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Text Contents 읽기(선형대수학 - Overtake class 강의자료2)

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(1)

2012 Spring semester

Linear Algebra

Overtake class

(Section 3.4 - Section 3.6)

(2)

(1) Subspaces of

R

n

Definition

W (= ∅) ⊂ Rn is called a subspace of Rn if it is closed under addition and scalar multiplication.

Examples (Subspaces ofRn)

Lines and planes through o of Rn {o} : zero space

Rn

For v1, . . . , vs ∈ Rn,

{x = t1v1+· · · + tsvs|t1, . . . , ts ∈ R} is a subspace of Rn.

(3)

(2) Span

Definition Let v1, . . . , vs ∈ Rn. span{v1, . . . , vs} := {t1v1+· · · + tsvs|t1, . . . , ts ∈ R} : thespan of v1, . . . , vs Examples

(o) = span{o}, Rn= span{e1, . . . , en}

span{v} : line through o of Rn span{v1, v2} : plane through o of Rn

(4)

(3) Solution space of a linear system

Theorem

Let A be an m×n matrix. Then {x ∈ Rn|Ax = o} is a subspace of Rn, which is called the solution spaceof the system. • R2

Homo. lin. Solution system space

(o) line through o

R2

• R3

Homo. lin. Solution system space

(o) line through o plane through o

(5)

(3) Solution space of a linear system (continued)

Theorem

(a) For an m × n matrix A,

{x ∈ Rn|Ax = o} = Rn ⇔ A = O.

(b) For m × n matrices A, B,

A = B ⇔ Ax = Bx, ∀x ∈ Rn. (Proof.)

(6)

(4) Linear independence

The geometric properties of a subspace x = t1v1+· · · + tsvs are affected by interrelationships among the vectors v1, . . . , vs.

Definition

S(= ∅) = {v1, . . . , vs} ⊂ Rn is said to be linearly indepen-dent if

c1v1+· · · + csvs = o ⇒ c1=· · · = cs = 0.

(7)

(4) Linear independence (continued)

Examples

v is linearly dependent if v = o; v is linearly independent if v= o.

For S(= ∅) ⊂ Rn, if o∈ S then S is linearly dependent.

Theorem

S = {v1, . . . , vs} ⊂ Rn (s ≥ 2) is linearly independent

⇔ At least one of the vectors in S is expressible as a linear combination of the other vectors in S.

(8)

(5) Linearly independence and Homogeneous linear

systems

Theorem

Ax = o has only the trivial solution

⇔ the column vectors of A are linearly independent.

Examples

Determine whether the given vectors are linearly independent or not.

v1 = (1, 2, 1), v2 = (2, 5, 0), v3 = (3, 3, 8) v1 = (1, 2, −1), v2 = (6, 4, 2), v3 = (4, −1, 5)

(9)

(5) Linearly independence and Homogeneous linear

systems (continued)

Example

v1 = (2, −4, 6), v2 = (0, 7, −5), v3 = (6, 9, 8), v4 = (5, 0, 1)

Theorem

(10)

(6) A unifying theorem

Theorem

For an n × n matrix A, TFAE: (a) RREF (A) = In.

(b) A = E1· · · Ek, Ei : elementary matrix. (c) A is invertible.

(d) Ax = o has only the trivial solution. (e) Ax = b is consistent for every b ∈ Rn.

(f) Ax = b has exactly one solution for every b ∈ Rn. (g) The column vectors of A are linearly independent. (h) The row vectors of A are linearly independent.

(11)

(7) The relationship between

Ax = b and Ax = o

For x0, v1, . . . , vs ∈ Rn, let W = span{v1, . . . , vs}.

x0+ W = {x0+ w|w ∈ W }

is called the translation of W by x0.

Theorem

If Ax = b is consistent (b = o) and W = {x|Ax = o} then {x|Ax = b} = x0+ W ={x0+ xh|Axh= o}

(12)

(7) The relationship between

Ax = b and Ax = o

(continued)

Solution sets inR2 Solution sets in R3 A point A point

A line A line

R2 A plane

R3 Theorem

Let A be an m × n matrix. Then Ax = o has only the trivial solution

⇔ Ax = b has at most one solution for every b ∈ Rm.

A nonhomogeneous linear system with more unknowns than equations is either inconsistent or has infinitely many solutions.

(13)

(8) Consistency of a linear system from the vector point

of view

Theorem

For an m × n matrix A,

Ax = b is consistent ⇔ b ∈ col(A) : column space of A.

Examples

Is w = (9, 1, 0) a linear combination of v1 = (1, 2, 3), v2 = (1, 4, 6), v3 = (2, −3, −5)?

(14)

(9) Hyperplanes, Geometric interpretation of solution

spaces

Definition For a1, . . . , an, b ∈ R, {(x1, . . . , xn)∈ Rn|a1x1+· · · + anxn= b} ={x ∈ Rn|a · x = b} (a = o) : ahyperplanein Rn

a:={x ∈ Rn|a · x = 0} : theorthogonal complementof a

Theorem

Let A be an m × n matrix. Then

(15)

(10) Diagonal matrix

D = ⎡ ⎢ ⎢ ⎢ ⎣ d1 O d2 . . . O dn ⎤ ⎥ ⎥ ⎥ ⎦, D−1 = ⎡ ⎢ ⎢ ⎢ ⎣ 1/d1 O 1/d2 . .. O 1/dn ⎤ ⎥ ⎥ ⎥ ⎦ Dk = ⎡ ⎢ ⎢ ⎢ ⎣ d1k O d2k . .. O dnk ⎤ ⎥ ⎥ ⎥

(true for k < 0 if D is invertible.) A diagonal matrix is invertible ⇔ its main diagonal entries are all nonzero.

(16)

(11) Triangular matrix

A square matrix in which all entries above (below, resp.) the main diagonal are zero is said to be lower (upper, resp.) triangular.

A is upper (lower, resp.) triangular ⇒ AT is lower (upper, resp.) triangular.



(upper triangular)=upper triangular, 

(lower triangular)=lower triangular

A triangular matrix is invertible ⇔ its main diagonal entries are all nonzero.

A is invertible upper (lower, resp.) triangular ⇒ A−1 is upper (lower, resp.) triangular.

(17)

(12) Symmetric and Skew-symmetric matrices

A square matrix A is said to be symmetricif AT = A, and skew-symmetric if AT =−A.

Let A, B be n × n symmetric matrices. Then AT, A + B, A − B, cA are symmetric. AB is symmetric ⇔ AB = BA

If A is an invertible symmetric matrix then A−1 is symmetric. [Note] For any m × n matrix A, AAT and ATA are symmetric.

(18)

(13) Inverting

I − A when A is nilpotent

Let A be an n × n matrix. x ∈ Rn is called the fixed point of

A if Ax = x i.e., (I − A)x = o.

Theorem

For a square matrix A, if Ak = O for some positive integer k then I − A is invertible and

(I − A)−1 = I + A + · · · + Ak−1.

A square matrix A is said to be nilpotentif Ak = O for some positive integer k.

(19)

(14) Inverting

I − A by power series

[Observation] For 0< x < 1,

lim

k→+∞(1− x)(1 + x + x

2+· · · + xk) = 1.

For a square matrix A,

(I − A)(I + A + A2+· · · ) = I.

Theorem

For an n × n matrix A = [aij], if ni=1|aij| < 1 (or nj=1|aij| <

1) for each j = 1, . . . , n (or i = 1, . . . , n) then I −A is invertible and

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