An Inner Product on a Vector Space V Is Rule for Combining Two Vectors F,G of V

Math 217 §5.5 Professor Karen Smith

Definition: An inner product on a vector space V is rule for combining two vectors f, g of V into a scalar hf, gi satisfying the following axioms:

1. hf, gi = hg, fi for all f, g ∈ V (symmetric property)

1 2. hf1 + f2, gi = hf1, gi + hf2, gi for all f1, f2, g ∈ V (linearity in ﬁrst argument ) 3. hcf, gi = chf, gi for all f, g ∈ V and scalars c (linearity in ﬁrst argument2)

4. hf, fi ≥ 0 for all f ∈ V with hf, fi = 0 if and only if f = 0 (positive deﬁnite property).

A vector space equipped with an inner product is called an inner product space.

Task: Your group is given one vector space, together with a rule for combining two vectors into a scalar. Investigate the following questions.

1. Is it an inner product space? Prove it, or adapt it to make it an inner product space.

2. Pick any non-zero vector in your inner product space. Find its length.

3. Find a pair of orthonormal vectors ~u1, ~u2 in your space. Let W be the space that they span. What is dim W ?

2 4. Compute the distance between ~u1 and ~u2 in your inner product. Also compute ||~u1 − ~u2|| . 5. What is the dimension of W ⊥? Find an element ~z in W ⊥. What is the closest vector in W to ~z? What is the projection of ~z onto W ?

6. Find an element ~q not in W ∪ W ⊥. Compute the projection of ~q to W . What is the closest vector in W to ~q?

Z 1 A. Let C0 be the space of continuous functions. Let hf, gi be deﬁned by fg dx. −1 2×2 T B. Let R be the space of 2 × 2 matrices. Deﬁne hA, Bi by trace (A B).

3 T C. Let V be R . Deﬁne h~x,~yi as the matrix product ~x A ~y where A is a diagonal matrix with 1 0 0 positive entries on the diagonal. For problems 2-6, take A = 0 2 0. 0 0 3

D. Let P2 be the space of polynomials of degree ≤ 2. Deﬁne hf, gi by f(−1)g(−1) + f(0)g(0) + f(1)g(1).

1 We also have linearity in the second argument: hf, g1 + g2i = hf, g1i + hf, g2i ∀f, g1, g2 ∈ V but this follows from symmetry. 2We also have linearity in the second argument: hf, cgi = chf, gi ∀f, g ∈ V and scalars c but this follows from symmetry. Solution note: A. 1). Yes, this is an inner product. We check the four properties: Z 1 Z 1 1. Symmetry: hf, gi = hg, fi because fg dx = gf dx. −1 −1 Z 1 Z 1 Z 1 2. Linearity: hf1 + f2, gi = hf1, gi + hf2, gi because (f1 + f2)g dx = f1g dx + f2g dx. −1 −1 −1 Z 1 Z 1 3. Linearity: hcf, gi = chf, gi because (cf)g dx = c fg dx. −1 −1 Z 1 4. Positive definite: hf, fi = f 2 dx ≥ 0 because f 2 ≥ 0 everywhere on the interval [−1, 1]. −1 The only way for this integral to be zero is for f = 0—this uses the fact that f is CONTIN- UOUS, so there must be some interval where f is strictly positive if it is non-zero. You do not have to prove this fact—you will learn how to do so in Math 451. s Z 1 √ 2). ||1|| = 1 dx = 2. −1 Z 1 3). Let f = 1 and g = x. Then fg dx = 0 since x is odd. So this is almost an −1 1/2 orthonormal basis: we need only scale each by its length. Compute√ ||x|| = hx, xi = 1 √ R 2 1/2 1/2 x √3 x [ −1 x dx] = [2/3] . So let ~u1 = 1/ 2 and ~u2 = ||x|| = . The dimension of √ 2 1 3 x W is 2. It has orthonormal basis is (~u1, ~u2) = ( √ , √ ). √ 2 2 4). The distance between is 2. 5). The dimension of W ⊥ is (very!) infinite. To find a vector z ∈ W ⊥, we need a R 1 R 1 function z satisfying −1 zdx = 0 and −1 xzdx = 0. One way to do this is to take ANY vector, then subtract its projection to W . Take g = 3x2. The component in √ 1 √ √1 R 2 √2 the ~u1 direction has length hg, 1/ 2i = −1 3x dx = = 2. The length of the 2 √ 2 3 R 1 3 3 component in the ~u2 direction has hg, ~u2i = √ 3x dx = 0 since x is odd. So √ 2 −1 || ⊥ 2 ⊥ g = 2~u1 + 0~u2 = 1. So g = g − 1 = 3x − 1 is vector in W . For any vector in W ⊥, including g = 3x2 − 1, the projection to W is zero. So the closest vector in W is zero as well. 6). For q, we can take the function g, it is neither in W nor W ⊥. Its projection onto W is hg, ~u i~u + hg, ~u i~u = [R 1 g~u dx]~u + [R 1 g~u dx]~u . In this case, for g = 3x2, √ 1 1 2 2 −1 1 1 −1 2 2 we get 2~u1 + 0~u2 = 1. Solution note: B. 1.) This is an inner product. Check the properties. T T Symmetry: trace(A B) = trace(B A) because if we write A = [~a1 ~a2] and B = T ~ ~ T ~a1 ~ ~ ~ ~ [b1 b2], then A B = T [b1 b2] which has diagonal entries ~a1 · b1 and ~a2 · b2. Simi- ~a2 T larly, B A has diagonal entries ~a1 ·~b1 and ~a2 ·~b2, so the trace is the same. T T T T T Linearity: (A1 + A2) B = (A1 + A2 )B = A1 B + A2 B, so these have the same trace. SImilarly, (cA)T B = cAT B so the traces are the same. T 2 2 Positive definite: A A has trace ~a1 · ~a1 + ~a2 · ~a2 = ||~a1|| + ||~a2|| > 0 unless both ~a1 and ~a2 = 0.√ 2. ||I2|| = 2. 3. E11,E22 is orthonormal. W has dimension 2. 0 1 √ 2). The length of is p(trace(AT A)) = p(trace(I )) = 2. 1 0 2 3). Let vectors E11 and E22 are orthonormal, as you can easily check by computing hE11,E11i = 1, hE11,E22i = 0 and hE22,E22i = 1. The space they span has dimension 2. s 1 0 √ 4). ||u − u || = trace( )2 = 2. 1 2 0 −1 ⊥ 5). The element E12 is in W . It’s projection to W is zero. Thus zero is the closest ⊥ vector in W to E12. The dimension of W is the dimension of the ambient space 2×2 R minus the dimension of W , or 4 − 2 = 2. 0 1 0 0 6). Let ~q = . The projection to W is . 1 1 0 1

Solution note: C. 1. For this we need A to have strictly positive entries on the diagonal. Otherwise, it is not an inner product because positive deﬁnitivity fails. Symmetry: ~xT A~y = ~yT A~x because transposing, (~yT A~x)T = ~xT AT~(yT )T = ~xT A~y. Linearity follows from basic properties of matrix arithmetic. T 2 2 2 T Positive Deﬁnite: ~x A~x = x1d1 + x2d2 + . . . xndn where ~x = [x1 x2 . . . xn] . This is non-negative because the di are all positive. It is 0 if and only if all xi = 0.   1 0 0 √ T 1/2 2. ||~e3|| = (~e3 0 2 0~e3) = 3. 0 0 3 3. ~e , √1 ~e . W has dimension 2. 1 2 2 √ 4). The distance between is 2.

⊥ ⊥ 5). The vector ~e3 ∈ W . For any vector in W , the projection to W is zero. So the closest vector in W is zero as well. W ⊥ has dimension 1, since dim W + dim W ⊥ = 3, 3 which is the dimension of the ambient space R . 6). Let q = I2. Then the projection onto W is E11. Solution note: D.

1. This is obviously symmetric: hf, gi = hg, fi. To check linearlity: hf + h, gi = (f + h)(−1)g(−1) + (f + h)(0)g(0) + (f + h)(1)g(1) = f(−1)g(−1) + f(0)g(0) + f(1)g(1) + h(−1)g(−1) + h(0)g(0) + h(1)g(1) = hf, gi + hh, gi. To check positive deﬁniteness, compute: hf, fi = (f(−1))2 + (f(0)2 + (f(1))20, which is non-negative, since it is a sum of squares. If hf, fi = 0 for some f = a + bx + cx2, then hf, fi = (c − b + a)2 + (a)+(a + b + c)2 = 0. Since squares are all positive, we conclude that (c − b + a) = a = (a + b + c). This means that a = 0, b + c = 0 and c − b = 0, so a = b − c = 0. So it is positive deﬁnite. √ 2. The length of the polynomial x is hx, xi1/2 = ((−1)2 + 02 + (1)2)1/2 = 2.

3. Note that h1, xi = (1)(−1) + (1)(0) + (1)(−1)2 = 0, so that 1 and x are perpendicular. So we can get√ an orthonormal basis by scaling each to be a unit vector. We already checked that ||x|| = 2, so √x is a unit vector. Note also h1, 1i = 12 + 12 + 1=3, so √1 is a unit vector. 2 3 Our orthonormal basis is ( √1 , √x ). 3 2

1/2 4. The distance between√ ~u1 and ~u1 is ||~u1 − ~u2|| = h~u1 − ~u2, ~u1 − ~u2i. Do the computation to get the distance is 2.

5. Since W has dimension 2, W ⊥ has complementary dimension, which is 1 here, because the 2 2 ⊥ ambient space P2 is dimension 3. The vector z = x − 3 is in W , which you can check by computing that both hu1, zi and hu2, zi are zero. I found this by picking any vector (I picked x2) and computing the projection onto W , then subtracting to get something in W ⊥.

6. The vector x2 is not in W ∪W ⊥. Its projection to W is hx2, ~u i~u +hg, ~u i~u = √2 ~u +0~u = 2 . 1 1 2 2 3 1 2 3