Dr. Neal, WKU
MATH 337
Euclidean Space/Absolute Value
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Euclidean Space (n-dimensional space): For each positive integer n , we let R denote
the set of all ordered n -tuples of the form
u = { u1 , . . . , un },
n
where ui ∈ ℜ for 1 ≤ i ≤ n . Elements in R are called vectors. However for n = 1 , we
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simply refer to them as real numbers or scalars and write ℜ rather than R .
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Henceforth, let u = { u1 , . . . , un } and v = { v1, . . . , vn } be vectors in R .
0. Sum and Scalar Multiplication: (i) We define the sum of vectors u and v by
u + v = {u1 + v1 , . . ., un + v n}.
(ii) For c ∈ ℜ , the scalar product c u is defined by c u = { c u1 , . . . , c un } .
We can then define the difference v − u by
v − u = v + (−1)u = {v1 − u1 , . . ., v n − un} ,
which gives the directional vector from u to v .
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With these operations of addition and scalar multiplication, R is a vector space over
the field of real numbers.
Properties of Addition
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1. If u , v ∈ R , then u + v ∈ R . (closure)
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2. (u + v) + w = u + (v + w) for all u , v , w ∈ R . (associative)
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3. u + v = v + u for all u , v ∈ R . (commutative)



n
n
0
0
+
v
=
v
=
v
+
0
4. There exists a “0” element in R
,
denoted
,
such
that
for all v ∈ R .

(additive identity). Specifically, 0 = {0, . . ., 0}, which we call the zero vector.

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5. For every v ∈ R , there exists an element − v such that v + (−v) = 0 = −v + v (additive
inverse) Specifically, –{ v1, . . . , vn } = {– v1, . . . , – vn } .
Properties of Scalar Multiplication
1.
2.
3.
4.
5.
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If v ∈ R and c is any scalar (i.e., real number), then c v ∈ R . (closure)
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c(d v) = (c d)v for all v ∈ R and all c , d ∈ ℜ . (associative)
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c (u + v) = cu + cv for all u , v ∈ R and all c ∈ ℜ . (scalar distributive)
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(c + d)v = c v + d v for all v ∈ R and all c , d ∈ ℜ . (vector distributive)
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For the scalar 1, 1v = v for all v ∈ R . (multiplicative identity)
Dr. Neal, WKU
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I. Norm: The norm (or length or magnitude) of a vector u ∈ R is defined by
u =
u12 + . . . + un2 .
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In R , the norm of the vector u = (x , y) is simply the length of the segment from
the origin (0, 0) to the point (x , y) , and is given by
x 2 + y2 .
Properties of Norm
(i)

u ≥ 0, and u = 0 if and only if u = 0 ;
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(ii) u = u12 + . . . + un 2 ;
(iii) c u = c × u for every scalar c ∈ ℜ .
II. Distance: The distance d(u, v) between vectors u and v is given by the length of the
vector from u to v :
d(u, v) = v − u =
(v1 − u1 )2 + . . . + (v n − un )2 .
Properties of Distance
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For all for all u , v ∈ R :
(i) d(u, v) ≥ 0
(ii) d(u, u) = 0 and d(u, v) = 0 if and only if u = v
(iii) d(u, v) = d(v,u) .
III. Dot Product: We define the dot product u ⋅ v (also called inner product) between
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vectors u and v in R by u ⋅ v = u1 v1 + . . . + un vn .
Properties of Dot Product
(a)
(b)
(c)
(d)
(e)
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u ⋅u = u12 + . . . + un 2 = u ≥ 0.
u = u⋅u
u⋅v = v⋅u
For any scalar c , (c u) ⋅v = c × (u ⋅ v) and u ⋅(cv) = c × (u ⋅ v) .
For another vector w , (u + v)⋅ w = (u ⋅ w) + (v ⋅ w) .
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Theorem (Cauchy-Schwarz Inequality). Let u and v be vectors in R . Then
u⋅ v ≤ u
v .
Dr. Neal, WKU
Proof. Let k be a scalar and consider the vector k u + v . Then
0 ≤ k u + v 2 = (k u + v) ⋅(k u + v)
= k 2 (u ⋅ u) + k(u ⋅v) + k(v ⋅ u) + (v ⋅v)
= u 2 k 2 + 2(u⋅ v)k + v 2 .
This expression defines a quadratic in k which is always non-negative; so the
quadratic has zero roots or just one root. Thus the discriminant “ b 2 − 4ac ” must be
2
2
less than or equal to 0, where a = u , b = 2(u ⋅ v) , and c = v . (If the discriminant
were positive, then the quadratic would have two roots and therefore would have to be
negative over some interval.)
2
2
2
2
Thus, b 2 − 4ac = 4(u⋅ v) 2 − 4 u v ≤ 0 , which implies that (u ⋅ v) 2 ≤ u v . By
taking square roots we obtain u⋅ v ≤ u v .
QED
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Theorem (Triangle Inequality). Let u and v be vectors in R . Then
u+v ≤ u + v .
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Proof: We simply expand u + v . The first inequality below arises from the fact that
the number u ⋅ v is less than or equal to its absolute value. The second inequality is
from Cauchy-Schwarz:
u + v 2 = (u + v) ⋅ (u + v ) = u 2 + v 2 + 2 (u ⋅ v)
≤ u 2 + v 2 + 2 u⋅v
≤ u 2+ v 2+2 u
v
= ( u + v )2 .
Because the square root function is strictly increasing, we maintain the inequality by
taking square roots. Thus, we obtain u + v ≤ u + v .
QED
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Note: For vectors u and v in R , the triangle
inequality states that length of the diagonal u + v
can be no more than the sum of the lengths of the
two sides u and v
(hence, the name triangle
inequality). The result can be generalized to the
distance between vectors as shown next.
u+v
u
v
Dr. Neal, WKU
Shortest Distance Between Points
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Theorem. Let u and v be vectors in R .
d(u, v) ≤ d(u, w) + d(w, v) .
For any other vector w , we have
Proof. We use the definition of distance and apply the triangle inequality:
d (u,v) = v − u
= (v − w) + (w − u)
≤ v− w + w −u
= d(w,v ) + d(u, w)
= d(u, w) + d (w, v).
QED
The result states that the distance directly from u to v is less than or equal to the
sum of the distances from u to w then from w to v . We note that if u , v , and w are
collinear with w between u and v , then d(u, v) will equal d(u, w) + d(w, v) .
Absolute Value
For x ∈ ℜ , we define the absolute value of x by
 x if x ≥ 0
x =
 − x if x < 0 .
We shall show that x is really the norm x when considering x to be an element
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of the Euclidean space ℜ = R . So we will be able to apply all the properties of norm
and distance to the absolute value function.
Properties of Absolute Value
(i) For all x ∈ ℜ , x ≥ 0 .
If x ≥ 0 , then x = x ≥ 0 . If x < 0 , then x = − x > 0 .
(ii) For all x ∈ ℜ , x ≤ x .
If x ≥ 0 , then x = x . If x < 0 , then x < 0 ≤ x .
(iii) x = 0 if and only if x = 0 . That is, 0 = 0 and x > 0 for x ≠ 0 .
(iv) For all x ∈ ℜ ,
x2 = x .
If x ≥ 0 , then x = x =
x 2 . If x < 0 , then
From (iv), we see that x =
as the absolute value in ℜ .
x 2 = (− x )(− x ) = (− x ) = x .
1
x 2 = x ; thus, the Euclidean norm of R is the same
Dr. Neal, WKU
(v) Cauchy-Schwarz: For all x, y ∈ ℜ , x y = x y .
Consider three cases. (i) If
x ≥ 0 and y ≥ 0 , then x y ≥ 0 and thus x y = x y = x y . (ii) If x < 0 and y < 0 , then
x y > 0 and x y = x y = (−1)x (−1)y = x y . (iii) If one of x, y is negative and the other
non-negative, then x y ≤ 0 ; thus, x y = − (x y) = x y .
(vi) − x = x
(vii)
Here we apply (iv) to obtain − x = − 1x = − 1 x = 1 x = x .
x 2 = x2
Here we have x
2
 x 2
=
(− x )2
if x ≥ 0
= x 2 for all x .
if x < 0
(viii) Triangle Inequality for Absolute Value: For all x, y ∈ ℜ ,
x+y ≤ x + y .
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A direct proof in ℜ is easier than the general proof in R . For all x, y ∈ ℜ ,
x + y 2 = ( x + y)2 = x 2 + 2 x y + y2 ≤ x 2 + 2 x y + y2
= x 2 + 2 x y + y 2 = ( x + y )2.
Because the square root function is strictly increasing, we maintain the inequality by
taking square roots. Thus, we obtain x + y ≤ x + y .
(ix) Distance in ℜ : For all x, y ∈ ℜ , we define the distance between x and y by
d (x, y ) = x − y
We shall use the triangle inequality repeatedly in ℜ as
x − y = x − z + z− y ≤ x − z + z− y .
That is, d (x, y ) ≤ d(x ,z ) + d(z, y) .
Theorem. For all x ∈ ℜ , x
n
n
= x
for all integers n ≥ 1.
Proof. We shall use induction on n . For n = 1 , we have x
For n = 2 , we have independently shown that x
have x
2
2
=x = x
2
n+1
= x = x
1
for all x ∈ ℜ .
2
2
= x . And because x ≥ 0 , we then
for all x ∈ ℜ .
Now assume that for all x ∈ ℜ , x
x
2
1
n
= x x = x
n
n
= x
x = x
n
n
for some particular n ≥ 1. Then
x = x
n+1
, for all x ∈ ℜ .
By mathematical induction, the result holds for all integers n ≥ 1.
Dr. Neal, WKU
Exercises
1. Prove that for all integers n ≥ 1 and all real numbers x1 , x2 , . . . , xn
(a) x1 + x2 + . . . + xn ≤ x1 + x2 + . . . + xn
(b)
x1 × x 2 × . . . × x n = x1 × x 2 × . . . × x n .
2. For all real numbers x and y , prove that
x − y
≤ x− y .
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3. Let u = { u1 , . . . , un } and v = { v1, . . . , vn } be vectors in R . Consider the individual
coordinates as points in ℜ . Prove that for 1 ≤ i ≤ n ,
(a)
(b) d (ui , vi ) ≤ d(u, v) .
ui ≤ u
4. Prove the following results about the absolute value:
(a) x − y = 0 if and only if x = y .
(b) x − y = y − x
€ 5. Suppose that x − y€< ε for all ε > 0 . Prove that
x = y.
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