Math 304–504
Linear Algebra
Lecture 27:
Inner product spaces.
Norm
The notion of norm generalizes the notion of length
of a vector in Rn .
Definition. Let V be a vector space. A function
α : V → R is called a norm on V if it has the
following properties:
(i) α(x) ≥ 0, α(x) = 0 only for x = 0 (positivity)
(ii) α(r x) = |r | α(x) for all r ∈ R (homogeneity)
(iii) α(x + y) ≤ α(x) + α(y) (triangle inequality)
Notation. The norm of a vector x ∈ V is usually
denoted kxk. Different norms on V are
distinguished by subscripts, e.g., kxk1 and kxk2.
Examples. V = Rn , x = (x1, x2, . . . , xn ) ∈ Rn .
• kxk∞ = max(|x1|, |x2|, . . . , |xn |).
• kxkp = |x1|p + |x2|p + · · · + |xn |p
In particular, kxk2 = |x|.
1/p
Examples. V = C [a, b], f : [a, b] → R.
• kf k∞ = max |f (x)|.
a≤x≤b
• kf kp =
Z
a
b
|f (x)|p dx
1/p
, p ≥ 1.
, p ≥ 1.
Inner product
The notion of inner product generalizes the notion
of dot product of vectors in Rn .
Definition. Let V be a vector space. A function
β : V × V → R, usually denoted β(x, y) = hx, yi, is
called an inner product on V if it is positive,
symmetric, and bilinear. That is, if
(i) hx, yi ≥ 0, hx, xi = 0 only for x = 0 (positivity)
(ii) hx, yi = hy, xi
(symmetry)
(iii) hr x, yi = r hx, yi
(homogeneity)
(iv) hx + y, zi = hx, zi + hy, zi (distributive law)
An inner product space is a vector space endowed
with an inner product.
Examples. V = Rn .
• hx, yi = x · y = x1y1 + x2y2 + · · · + xn yn .
• hx, yi = d1 x1y1 + d2x2 y2 + · · · + dn xn yn ,
where d1, d2 , . . . , dn > 0.
Example. V = Pn , polynomials of degree < n.
• hp, qi = p(x1)q(x1) + p(x2)q(x2) + · · · + p(xn )q(xn ),
where x1, x2, . . . , xn are distinct points on R.
We have hp, pi = 0 =⇒ p = 0 since a nonzero
polynomial of degree less than n cannot have n
roots.
Examples. V = C [a, b].
Z b
• hf , g i =
f (x)g (x) dx.
a
• hf , g i =
Z
b
f (x)g (x)w (x) dx,
a
where w is bounded, piecewise continuous, and
w > 0 everywhere on [a, b].
w is called the weight function.
Theorem Suppose hx, yi is an inner product on a
vector space V . Then
hx, yi2 ≤ hx, xihy, yi for all x, y ∈ V .
Proof: For any t ∈ R let vt = x + ty. Then
hvt , vt i = hx, xi + 2thx, yi + t 2 hy, yi.
The right-hand side is a quadratic polynomial in t
(provided that y 6= 0). Since hvt , vt i ≥ 0 for all t,
the discriminant D is nonpositive. But
D = 4hx, yi2 − 4hx, xihy, yi.
Cauchy-Schwarz Inequality:
p
p
|hx, yi| ≤ hx, xi hy, yi.
Cauchy-Schwarz Inequality:
p
p
|hx, yi| ≤ hx, xi hy, yi.
Corollary 1 |x · y| ≤ |x| |y| for all x, y ∈ Rn .
Equivalently, for all xi , yi ∈ R,
(x1y1 + · · · + xn yn )2 ≤ (x12 + · · · + xn2)(y12 + · · · + yn2).
Corollary 2 For any f , g ∈ C [a, b],
Z b
2 Z b
Z
|f (x)|2 dx ·
f (x)g (x) dx ≤
a
a
b
a
|g (x)|2 dx.
Norms induced by inner products
Theorem Suppose hx, yi is an
pinner product on a
vector space V . Then kxk = hx, xi is a norm.
Proof: Positivity is obvious. Homogeneity:
p
p
p
kr xk = hr x, r xi = r 2hx, xi = |r | hx, xi.
Triangle inequality (follows from Cauchy-Schwarz’s):
kx + yk2 = hx + y, x + yi
= hx, xi + hx, yi + hy, xi + hy, yi
≤ hx, xi + |hx, yi| + |hy, xi| + hy, yi
≤ kxk2 + 2kxk kyk + kyk2 = (kxk + kyk)2.
Examples. • The length of a vector in Rn ,
p
|x| = x12 + x22 + · · · + xn2,
is the norm induced by the dot product
x · y = x1 y1 + x2 y2 + · · · + xn yn .
• The norm kf k2 =
Z
a
b
2
|f (x)| dx
1/2
on the
vector space C [a, b] is induced by the inner product
Z b
hf , g i =
f (x)g (x) dx.
a
Angle
Since |hx, yi| ≤ kxk kyk, we can define the angle
between nonzero vectors in any vector space with
an inner product (and induced norm):
hx, yi
∠(x, y) = arccos
.
kxk kyk
Then hx, yi = kxk kyk cos ∠(x, y).
In particular, vectors x and y are orthogonal
(denoted x ⊥ y) if hx, yi = 0.
Problem. Find the angle between functions
f1 (x) = x and f2 (x) = x 2 in the inner space C [0, 1]
with the inner productZ
1
hf , g i =
hf1 , f2 i =
Z
1
1
x · x 2 dx = ,
4
1
1
x dx = ,
3
0
f (x)g (x) dx.
0
1
1
(x 2 )2 dx = .
5
0
0
√
1/4
hf1 , f2 i
15
p
=p
cos ∠(f1 , f2 ) =
=
kf1 k kf2k
4
1/3 1/5
p
sin ∠(f1 , f2 ) = 1 − cos2 ∠(f1 , f2 ) = 1/4
hf1 , f1 i =
Z
2
∠(f1 , f2 ) = arcsin 14
hf2 , f2 i =
Z
x+y
x
y
Pythagorean Theorem:
x ⊥ y =⇒ kx + yk2 = kxk2 + kyk2
Proof:
kx + yk2 = hx + y, x + yi
= hx, xi + hx, yi + hy, xi + hy, yi
= hx, xi + hy, yi = kxk2 + kyk2.
y
x−y
x
x+y
x
y
Parallelogram Identity:
kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2
Proof: kx+yk2 = hx+y, x+yi = hx, xi + hx, yi + hy, xi + hy, yi.
Similarly, kx−yk2 = hx, xi − hx, yi − hy, xi + hy, yi.
Then kx+yk2 + kx−yk2 = 2hx, xi + 2hy, yi = 2kxk2 + 2kyk2.
Example. Norms on Rn , n ≥ 2:
• kxk∞ = max(|x1|, |x2|, . . . , |xn |),
1/p
, p ≥ 1.
• kxkp = |x1|p + |x2|p + · · · + |xn |p
Theorem The norms kxk∞ and kxkp , p 6= 2 do
not satisfy the Parallelogram Identity. Hence they
are not induced by any inner product on Rn .
Proof: A counterexample to the Parallelogram Identity is
provided by vectors e1 = (1, 0, 0, . . . , 0) and e2 = (0, 1, 0, . . . , 0).
ke1 k∞ = ke2 k∞ = 1, ke1 kp = ke2 kp = 1 for any p ≥ 1.
ke1 + e2 k∞ = ke1 − e2 k∞ = 1,
ke1 + e2 kp = ke1 − e2 kp = 21/p for any p ≥ 1.
Thus 2ke1 k2∞ + 2ke2 k2∞ = 2ke1 k2p + 2ke2k2p = 4.
On the other hand, ke1 + e2 k2∞ + ke1 − e2 k2∞ = 2 6= 4 and
ke1 + e2 k2p + ke1 − e2 k2p = 2(21/p )2 = 21+2/p 6= 4 unless p = 2.