set. A metric on X is a function d : X × X → R such that (i) d(x, y) ≥ 0 for every x, y ∈ X. (ii) d(x, y) = 0 if and only if x = y. (iii) d(x, y) = d(y, x) for every x, y ∈ X. (iv) (Triangle inequality) d(x, z) ≤ d(x, y)+d(y, z), for all x, y, z ∈ X.

Theorem 3.1. Let V be an inner product space over R. Then, d(x, y) = ||x − y||, for x, y ∈ V defines a metric on V . Chapter 5. Inner Product Spaces and Linear Operators 5

Proof. Properties (i) - (iii) follow from the definition. We prove the triangle inequality for d. d(x, z) = ||x − z|| = ||(x − y) + (y − z)|| ≤ ||(x − y)|| + ||(y − z)|| = d(x, y) + d(y, z).