Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

WEN-CHINGLIEN Advanced Calculus (II)

Ch8: Euclidean Spaces

8.1: Algebraic Structure of R

Definition (8.1)

Letx = (x1, . . . ,xn) ,y = (y1, . . . ,yn) ∈Rⁿ be vectors and α ∈R be a scalar.

(i) The sum ofx and y is the vector

x + y := (x1+y1,x2+y2, . . . ,xn+yn) (ii) The difference ofx and y is the vector

x − y := (x1− y1,x2− y2, . . . ,xn− yn)

Definition (8.1)

(iii) The product of a scalar α and a vectorx is the vector αx := (αx1, αx2, . . . , αxn)

(iv) The (Euclidean) dot product (or scalar product or inner product ) ofx and y is the scalar

x · y := x1y1+x2y2+ · · · +xnyn

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.2)

Letx, y, z∈ Rⁿand 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.

Definition (8.3)

Leta 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 thata = tb.

(ii)a and b are said to be orthogonal if and only if a · b = 0.

WEN-CHINGLIEN Advanced Calculus (II)

Definition (8.4) Letx ∈ Rⁿ.

(i) The (Euclidean) norm (or magnitude) ofx is the scalar

kxk :=

v u u t

n

X

k =1

|x_k|².

(ii) The `¹− norm (read L-one-norm) of x is the scalar

kxk₁:=

n

X

k =1

|xk|.

(iii) The sup − norm ofx is the scalar kxk_∞:=max {|x1|, . . . , |xn|}.

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0,substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0, substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0,substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0, substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0, substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0, substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0, substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.5 Cauchy-Schwarz Inequality) Ifx, y ∈ Rⁿ, then

|x · y| ≤ kxk kyk . Proof.

The Cauchy-Schwarz Inequality for trivial wheny = 0. If y6=0, substitute t =(x · y)/ kyk² into

0 ≤ kx − tyk² = (x−ty)·(x−ty) = kxk²−2t(x · y)+t²kyk² to obtain

0 ≤ kxk²− t(x · y) = kxk²− (x · y)² kyk² .

Proof.

It follow that 0 ≤ kxk²− (x · y)/ kyk². Solving this inequality for (x · y)², we conclude that

(x · y)²≤ kxk²kyk².

WEN-CHINGLIEN 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≤ kxk²kyk².

Theorem (8.6) Letx, 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).

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.6) Letx, 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).

Theorem (8.6) Letx, 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).

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.6) Letx, 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).

Theorem (8.6) Letx, 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).

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.6) Letx, 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).

Theorem (8.6) Letx, 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).

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.6) Letx, 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).

Theorem (8.6) Letx, 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).

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.6) Letx, 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).

Theorem (8.6) Letx, 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).

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.6) Letx, 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).

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

WEN-CHINGLIEN Advanced Calculus (II)

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

WEN-CHINGLIEN Advanced Calculus (II)

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

WEN-CHINGLIEN Advanced Calculus (II)

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

Remark (8.7) Letx ∈ Rⁿ.Then (i) |xj| ≤ kxk ≤√

n kxk_∞for each j = 1,2,. . . ,n, (ii)kxk ≤ kxk₁.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|x_j|² ≤ kxk² =x₁²+ . . . +x_n²≤ n( max

1≤`≤n|x<sub>`</sub>|)²=n kxk²_∞

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(ii) Observe that

(|x1| + · · · + |x_n|)² = |x1|²+ · · · + |xn|²+2 X

(i,j∈A)

|x_i||x_j|

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| + . . . + |x_n|)²

Proof.

(ii) Observe that

(|x1| + · · · + |x_n|)² = |x1|²+ · · · + |xn|²+2 X

(i,j∈A)

|x_i||x_j|

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| + . . . + |x_n|)²

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(ii) Observe that

(|x1| + · · · + |x_n|)² = |x1|²+ · · · + |xn|²+2 X

(i,j∈A)

|x_i||x_j|

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| + . . . + |x_n|)²

Definition (8.8)

The cross product of two vectorsx = (x1,x2,x3)and y = (y₁,y2,y3)in R³ is the vector defined by

x × y := (x2y3− x₃y2,x3y1− x₁y3,x1y2− x₂y1).

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:

det





i j k

x1 x2 x3

y1 y2 y3



.

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

WEN-CHINGLIEN Advanced Calculus (II)

Theorem (8.9)

Letx, y, z ∈ R³be vectors and α be a scalar.Then (i) x × x = 0, x × y = −y × x,

(ii) (αx) × y = α(x × y) = x × (αy), (iii) x · (y + z) = (x × y) + (x × z), (iv ) (x × y) · z = x · (y × z) = det





x1 x2 x3

y1 y2 y3

z1 z2 z3



, (v ) x × (y × z) = (x · z)y − (x · y)z,

(vi) kx × yk² = (x · x)(y · y) − (x · y)². (vii) Moreover, ifx × y 6= 0, then the vector x × y is orthogonal tox and y.

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= (x2y3−x₃y2)z1+(x3y1−x₁y3)z2+(x1y2−x₂y1)z3

=x1(y2z3− y3z2) +x2(y3z1− y1z3) +x3(y1z2− y2z1).

Since this last expression is both the scalarx · (y × z) and the value of the determinant on the right side of (iv)

(expanded along the first row), this verifies (iv).

WEN-CHINGLIEN 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 = (x₂y3−x₃y2)z1+(x3y1−x₁y3)z2+(x1y2−x₂y1)z3

=x1(y2z3− y3z2) +x2(y3z1− y1z3) +x3(y1z2− y2z1).

Since this last expression is both the scalarx · (y × z) and the value of the determinant on the right side of (iv)

(expanded along the first row), this verifies (iv).

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 = (x₂y3−x₃y2)z1+(x3y1−x₁y3)z2+(x1y2−x₂y1)z3

=x1(y2z3− y3z2) +x2(y3z1− y1z3) +x3(y1z2− y2z1).

Since this last expression is both the scalarx · (y × z) and the value of the determinant on the right side of (iv)

(expanded along the first row), this verifies (iv).

WEN-CHINGLIEN 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 = (x₂y3−x₃y2)z1+(x3y1−x₁y3)z2+(x1y2−x₂y1)z3

=x1(y2z3− y3z2) +x2(y3z1− y1z3) +x3(y1z2− y2z1).

Since this last expression is both the scalarx · (y × z) and the value of the determinant on the right side of (iv)

(expanded along the first row), this verifies (iv).

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

WEN-CHINGLIEN Advanced Calculus (II)

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

Proof.

(v) Sincex × (y × z)

= (x1,x2,x3) × (y2z3− y₃z2,y3z1− y₁z3,y1z2− y₂z1), the component ofx × (y × z) is

x2y1z2− x2y2z1− x3y3z1+x3y1z3

= (x1z1+x2z2+x3z3)y1− (x₁z1+x2z2+x3z3)z1. this proves that the first conponents ofx × (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.

WEN-CHINGLIEN Advanced Calculus (II)

Remark (8.10)

Letx and y be nonzero vectors in R³and θ be the angle betweenx and y. Then kx × yk = kxk kyk sin θ.

Proof.

By theorem 8.9(vi) and cos θ = _kakkbk^a·b ,

kx × yk² = (kxk kyk)²− (kxk kyk cos θ)²

= (kxk kyk)²(1 − cos²θ) = (kxk kyk)²sin²θ.

Remark (8.10)

Letx and y be nonzero vectors in R³and θ be the angle betweenx and y. Then kx × yk = kxk kyk sin θ.

Proof.

By theorem 8.9(vi) and cos θ = _kakkbk^a·b ,

kx × yk² = (kxk kyk)²− (kxk kyk cos θ)²

= (kxk kyk)²(1 − cos²θ) = (kxk kyk)²sin²θ.

WEN-CHINGLIEN Advanced Calculus (II)

Remark (8.10)

Letx and y be nonzero vectors in R³and θ be the angle betweenx and y. Then kx × yk = kxk kyk sin θ.

Proof.

By theorem 8.9(vi) and cos θ = _kakkbk^a·b ,

kx × yk² = (kxk kyk)²− (kxk kyk cos θ)²

= (kxk kyk)²(1 − cos²θ) = (kxk kyk)²sin²θ.

Remark (8.10)

Letx and y be nonzero vectors in R³and θ be the angle betweenx and y. Then kx × yk = kxk kyk sin θ.

Proof.

By theorem 8.9(vi) and cos θ = _kakkbk^a·b ,

kx × yk² = (kxk kyk)²− (kxk kyk cos θ)²

= (kxk kyk)²(1 − cos²θ) = (kxk kyk)²sin²θ.

WEN-CHINGLIEN Advanced Calculus (II)

Thank you.