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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch8: Euclidean Spaces

### 8.1: Algebraic Structure of R

n

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)

(3)

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

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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.

(5)

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.

(6)

Definition (8.4) Letx ∈ Rn.

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

kxk :=

v u u t

n

X

k =1

|xk|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|}.

(7)

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 .

(8)

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 .

(9)

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 .

(10)

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 .

(11)

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 .

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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 .

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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 .

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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 .

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Proof.

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

(x · y)2≤ kxk2kyk2.

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Proof.

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

(x · y)2≤ kxk2kyk2.

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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).

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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).

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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).

(20)

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).

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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).

(22)

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).

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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).

(24)

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).

(25)

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).

(26)

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).

(27)

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).

(28)

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).

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

(33)

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

(34)

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

(35)

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

n kxkfor each j = 1,2,. . . ,n, (ii)kxk ≤ kxk1.

Proof.

(i) Let 1 ≤ j ≤ n. By definition,

|xj|2 ≤ kxk2 =x12+ . . . +xn2≤ n( max

1≤`≤n|x`|)2=n kxk2

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Proof.

(ii) Observe that

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

(i,j∈A)

|xi||xj|

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

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Proof.

(ii) Observe that

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

(i,j∈A)

|xi||xj|

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

(38)

Proof.

(ii) Observe that

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

(i,j∈A)

|xi||xj|

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

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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

.

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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.

(41)

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.

(42)

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.

(43)

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.

(44)

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.

(45)

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.

(46)

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.

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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).

(48)

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).

(49)

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).

(50)

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).

(51)

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.

(52)

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.

(53)

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.

(54)

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.

(55)

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.

(56)

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.

(57)

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.

(58)

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.

(59)

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.

(60)

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.

(61)

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.

(62)

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 = kakkbka·b ,

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

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

(63)

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 = kakkbka·b ,

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

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

(64)

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 = kakkbka·b ,

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

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

(65)

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 = kakkbka·b ,

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

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

(66)

## Thank you.

Department of Mathematics National Cheng Kung

W EN -C HING L IEN Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

W EN -C HING L IEN Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

On the other hand, by (22), the triangle inequality, and the definition of the operator norm... On the other hand, by (22), the triangle inequality, and the definition of the

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

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Every convergent sequence is bounded.. W EN -C HING L IEN Advanced

Department of Mathematics National Cheng Kung

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W EN -C HING L IEN Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

W EN -C HING L IEN Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung University.. 2009.. It follows that.. It follows that.. It follows that.. It follows that.. It follows that.. It follows that.. It

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