Theorem 3.3: rank(T

Page 1

Chapter 3 (2)

Review: Elementary matrix operation and elementary matrix

⌅ w = T (v) turns into y = Ax, a system of linear equations.

type 1:

0

@1 0 0 0 0 1 0 1 0

1

A type 2:

0

@-2 0 0 0 1 0 0 0 1

1

A type 3:

0

@1 0 0 0 1 -2 0 0 1

1 A

횵 ^! ;

Page 2

Review: Rank and inverse of matrix

⌅ ⌅ rank of a matrix A, rank(A): rank(L_A)

⌅ ⌅ Theorem 3.3: rank(T ) = rank([T ] )

⌅ ⌅ Theorem 3.4: A 2 Mm⇥n; P 2 Mm⇥m; Q 2 Mn⇥n; P and Q are invertible. Then

rank(A) = rank(P A) = rank(AQ) = rank(P AQ) ⇤

⌅ ⌅ Theorem 3.5: rank(A) = dim(column space of A). In other words, the rank of a matrix A is the maximum number of linearly indepen- dent columns of A.

⌅ ⌅ Theorem 3.6: An m ⇥ n matrix A of rank r can be transformed into D =

✓ I_r O₁ O₂ O₃

◆

by elementary row and column operations, where O_i are zero matrices.

res www.op^.

一潚

, column ^{de .} _Op ^.

Y ^{= AK = E7G}di ^e _span ³

ois

=R ( LA)

Page 3

⌅ ⌅ Corollary 3.6.1: D =

✓ I_r O₁ O₂ O₃

◆

= BAC for some invertible matri- ces B and C. This is from that D = E_pE_{p 1} · · · E1AG₁G₂ · · · G^p, where E_i and G_j are elementary matrices.

⌅ ⌅ matrix inversion by Gaussian elimination:

(A|Iⁿ) ! (Iⁿ|B) by eros [not by ecos]

) A = B ¹

⌅ ⌅ We can find the inverse of a linear transformation T : V ! V through computing the inverse of its matrix representation in a basis .

T ! [T ] ! [T ] ¹ ! [T ¹] ! T ¹

○

^Singer

B ⁼ A

t.int

Page 4

System of linear equations

⌅ ⌅ system of linear equations:

a₁₁x₁ + a₁₂x₂ + · · · + a1nx_n = b₁ a₂₁x₁ + a₂₂x₂ + · · · + a2nx_n = b₂

· · ·

a_m1x₁ + a_m2x₂ + · · · + a^mnx_n = b_m A =

0

@ a₁₁ · · · a1n

... ...

a_m1 · · · a^mn 1

A, x = 0

@x₁ ...

x_n 1

A, b = 0

@b₁ ...

b_m 1

A ) Ax = b ⇤

⌅ Recall that this can be considered an n-tuple-matrix representation of w = T (v).

any ^Linear Transformation

Page 5

⌅ solution to the system: s such that As = b ⇤

⌅ solution set to the system: {s 2 Fⁿ : As = b} ⇤

⌅ ⌅ example: x₁ + x₂ + x₃ = 0, x₁ + x₂ x₃ = 2 )

✓1 1 1 1 1 1

◆ 0

@x₁ x₂ x₃

1

A = ✓ 0 2

◆

) x1 + x₂ = 1, x₃ = 1 ) solution set = {(x1, x₂, 1)^t : x₁ + x₂ = 1}

f = {(t, 1 t, 1)^t : t 2 IR}

f =

(0

@ 0 1

1 1

A + t 0

@ 1 1 0

1

A: t 2 IR )

The last two are parametric expressions.

?

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tf

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김 ⁼ 1 ^-E

Not ^vs

coat

b_EN

SE V

Et

NA⁾ _.

23.5

Page 6

⌅ ⌅ homogeneous system: Ax = 0 ⇤

⌅ nonhomogeneous system: Ax = b, b 6= 0 ⇤

⌅ A homogeneous system Ax = 0 always has a solution x = 0, called the trivial solution.

⌅ ⌅ Theorem 3.8: Ax = 0.A 2 Mm⇥n, and K is the solution set.

Then K = N(L_A) ⇤

30, ^et

/

-

t.cl

시에Space

Page 7

⌅ The solution set K of an m ⇥ n homogeneous system is a subspace of F_n of dimension

n - rank(L_A) = n - rank(A).

⌅ ⌅ Corollary 3.8: A 2 Mm⇥n, m < n ) Ax = 0 has a nonzero solution. ⇤

proof: L_A : Fⁿ ! F^m

) rank(LA) = rank(A)  m

) dim(K) = nullity(LA) = n - rank(L_A) n m > 0 ⇤ 仁{ 桁些

_f

Page 8

⌅ ⌅ Theorem 3.9: K is the solution set of Ax = b; K_H is the solution set of Ax = 0; and s 2 K.

Then K = {s} + KH = {s + k : k 2 KH}. ⇤ proof: (i) w 2 K

) A(w s) = Aw As = b b = 0 ) w s 2 KH

) w = s + (w s) 2 {s} + KH

) K ✓ {s} + KH

(ii) w 2 {s} + K_H ) w = s + k, k 2 KH

) Aw = As + Ak = b + 0 = b ) w 2 K ) {s} + KH ✓ K ⇤

ritotftrbF0KH-EKF-stK@0tlineofpfAi.B

WEA ⇒ ^WEB _, ^WEB ⇒ ^W Et

Page 9

⌅ example: x₁ + x₂ + x₃ = 0, x₁ + x₂ x₃ = 2

(0, 1, 1)^t is a solution to the nonhomogeneous system.

homogeneous system: x₁ + x₂ + x₃ = 0, x₁ + x₂ x₃ = 0 ) x1 + x₂ = 0, x₃ = 0

) KH = {(t, t, 0)^t : t 2 IR} = {t(1, 1, 0)^t : t 2 IR}

) K = {(0, 1, 1)^t} + KH =

(0

@ 0 1

1 1

A + t 0

@ 1 1 0

1

A: t 2 IR )

(3, 2, 1)^t is another solution to the nonhomogeneous sys.

) K = {(3, 2, 1)^t} + KH =

(0

@ 3 2 1

1

A + t 0

@ 1 1 0

1

A: t 2 IR )

⇒ 죄= t _, ^{X2= - t ,} 石二⁰

€

41

E)

Page 10

⌅ example: x₁ 2x₂ + x₃ = 4

(4, 0, 0)^t is a solution to the nonhomogeneous system.

) homogeneous system: x1 2x₂ + x₃ = 0 ) x3 = x₁ + 2x₂

) x1 = t₁, x₂ = t₂, x₃ = t₁ + 2t₂: parameterization ) KH = {t1(1, 0, 1)^t + t₂(0, 1, 2)^t : t₁, t₂ 2 IR}

) K = {(4, 0, 0)^t} + KH

f = {(4, 0, 0)^t + t₁(1, 0, 1)^t + t₂(0, 1, 2)^t : t₁, t₂ 2 IR}

쟜쓺

TS ⁼ 2t 一打

徽喇

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Not Vector Space

Page 11

⌅ ⌅ Theorem 3.10: A is invertible

f , Ax = b has only one solution A ¹b. ⇤

⌅ ⌅ Theorem 3.11: Ax = b has a solution f , rank(A) = rank(A|b). ⇤

proof: Ax = b has a solution s = (s₁, · · · , sⁿ)^t. , b 2 R(LA) [b = As = L_A(s)]

, b 2 span({a1, · · · , aⁿ}) = column space

f [a_i is the ith column of A; a₁s₁ + · · · + aⁿs_n = b] , span({a1, · · · , aⁿ}) = span({a1, · · · , aⁿ, b}) ⇤

⌅ example: x₁ + x₂ = 0, x₁ + x₂ = 1 A =

✓1 1 1 1

◆

, (A|b) =

✓1 1 0 1 1 1

◆

rank(A) = 1 6= rank(A|b) = 2 ) no solution

(^{a and}叫剡

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b= AS ⁼

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⇒ ^te column 수 같은 _{.ci b} ^든

of

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虹一

Page 12

⌅ example: x₁ + x₂ = 0, x₁ x₂ = 1 A =

✓1 1 1 1

◆

, (A|b) =

✓1 1 0 1 1 1

◆

rank(A) = rank(A|b) = 2 ) solution

Page 13

⌅ ⌅ equivalent systems of linear equations: systems with the same solution set ⇤

⌅ ⌅ Theorem 3.13: C is invertible

) CAx = Cb is equivalent to Ax = b. ⇤ proof: K is the solution set for Ax = b;

K⁰ is the solution set for CAx = Cb.

(i) w 2 K ) Aw = b ) CAw = Cb ) w 2 K⁰ ) K ✓ K⁰

(ii) w 2 K⁰ ) CAw = Cb ) C ¹CAw = C ¹Cb ) Aw = b ) w 2 K ) K⁰ ✓ K ⇤

Show

談威

Page 14

⌅ ⌅ Corollary 3.13: (A|b) ! (A⁰|b⁰) by eros ) Ax = b and A⁰x = b⁰ are equivalent. ⇤

proof: C = E_p · · · E1, where E_i is and elementary matrix.

(A⁰|b⁰) = C(A|b) = (CA|Cb) ) A⁰ = CA, b⁰ = Cb

) CAx = Cb is equivalent to Ax = b. [Thm 3.13] ⇤

⌅ ⌅ example:

x₁ + 2x₂ x₃ = 1f x₁ = 4 2x₁ + 2x₂ + x₃ = 1f ) x2 = 3 3x₁ + 5x₂ 2x₃ = 1f x₃ = 1 0

@1 2 1 1 2 2 1 1

3 5 2 1

1 A ^3,3!

0

@1 2 1 1

0 2 3 3

0 1 1 2

1 A !¹

0

@1 2 1 1

0 1 1 2

0 2 3 3

1 A

- 2 X

+ELn

Page 15

!2

0

@1 2 1 1

0 1 1 2

0 2 3 3

1 A ^3,3!

0

@1 0 1 3

0 1 1 2

0 0 1 1

1 A ^3,3!

0

@1 0 0 4 0 1 0 3 0 0 1 1

1 A

) 0

@1 0 0 0 1 0 0 0 1

1

A x = 0

@ 4 3 1

1

A ( EAx = Eb)

) x = 0

@ 4 3 1

1 A

(

1끄

Page 16

⌅ ⌅ example:

3x₁+ 2x₂+ 3x₃ 2x₄ = 1f x₁+ x₃ = 1f x₁ = t, x₃ = 1 t x₁ + x₂ + x₃ = 3f ! x2 = 2f ! x2 = 2

x₁ + 2x₂ + x₃ x₄ = 2f x₄ = 3f x₄ = 3

) 0 BB

@ x₁ x₂ x₃ x₄

1 CC A =

0 BB

@ 0 2 1 3

1 CC

A + t 0 BB

@ 1 0

1 0

1 CC A 2

(0 BB

@ 0 2 1 3

1 CC A

)

+ K_H

0

@3 2 3 2 1 1 1 1 0 3 1 2 1 1 2

1 A !¹

0

@1 1 1 0 3 3 2 3 2 1 1 2 1 1 2

1 A ^3,3!

0

@1 1 1 0 3

0 1 0 2 8

0 1 0 1 1

1 A

!2

0

@1 1 1 0 3 0 1 0 2 8 0 1 0 1 1

1 A ^3,3!

0

@1 0 1 2 5 0 1 0 2 8 0 0 0 3 9

1 A !²

0

@1 0 1 2 5 0 1 0 2 8 0 0 0 1 3

1 A

9

8

15

^品

Page 17

!3,3

0

@1 0 1 0 1 0 1 0 0 2 0 0 0 1 3

1 A

) 0

@1 0 1 0 0 1 0 0 0 0 0 1

1 A

0 BB

@ x₁ x₂ x₃ x₄

1 CC A =

0

@1 2 3

1

A ( EAx = Eb)

) x1 + x₃ = 1f x₁ = t, x₃ = 1 t f x₂ = 2f ! x2 = 2

f x₄ = 3f x₄ = 3

) 0 BB

@ x₁ x₂ x₃ x₄

1 CC A =

0 BB

@ 0 2 1 3

1 CC

A + t 0 BB

@ 1 0

1 0

1 CC A 2

(0 BB

@ 0 2 1 3

1 CC A

)

+ K_H

Page 18

⌅ ⌅ reduced row echelon form, rref:

1. Nonzero row precedes all-zero rows.

2. The first nonzero entry in a row is the only nonzero entry in its column (key column, pivot column).

3. The first nonzero entry in a row is 1 and moves right as we go down.

⌅ It is obtained by only elementary row operations.

⌅ If a matrix is invertible, its rref is the identity matrix.

⌅ ⌅ example: rref ^↳ 屯們

Page 19

⌅ ⌅ Gaussian elimination: conversion of a matrix into a rref by eros.

⇤

⌅ finding the rref: A ^Ge! B, B = rref(A)

⌅ inverting a matrix: (A|I) ^Ge! (I|A ¹) [invertible]

⌅ solving a system of linear equations:

(A|b) ^Ge! (I|s) [unique]

(A|b) ^Ge! (B|s) [general solution using null vectors]

→ B

K-군

WAY

If

-

Page 20

⌅ ⌅ Theorem 3.15: Ax = b, A 2 Mm⇥n, with the solution set K;

rank(A) = rank(A|b); (A|b) is in rref; and (A|b) has r nonzero row.

1. rank(A) = rThen

2. The general solution has the form of f s = s₀ + t₁u₁ + · · · + t^{n r}u_{n r}

f ) s0 2 K and {u1, · · · , un r} is a basis for KH. ⇤ proof:“1”: rref ) Nonzero rows are linearly independent.

f ) r = rank(A|b) = rank(A)

“2”: Let t₁ = · · · = t^{n r} = 0 ) s = s0 ) s0 2 K ) KH = {s0} + K [Thm 3.9]

f = {t1u₁+· · ·+tn ru_{n r} : t_i 2 F } = span({u1, · · · , un r}) ) {u1, · · · , u^{n r}} is a basis for KH (* rank(A) = r and so dim(K_H) = n r). ⇤

⇒

nat

:

particle

?

Sole ^- Null _Space

Page 21

⌅ ⌅ Theorem 3.16: A 2 Mm⇥n; rank(A) = r > 0; and B = rref(A).

Then

1. B has r nonzero row.

2. B has r key columns that are e₁, · · · , e^r.

3. The key columns of A are linearly independent.

4. If column k of B is a linear combination of the key columns of B, then column

f k of A is the same linear combination of the key columns of A.

f b_k = d₁b_j1 + · · · + d^rb_jr ) ak = d₁a_j1 + · · · + d^ra_jr ⇤

A

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Page 22

⌅ ⌅ Corollary 3.16: The rref of a matrix is unique. ⇤ proof:

if A has two rrefs B and C, then either

(1) they have different sets of key columns, or (2) they have the same set of key columns.

(1) ) contradicts to Thm 3.16 - 3, because

by Thm 3.16 - 3, the set of key columns of A is the set of linearly independent columns, and

the set of linearly independent vectors with the lowest indices is unique.

(2) ) b_k 6= c_k for some k [non-key column k]

f ) b_k and c_k are the same lin comb of the key columns f ) contradiction [Thm 3.16 - 4] ⇤

une

④ aaa.at 斷門

① ② ③ ①② ③

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Page 23

⌅ ⌅ example: finding linearly independent vectors using rref:

v₁ = 2 + x + 2x² + 3x³ v₂ = 4 +2x + 4x²+ 6x³ v₃ = 6 +3x + 8x²+ 7x³ v₄ = 2 + x + 5x³

v₅ = 4 + x + 9x³

! 0 BB

@

2 4 6 2 4 1 2 3 1 1 2 4 8 0 0 3 6 7 5 9

1 CC A !

0 BB

@

1 2 0 4 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0

1 CC A

) v1, v₃, v₅ are linearly independent. [key columns]

Of course there are other linearly independent sets, such as {v1, v₄, v₅} and {v2, v₃, v₅}

Page 24

⌅ ⌅ example: extending linearly independent vectors into a basis using rref:

0 BB

@

2 6 0 1 3 2 2 8 4 3 7 6

1 CC A !

0 BB

@

2 6 0 1 0 0 0 1 3 2 0 1 0 0 2 8 4 0 0 1 0 3 7 6 0 0 0 1

1 CC A !¹

0 BB

@

1 3 2 0 1 0 0 2 6 0 1 0 0 0 2 8 4 0 0 1 0 3 7 6 0 0 0 1

1 CC A

3,3,3

!

0 BB

@

1 3 2 0 1 0 0 0 0 4 1 2 0 0 0 2 0 0 2 1 0 0 2 0 0 3 0 1

1 CC A

1,2! 0 BB

@

1 3 2 0 1 0 0 0 1 0 0 1 ¹₂ 0 0 0 4 1 2 0 0 0 2 0 0 3 0 1

1 CC A

4th

Key

Page 25

!3,3

0 BB

@

1 0 2 0 4 ³₂ 0 0 1 0 0 1 ¹₂ 0 0 0 4 1 2 0 0 0 0 0 0 5 1 1

1 CC A !²

0 BB B@

1 0 2 0 4 ³₂ 0 0 1 0 0 1 ¹₂ 0 0 0 1 ¹₄ ¹₂ 0 0 0 0 0 0 5 1 1

1 CC CA

!3,2

0 BB B@

1 0 0 ¹₂ 3 ³₂ 0 0 1 0 0 1 ¹₂ 0 0 0 1 ¹₄ ¹₂ 0 0 0 0 0 0 1 ¹₅ ¹₅

1 CC CA

3,3,3

!

0 BB B@

1 0 0 ¹₂ 0 ₁₀^{9 3}₅ 0 1 0 0 0 ₁₀^{3 1}₅ 0 0 1 ¹₄ 0 ₁₀¹ ₁₀¹ 0 0 0 0 1 ¹₅ ¹₅

1 CC CA

) (0, 1, 0, 0)^t can be added to form a basis.

⌅ This method of extending a set into a basis is especially useful in finding a basis using a computer program.

4-th

Key

Page 1

Chapter 4 Review

⌅ ⌅ Corollary 3.6.1: D =

✓ I_r O₁ O₂ O₃

◆

= BAC for some invertible matri- ces B and C. This is from that D = E_pE_{p 1} · · · E1AG₁G₂ · · · Gp, where E_i and G_j are elementary matrices.

⌅ ⌅ matrix inversion by Gaussian elimination:

(A|Iⁿ) ! (Iⁿ|B) by eros [not by ecos]

) A = B ¹ E(A|Iⁿ) = (I_n|B) Question:

(I|A)G = (B|I)!IG = B, AG = I ! AB = I ! A = B ¹

✓A I

◆

G =

✓I B

◆

!IG = B, AG = I ! AB = I ! A = B ¹

Page 2

⌅ ⌅ homogeneous system: Ax = 0.

⌅ nonhomogeneous system: Ax = b, b 6= 0.

⌅ ⌅ Theorem 3.8: Ax = 0. A 2 Mm⇥n, and K is the solution set.

Then K = N(L_A).

⌅ ⌅ Theorem 3.9: K is the solution set of Ax = b; K_H is the solution set of Ax = 0; and s 2 K.

Then K = {s} + KH = {s + k : k 2 KH}.

⌅ example: x₁ 2x₂ + x₃ = 4

(4, 0, 0)^t is a solution to the nonhomogeneous system.

) homogeneous system: x1 2x₂ + x₃ = 0 ) x3 = x₁ + 2x₂

) x1 = t₁, x₂ = t₂, x₃ = t₁ + 2t₂: parameterization ) KH = {t1(1, 0, 1)^t + t₂(0, 1, 2)^t : t₁, t₂ 2 IR}

) K = {(4, 0, 0)^t} + KH

f = {(4, 0, 0)^t + t₁(1, 0, 1)^t + t₂(0, 1, 2)^t : t₁, t₂ 2 IR}

Page 3

⌅ ⌅ Theorem 3.10: A is invertible

f , Ax = b has only one solution A ¹b.

⌅ ⌅ Theorem 3.11: Ax = b has a solution f , rank(A) = rank(A|b).

⌅ ⌅ Theorem 3.13: C is invertible

) CAx = Cb is equivalent to Ax = b.

⌅ ⌅ Corollary 3.13: (A|b) ! (A⁰|b⁰) by eros ) Ax = b and A⁰x = b⁰ are equivalent.

⌅ ⌅ reduced row echelon form, rref:

v₁ = 2+x+2x²+3x³ v₂ = 4+2x+4x²+6x³ v₃ = 6+3x+8x²+7x³ v₄ = 2 + x + 5x³

v₅ = 4 + x + 9x³

! 0 BB

@

2 4 6 2 4 1 2 3 1 1 2 4 8 0 0 3 6 7 5 9

1 CC A !

0 BB

@

1 2 0 4 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0

1 CC A

) v1, v₃, v₅ are linearly independent. [key columns]

Page 4

⌅ ⌅ Gaussian elimination: conversion of a matrix into a rref by eros. ⇤

⌅ finding the rref: A ^Ge! B, B = rref(A)

⌅ inverting a matrix: (A|I) ^Ge! (I|A ¹) [invertible]

⌅ solving a system of linear equations:

(A|b) ^Ge! (I|s) [unique]

(A|b) ^Ge! (B|s) [general solution using null vectors]

⌅ ⌅ example:

0

@3 2 3 2 1 1 1 0 1 2 1 1

1 A

0 BB

@ x₁ x₂ x₃ x₄

1 CC A =

0

@1 3 2

1 A

Page 5

0

@3 2 3 2 1 1 1 1 0 3 1 2 1 1 2

1

A ^eroc! 0

@1 0 1 0 1 0 1 0 0 2 0 0 0 1 3

1 A

) 0

@1 0 1 0 0 1 0 0 0 0 0 1

1 A

0 BB

@ x₁ x₂ x₃ x₄

1 CC A =

0

@1 2 3

1

A ( (rref)EAx = Eb)

) x1 + x₃ = 1f x₁ = t, x₃ = 1 t f x₂ = 2f ! x2 = 2

f x₄ = 3f x₄ = 3

) 0 BB

@ x₁ x₂ x₃ x₄

1 CC A =

0 BB

@ 0 2 1 3

1 CC

A + t 0 BB

@ 1 0

1 0

1 CC A 2

(0 BB

@ 0 2 1 3

1 CC A

)

+ K_H

Page 6

⌅ ⌅ Theorem 3.15: Ax = b, A 2 Mm⇥n, with the solution set K;

rank(A) = rank(A|b); (B|s) is in rref; and (A⁰|b⁰) has r nonzero row. Then

1. rank(A) =rank(A⁰)= r

2. The general solution has the form of f s = s₀ + t₁u₁ + · · · + t^{n r}u_{n r}

⌅ ⌅ Theorem 3.16: A 2 M^m_⇥n; rank(A) = r > 0; and B = rref(A).

1. B has r nonzero row.Then

2. B has r key columns that are e₁, · · · , e^r.

3. The key columns of A are linearly independent.

4. If column k of B is a linear combination of the key columns of B, then column

f k of A is the same linear combination of key columns of A.

f b_k = d₁b_j1 + · · · + d^rb_jr ) ak = d₁a_j1 + · · · + d^ra_jr

Page 7

Determinant

⌅⌅ Determinant is a number that represents a square matrix.

⌅ determinant of a 2⇥2 matrix: det

✓ a b c d

◆

= ad bc.

⌅ det(A + B) 6= det(A) + det(B), eg, det

✓ ✓ 1 0 0 1

◆ +

✓ 1 0 0 1

◆ ◆

= 4 6= 2.

det(cA) 6= c det(A), eg, det

✓ 3

✓ 1 0 0 1

◆ ◆

= 9 6= 3.

) Determinant is not linear.

⌅ The determinant is a linear function of one row, when the other rows are held fixed, which is explained in Theorem 4.1. This is true for any square matrix, but Theorem 4.1 illustrates the case of 2⇥2 matrix.

Page 8

⌅⌅ Theorem 4.1: u₁, u₂, v are row vectors of 2⇥2 matrices; k1, k₂ are scalars. Then

det

✓ k₁u₁ + k₂u₂ v

◆

= k₁ det

✓ u₁ v

◆

+ k₂ det

✓ u₂ v

◆

det

✓ v

k₁u₁ + k₂u₂

◆

= k₁ det

✓ v u₁

◆

+ k₂ det

✓ v u₂

◆

⌅

⌅⌅ Theorem 4.2: For a 2⇥2 matrix A =

✓ A₁₁ A₁₂ A₂₁ A₂₂

◆ , 1. det(A) 6= 0 , A is invertible.

2. A is invertible , A ¹ = _det(A)¹

✓ A₂₂ A₁₂ A₂₁ A₁₁

◆

⌅ proof: “2” can be verified by multiplying A ¹ to A.

“1”: “)”: det(A) 6= 0 ) A ¹ is given by “2”.

122번^{AnAn -- AutoAn A리}

= de LA)

牆箚

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二工

Page 9

“(”: A is invertible ) rank(A) = 2 ) A11 6= 0 or A21 6= 0 (i) A₁₁ 6= 0 ) A !³ A₁₁ A₁₂

0 A₂₂ ^A12_A^A21

11

!

) A22 A₁₂A₂₁ A₁₁ 6=

0 ) det(A)6= 0

(ii) A₂₁ 6= 0 ) A !³ 0 A₁₂ ^A11_A^A22 A₂₁ A₂₂ 21

!

) A12 A₁₁A₂₂ A₂₁ 6=

0 ) det(A)6= 0 ⌅

淵門 ^만

偕劉

nar ^An

^An )

= 0 A22^-

TF

Page 10

⌅⌅ The area of the parallelogram determined by 2-dim vectors

✓ a b

◆

and

✓ c d

◆

is det

✓ a b c d

◆

= det

✓ a c b d

◆ .

⌅⌅ determinant of A 2 Mⁿ_⇥n, direct definition:

det(A)= P

⇡sgn(⇡)A_1⇡(1)A_2⇡(2) · · · A_n⇡(n), where

⌅ the sum is taken over all permutations ⇡

⌅ k(⇡): the total number of inversions in ⇡

⌅ sgn(⇡) = ( 1)^k(⇡). ⌅

1세

Ain't

^:i剖

A. _? ^{A2? - - - An2}

1 ^{2 ._ . 7}

2 f ^{. . . .} n ² 4 ^{-- 1}

:

Page 11

⌅ example:✓ A₁₁ A₁₂ A₂₁ A₂₂

◆ )

det(A)= A₁₁A₂₂ A₁₂A₂₁

⌅ example:0

@ A₁₁ A₁₂ A₁₃ A₂₁ A₂₂ A₂₃ A₃₁ A₃₂ A₃₃

1 A )

det(A)= A₁₁A₂₂A₃₃ A₁₁A₂₃A₃₂ A₁₂A₂₁A₃₃+A₁₂A₂₃A₃₁+ A₁₃A₂₁A₃₂ A₁₃A₂₂A₃₁

Page 12

⌅⌅ determinant of A 2 Mⁿ_⇥n, recursive definition:

1. n = 1 ) A = (A11) ) det(A)= A11

2. n 2 ) det(A) = P_n

j=1 A_1jc_1j[“1st” row], where c_ij is the cofactor given by c_ij = ( 1)^i+jdet( ˜A_ij), and ˜A_ij is the (n 1) ⇥ (n 1) matrix obtained from A by removing i-th row and j-th

column, ie,

Fth column

2 Cm ^{) 시머)}

← Eth ^ww

Page 13

⌅ example:

det 0

@ 1 2 3 4 5 6 7 8 9

1

A = 1·( 1)¹⁺¹·det

✓ 5 6 8 9

◆

+2·( 1)¹⁺²·det

✓ 4 6 7 9

◆

+3 · ( 1)¹⁺³·det

✓ 4 5 7 8

◆

= (45 48) 2(36 42) + 3(32 35) = 0

⌅ det(I)=1

proof: det(I_n) = ( 1)¹⁺¹det(I_{n 1}) = · · · = ( 1)¹⁺¹det(I₂) = 1

Page 14

⌅⌅ The volume of the parallelepiped determined by 3-dim vectors 0

@ a b c

1 A,

0

@ d e f

1

A and 0

@ g h

i 1 A is

det 0

@ a b c d e f g h i

1

A = det 0

@ a d g b e h c f i

1 A .

⌅⌅ Theorem 4.3: The determinant is a linear function of one row, when the other rows are held fixed. That is,

det 0 BB BB

@

a₁ ...

ku + lv ...

a_n

1 CC CC

A = k det 0 BB BB

@ a₁

...

u...

a_n 1 CC CC

A + l det 0 BB BB

@ a₁

...

v...

a_n 1 CC CC A. ⌅

Page 15

proof: (i) If i = 1,

det 0 BB

@

ku + lv a₂

...

a_n

1 CC

A = P_n

j=1(ku_j + lv_j)c_1j [def]

= k P_n

j=1 u_jc_1j + l P_n

j=1 v_jc_1j

= k det 0 BB

@ u a₂

...

a_n 1 CC

A + ldet 0 BB

@ v a₂

...

a_n 1 CC A.

(ii) If i > 1, we use induction in n. The theorem is already proved for n = 1, 2.

(iii) Assume that it holds for (n 1) ⇥ (n 1) matrices.

Page 16

(iv) det 0 BB BB

@

a₁ ...

ku + lv ...

a_n

1 CC CC

A = P_n

j=1 A_1jc_1j

c_1j = ( 1)^1+jdet( ˜A_1j)

= ( 1)^1+j[kdet( ˜A^(u)_1j ) + ldet( ˜A^(v)_1j ) [assumption]

= kc^(u)_1j + lc^(v)_1j

det 0 BB BB

@

a₁ ...

ku + lv ...

a_n

1 CC CC

A = P_n

j=1 A_ijc_1j = k P_n

j=1 A_1jc^(u)_1j +l P_n

j=1 A_1jc^(v)_1j

h

Page 17

=kdet 0 BB BB

@ a₁

...

u...

a_n 1 CC CC

A + ldet 0 BB BB

@ a₁

...

v...

a_n 1 CC CC A ⌅

⌅⌅ Corollary 4.3: If A has an all-zero row, then det(A)= 0. ⌅ Proof: Set k = l = 0 in Theorem 4.3. ⌅

Page 18

⌅⌅ Lemma 4.4:

det(A) = det 0 BB BB

@

a₁ ...

0 · · · 010 · · · 0 ...

a_n

1 CC CC

A = c_ik = ( 1)^i+kdet( ˜A_ik), where A_ik = 1. ⌅

proof:

(i) If i = 1, det 0 BB

@

0 · · · 010 · · · 0 a₂

...

a_n

1 CC

A = c^1k = ( 1)^1+kdet( ˜A_1k) [def]

(ii) If i > 1, we use induction in n. The lemma holds for n = 2.

(iii) Assume that it holds for (n 1) ⇥ (n 1) matrices.

-

Page 19

(iv) det(A)= P_n

j=1 A_1j( 1)^1+jdet( ˜A_1j) [def]

If j = k, ˜A_1j has all-zero row, and det( ˜A_1j) = 0. [Corol 4.3]

If j 6= k, either j < k or j > k.

det( ˜A_1j) = 8<

:

( 1)(i 1)+(k 1)det(( ˜˜A_1j)_{(i 1)(k 1)}), j < k

( 1)^{(i 1)+k}det(( ˜˜A_1j)_{(i 1)k}), j > k [(iii)]

det(A)= P

j<k A_1j( 1)^1+j( 1)(i 1)+(k 1)det(( ˜˜A_1j)_{(i 1)(k 1)}) + P

j>k A_1j( 1)^1+j( 1)^{(i 1)+k}det(( ˜˜A_1j)_{(i 1)k})

i .tt

ti^{" r}

^to

Page 20

= ( 1)^i+k[P

j<k A_1j( 1)^1+jdet(( ˜˜A_1j)_{(i 1)(k 1)}) + P

j>k A_1j( 1)^{1+(j 1)}det(( ˜˜A_1j)_{(i 1)k})]

= ( 1)^i+k[P

j<k A_1j( 1)^1+jdet(( ˜˜A_ik)_1j) + P

j>k A_1j( 1)^{1+(j 1)}det(( ˜˜A_ik)_{1(j 1)})]

= ( 1)^i+kdet( ˜A_ik) [def] ⌅ A와

AT

"

쏶는

.io ^j 潑茁 _{; !} ;

El^{k j}

凹一 嘲

(K)

Page 21

⌅⌅ Theorem 4.4: det(A) = P_n

j=1( 1)^i+jA_ijdet( ˜A_ij) ⌅

proof: The ith row is a_i = A_i1e₁ + A_i2e₂ + · · · + Aine_n, where e_j = (0, · · · , 0, 1, 0, · · · , 0) with 1 in the j-th place.

By Thm 4.3,

det(A)= A_i1det 0 BB BB

@

a₁ ...

10 · · · 0 ...

a_n

1 CC CC

A + · · · + Aindet 0 BB BB

@

a₁ ...

0 · · · 01 ...

a_n

1 CC CC A By Lemma 4.4,

det(A)= A_i1( 1)ⁱ⁺¹det( ˜A_i1)+ · · · + Ain( 1)^i+nodet( ˜A_in) ⌅

=

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며