MATH 304
Linear Algebra
Lecture 27:
Norms and inner products.
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 |).
Positivity and homogeneity are obvious.
The triangle inequality:
|xi + yi | ≤ |xi | + |yi | ≤ maxj |xj | + maxj |yj |
=⇒ maxj |xj + yj | ≤ maxj |xj | + maxj |yj |
• kxk1 = |x1| + |x2 | + · · · + |xn |.
Positivity and homogeneity are obvious.
The triangle inequality: |xi + yi | ≤ |xi | + |yi |
P
P
P
=⇒
|x
+
y
|
≤
|x
|
+
j
j j
j j
j |yj |
Examples. V = Rn , x = (x1, x2, . . . , xn ) ∈ Rn .
1/p
, p > 0.
• kxkp = |x1|p + |x2|p + · · · + |xn |p
Theorem kxkp is a norm on Rn for any p ≥ 1.
Remark. kxk2 = Euclidean length of x.
Definition. A normed vector space is a vector
space endowed with a norm.
The norm defines a distance function on the normed
vector space: dist(x, y) = kx − yk.
Then we say that a sequence x1 , x2, . . . converges
to a vector x if dist(x, xn ) → 0 as n → ∞.
Unit circle: kxk = 1
kxk = (x12 + x22 )1/2
1/2
kxk = 21 x12 + x22
kxk = |x1 | + |x2|
kxk = max(|x1 |, |x2|)
black
green
blue
red
Examples. V = C [a, b], f : [a, b] → R.
• kf k∞ = max |f (x)|.
a≤x≤b
• kf k1 =
• kf kp =
Z
b
|f (x)| dx.
a
Z
a
b
p
|f (x)| dx
1/p
, p > 0.
Theorem kf kp is a norm on C [a, b] for any 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, xi ≥ 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.
• hx, yi = (Dx) · (Dy),
where D is an invertible n×n matrix.
Remarks. (a) Invertibility of D is necessary to show
that hx, xi = 0 =⇒ x = 0.
(b) The second example is a particular case of the
1/2
1/2
1/2
third one when D = diag(d1 , d2 , . . . , dn ).
Counterexamples. V = R2 .
• hx, yi = x1 y1 − x2y2.
Let v = (1, 2), then hv, vi = 12 − 22 = −3.
hx, yi is symmetric and bilinear, but not positive.
• hx, yi = 2x1y1 + x1 x2 + 2x2y2 + y1y2 .
v = (1, 1), w = (1, 0) =⇒ hv, wi = 3, h2v, wi = 8.
hx, yi is positive and symmetric, but not bilinear.
• hx, yi = x1 y1 + x1y2 − x2 y1 + x2y2.
v = (1, 1), w = (1, 0) =⇒ hv, wi = 0, hw, vi = 2.
hx, yi is positive and bilinear, but not symmetric.
Problem. Find an inner product on R2 such that
he1 , e1 i = 2, he2 , e2i = 3, and he1 , e2i = −1,
where e1 = (1, 0), e2 = (0, 1).
Let x = (x1, x2), y = (y1, y2) ∈ R2 .
Then x = x1 e1 + x2e2 , y = y1e1 + y2 e2.
Using bi-linearity, we obtain
hx, yi = hx1 e1 + x2e2 , y1e1 + y2 e2i
= x1 he1 , y1e1 + y2 e2i + x2he2 , y1e1 + y2e2 i
= x1y1he1 , e1i + x1 y2he1 , e2i + x2y1he2 , e1i + x2y2he2 , e2i
= 2x1y1 − x1y2 − x2y1 + 3x2y2.
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.