Advanced Calculus (II)
W EN -C HING L IEN
Department of Mathematics
National Cheng Kung University
2009
W EN -C HING L IEN
Advanced Calculus (II)
Ch8: Euclidean Spaces
8.1: Algebraic Structure of Rn
Definition (8.1)
Let x = (x1 , . . . , xn ) , y = (y1 , . . . , yn ) ∈ Rn be vectors and
α ∈ R be a scalar.
(i) The sum of x and y is the vector
x + y := (x1 + y1 , x2 + y2 , . . . , xn + yn )
(ii) The difference of x and y is the vector
x − y := (x1 − y1 , x2 − y2 , . . . , xn − yn )
W EN -C HING L IEN
Advanced Calculus (II)
Definition (8.1)
(iii) The product of a scalar α and a vector x is the vector
αx := (αx1 , αx2 , . . . , αxn )
(iv) The (Euclidean) dot product (or scalar product or
inner product ) of x and y is the scalar
x · y := x1 y1 + x2 y2 + · · · + xn yn
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.2)
Let x, y, z∈ Rn and α, β ∈ R. Then
α0 = 0,
0x = 0,
1x = x,
α(βx) = β(αx) = (αβ)x,
α(x · y) = (αx) · y = x · (αy),
α(x + y) = αx + αy,
0 + x = x,
x − x = 0,
0 · x = 0,
x + (y + z) = (x + y) + z,
x + y = y + x,
x · y = y · x,
x · (y + z) = x · y + x · z.
W EN -C HING L IEN
Advanced Calculus (II)
Definition (8.3)
Let a and b be nonzero vectors in Rn .
(i) a and b are said to be parallel if and only if these is a
scalar t ∈ R such that a = tb.
(ii) a and b are said to be orthogonal if and only if
a · b = 0.
W EN -C HING L IEN
Advanced Calculus (II)
Definition (8.4)
Let x ∈ Rn .
(i) The (Euclidean) norm (or magnitude) of x is the scalar
v
u n
uX
kxk := t
|xk |2 .
k=1
(ii) The `1 − norm (read L-one-norm) of x is the scalar
kxk1 :=
n
X
|xk |.
k=1
(iii) The sup − norm of x is the scalar
kxk∞ := max{|x1 |, . . . , |xn |}.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Theorem (8.5 Cauchy-Schwarz Inequality)
If x, y ∈ Rn , then
|x · y| ≤ kxk kyk .
Proof.
The Cauchy-Schwarz Inequality for trivial when y = 0. If
y6=0, substitute t =(x · y)/ kyk2 into
0 ≤ kx − tyk2 = (x−ty)·(x−ty) = kxk2 −2t(x · y)+t 2 kyk2
to obtain
0 ≤ kxk2 − t(x · y) = kxk2 −
W EN -C HING L IEN
(x · y)2
kyk2
Advanced Calculus (II)
.
Proof.
It follow that 0 ≤ kxk2 − (x · y)/ kyk2 . Solving this
inequality for (x · y)2 , we conclude that
(x · y)2 ≤ kxk2 kyk2 .
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
It follow that 0 ≤ kxk2 − (x · y)/ kyk2 . Solving this
inequality for (x · y)2 , we conclude that
(x · y)2 ≤ kxk2 kyk2 .
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.6)
Let x, y ∈ Rn .Then
(i) kxk ≥ 0 with equality only when x = 0.
(ii) kαxk = |α| kxk for all scalars α.
(iii) [triangle inequalities].
kx + yk ≤ kxk + kyk and kx − yk ≥ kxk − kyk.
Proof.
To prove (iii), observe that by Definition 8.4, Theorem 8.2,
and the Cauchy-Schwarz Inequality,
kx + yk2 = (x + y) · (x + y) = x · x + 2x · y + y · y
= kxk2 + 2x · y + kyk2 ≤ kxk2 + 2 kxk kyk + kyk2
= (kxk + kyk)2
This establishes the first inequality in (iii). By modifying
the proof of Theorem 1.7, we can also establish the
second inequality in (iii).
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.7)
Let x ∈ Rn .Then√
(i) |xj | ≤ kxk ≤ n kxk∞ for each j = 1,2,. . . ,n,
(ii)kxk ≤ kxk1 .
Proof.
(i) Let 1 ≤ j ≤ n. By definition,
|xj |2 ≤ kxk2 = x12 + . . . + xn2 ≤ n( max |x` |)2 = n kxk2∞
1≤`≤n
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(ii) Observe that
(|x1 | + · · · + |xn |)2 = |x1 |2 + · · · + |xn |2 + 2
X
|xi ||xj |
(i,j∈A)
where
A = {(i, j) : 1 ≤ i, j ≤ n and i < j}.Since
P
(i,j)∈A |xi ||xj | ≥ 0, we conclude that
kxk2 = |x1 |2 + . . . + |xn |2 ≤ (|x1 | + . . . + |xn |)2
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(ii) Observe that
(|x1 | + · · · + |xn |)2 = |x1 |2 + · · · + |xn |2 + 2
X
|xi ||xj |
(i,j∈A)
where
A = {(i, j) : 1 ≤ i, j ≤ n and i < j}.Since
P
(i,j)∈A |xi ||xj | ≥ 0, we conclude that
kxk2 = |x1 |2 + . . . + |xn |2 ≤ (|x1 | + . . . + |xn |)2
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(ii) Observe that
(|x1 | + · · · + |xn |)2 = |x1 |2 + · · · + |xn |2 + 2
X
|xi ||xj |
(i,j∈A)
where
A = {(i, j) : 1 ≤ i, j ≤ n and i < j}.Since
P
(i,j)∈A |xi ||xj | ≥ 0, we conclude that
kxk2 = |x1 |2 + . . . + |xn |2 ≤ (|x1 | + . . . + |xn |)2
W EN -C HING L IEN
Advanced Calculus (II)
Definition (8.8)
The cross product of two vectors x = (x1 , x2 , x3 ) and
y = (y1 , y2 , y3 ) in R 3 is the vector defined by
x × y := (x2 y3 − x3 y2 , x3 y1 − x1 y3 , x1 y2 − x2 y1 ).
using the usual basis i = e1 , j = e2 , k = e3 , and the
determinant operator (see Appendix C),we can give the
cross product a more easily remembered form:


i
j k
det  x1 x2 x3  .
y1 y2 y3
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Theorem (8.9)
Let x, y, z ∈ R3 be vectors and α be a scalar.Then
(i)
x × x = 0, x × y = −y × x,
(ii)
(iii)
(iv )
(v )
(vi)
(αx) × y = α(x × y) = x × (αy),
x · (y + z) = (x × y) + (x 
× z),

x1 x2 x3
(x × y) · z = x · (y × z) = det  y1 y2 y3  ,
z1 z2 z3
x × (y × z) = (x · z)y − (x · y)z,
kx × yk2 = (x · x)(y · y) − (x · y)2 .
(vii) Moreover, if x × y 6= 0, then the vector x × y is
orthogonal to x and y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
These properties follow immediately from the definition.
We will prove properties (iv),(v),and (vii) and leave the
rest as an exercise.
(iv) Notice that by definition,
(x×y)·z = (x2 y3 −x3 y2 )z1 +(x3 y1 −x1 y3 )z2 +(x1 y2 −x2 y1 )z3
= x1 (y2 z3 − y3 z2 ) + x2 (y3 z1 − y1 z3 ) + x3 (y1 z2 − y2 z1 ).
Since this last expression is both the scalar x · (y × z) and
the value of the determinant on the right side of (iv)
(expanded along the first row), this verifies (iv).
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
These properties follow immediately from the definition.
We will prove properties (iv),(v),and (vii) and leave the
rest as an exercise.
(iv) Notice that by definition,
(x×y)·z = (x2 y3 −x3 y2 )z1 +(x3 y1 −x1 y3 )z2 +(x1 y2 −x2 y1 )z3
= x1 (y2 z3 − y3 z2 ) + x2 (y3 z1 − y1 z3 ) + x3 (y1 z2 − y2 z1 ).
Since this last expression is both the scalar x · (y × z) and
the value of the determinant on the right side of (iv)
(expanded along the first row), this verifies (iv).
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
These properties follow immediately from the definition.
We will prove properties (iv),(v),and (vii) and leave the
rest as an exercise.
(iv) Notice that by definition,
(x×y)·z = (x2 y3 −x3 y2 )z1 +(x3 y1 −x1 y3 )z2 +(x1 y2 −x2 y1 )z3
= x1 (y2 z3 − y3 z2 ) + x2 (y3 z1 − y1 z3 ) + x3 (y1 z2 − y2 z1 ).
Since this last expression is both the scalar x · (y × z) and
the value of the determinant on the right side of (iv)
(expanded along the first row), this verifies (iv).
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
These properties follow immediately from the definition.
We will prove properties (iv),(v),and (vii) and leave the
rest as an exercise.
(iv) Notice that by definition,
(x×y)·z = (x2 y3 −x3 y2 )z1 +(x3 y1 −x1 y3 )z2 +(x1 y2 −x2 y1 )z3
= x1 (y2 z3 − y3 z2 ) + x2 (y3 z1 − y1 z3 ) + x3 (y1 z2 − y2 z1 ).
Since this last expression is both the scalar x · (y × z) and
the value of the determinant on the right side of (iv)
(expanded along the first row), this verifies (iv).
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Proof.
(v) Since x × (y × z)
= (x1 , x2 , x3 ) × (y2 z3 − y3 z2 , y3 z1 − y1 z3 , y1 z2 − y2 z1 ),
the component of x × (y × z) is
x2 y1 z2 − x2 y2 z1 − x3 y3 z1 + x3 y1 z3
= (x1 z1 + x2 z2 + x3 z3 )y1 − (x1 z1 + x2 z2 + x3 z3 )z1 .
this proves that the first conponents of x × (y × z) and
(x · z)y − (x · y)z are equal. A similar argument shows that
the second and third components are also equal.
(vii) By parts (i) and (iv),
(x × y) · x= −(y × x) · x= −y · (x × x)= −y · 0 = 0.
Thus (x × y) is orthogonal to x. A similar calculation
shows that (x × y) is orthogonal to y.
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.10)
Let x and y be nonzero vectors in R3 and θ be the angle
between x and y. Then kx × yk = kxk kyk sin θ.
Proof.
By theorem 8.9(vi) and cos θ =
a·b
,
kakkbk
kx × yk2 = (kxk kyk)2 − (kxk kyk cos θ)2
= (kxk kyk)2 (1 − cos2 θ) = (kxk kyk)2 sin2 θ.
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.10)
Let x and y be nonzero vectors in R3 and θ be the angle
between x and y. Then kx × yk = kxk kyk sin θ.
Proof.
By theorem 8.9(vi) and cos θ =
a·b
,
kakkbk
kx × yk2 = (kxk kyk)2 − (kxk kyk cos θ)2
= (kxk kyk)2 (1 − cos2 θ) = (kxk kyk)2 sin2 θ.
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.10)
Let x and y be nonzero vectors in R3 and θ be the angle
between x and y. Then kx × yk = kxk kyk sin θ.
Proof.
By theorem 8.9(vi) and cos θ =
a·b
,
kakkbk
kx × yk2 = (kxk kyk)2 − (kxk kyk cos θ)2
= (kxk kyk)2 (1 − cos2 θ) = (kxk kyk)2 sin2 θ.
W EN -C HING L IEN
Advanced Calculus (II)
Remark (8.10)
Let x and y be nonzero vectors in R3 and θ be the angle
between x and y. Then kx × yk = kxk kyk sin θ.
Proof.
By theorem 8.9(vi) and cos θ =
a·b
,
kakkbk
kx × yk2 = (kxk kyk)2 − (kxk kyk cos θ)2
= (kxk kyk)2 (1 − cos2 θ) = (kxk kyk)2 sin2 θ.
W EN -C HING L IEN
Advanced Calculus (II)
Thank you.
W EN -C HING L IEN
Advanced Calculus (II)