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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch8: Euclidean Spaces

### 8.2: Planes and Linear Transformations

Definition (8.12)

A function T :Rn → Rm is said to be linear (notation:

T ∈ L(Rn;Rm)) if and only if it satisfies

T (x + y) = T (x) + T (y) and T (αx) = αT (x) for allx, y ∈ Rn and all scalars α.

Notations:x = (x1,x2, . . . ,xn) ∈Rn. [x] = [x1x2 · · · xn]

(3)

Remark (8.13)

Ifx, y ∈ Rn and α is a scalar, then

[x + y] = [x] + [y], [x · y] = [x][y]T, and [αx] = α[x].

(4)

Remark (8.14)

Let B = [bij]be an m × n matrix whose entries and real numbers, and lete1, . . . ,en represent the usual basis of Rn. If

(6) T (x) = Bx, x ∈ Rn,

then T is a linear function fromRntoRm and

(7) T (ej) = (b1j,b2j, . . . ,bmj), j = 1, 2, . . . , n.

(5)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence,suppose that T ∈ L(Rn;Rm). Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(6)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14.To prove existence, suppose that T ∈ L(Rn;Rm).Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(7)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence,suppose that T ∈ L(Rn;Rm). Define B by (7).Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(8)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence, suppose that T ∈ L(Rn;Rm).Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(9)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence, suppose that T ∈ L(Rn;Rm). Define B by (7).Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej)=

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(10)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence, suppose that T ∈ L(Rn;Rm). Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(11)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence, suppose that T ∈ L(Rn;Rm). Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej)=

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(12)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence, suppose that T ∈ L(Rn;Rm). Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(13)

Theorem (8.15)

For each T ∈ L(Rn;Rm)there is a matrix B = [bij]m×n such that (6) holds. Moreover, the matrix B is unique.

Specifically, for each fixed T there is only one B that satisfies (6), and the entries of that B are defined by (7).

Proof.

Uniqueness has been established in Remark 8.14. To prove existence, suppose that T ∈ L(Rn;Rm). Define B by (7). Then

T (x) = T

n

X

j=1

xjej

!

=

n

X

j=1

xjT (ej) =

n

X

j=1

xj(b1j,b2j, . . . ,bmj)

=

n

X

j=1

xjb1j,

n

X

j=1

xjb2j, . . . ,

n

X

j=1

xjbmj

!

=Bx.

(14)

Definition (8.16)

Let T ∈ L(Rn;Rm). The operator norm of T is the extended real number

kT k := inf {C > 0 : kT (x)k ≤ C kxk for all x ∈ Rn} one interesting corollary of Theorem 8.15 is that the operator norm of a linear function is always finite.

(15)

Theorem (8.17)

Let T ∈ L(Rn;Rm). Then the operator norm of T is finite, and satisfies

(8) kT (x)k ≤ kT k kxk

for allx ∈ Rn. Proof.

Let B be the m × n matrix that represents T , and suppose that the rows of T are given byb1, . . . ,bm.By the

definition of matrix multiplication and our identification of Rm with m × 1 matrices,

T (x) = (b1· x, . . . , bm· x).

(16)

Theorem (8.17)

Let T ∈ L(Rn;Rm). Then the operator norm of T is finite, and satisfies

(8) kT (x)k ≤ kT k kxk

for allx ∈ Rn. Proof.

Let B be the m × n matrix that represents T , and suppose that the rows of T are given byb1, . . . ,bm.By the

definition of matrix multiplication and our identification of Rm with m × 1 matrices,

T (x) = (b1· x, . . . , bm· x).

(17)

Theorem (8.17)

Let T ∈ L(Rn;Rm). Then the operator norm of T is finite, and satisfies

(8) kT (x)k ≤ kT k kxk

for allx ∈ Rn. Proof.

Let B be the m × n matrix that represents T , and suppose that the rows of T are given byb1, . . . ,bm.By the

definition of matrix multiplication and our identification of Rm with m × 1 matrices,

T (x) = (b1· x, . . . , bm· x).

(18)

Proof.

If B = O,then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2

and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(19)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2

and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(20)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2

and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(21)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2

and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(22)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2

and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(23)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(24)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(25)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(26)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0),it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(27)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(28)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0),it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(29)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(30)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(31)

Proof.

If B = O, then kT k = 0 and (8) is an equality.If B 6= O, then by the Cauchy-Schwarz Inequality, the square of the Euclidean norm of T (x) satisfies

kT (x)k2 = (b1· x)2+ . . . + (bm· x)2

≤ (kb1k kxk)2+ . . . + (kbmk kxk)2

≤ m · max{kbjk2:1 ≤ j ≤ m} kxk2 =:C kxk2 and C > 0.Thus the set defining kT k is nonempty. Since it is bounded below (by 0), it follows from the Completeness Axiom that kT k exists and is finite.In particular, there are Ck >0 such that Ck ↓ kT k and kT (x)k ≤ Ckkxk for all x ∈ Rn.Taking the limit of this last inequality as k → ∞, we obtain (8).

(32)

## Thank you.

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

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

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

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