1.3 Subspaces in R

1.3 Subspaces in R

R⁰

{0} R¹

0

R² R³

· · ·

Example. Let V ={a(2, 1) ∈ R²|a ∈ R }

V satisfies the 10 properties of the vector space Rⁿ Let u, v, w ∈ V, a, b, c ∈ R

Write u = a(2, 1), v = b(2, 1), w = c(2, 1),

(1) u + v = a(2, 1) + b(2, 1) = (a + b)(2, 1), (a + b ∈ R)

∈ V (3) Take 0(2, 1) ∈ V

then 0(2, 1) = 0 = (0, 0) ∈ V for u ∈ V , u + 0 = u (항등원) (4) for u = a(2, 1) ∈ V

Since −a ∈ R, −u = −a(2, 1) ∈ V So, u + (−u) = 0

(2),(5)-(10) : satisfy!

Definition 1.3.1.

Let W ∈ Rⁿ

W : subspace of Rⁿ, (W < Rⁿ) if

W satisfies the 10-properties of Rⁿ with the same operation +, ·

(W, Rⁿ에서 사용하는 +, · 은 같은 것이다)

Theorem 1.3.2.

Let W ⊂ Rⁿ

Then W : subspace of Rⁿ

⇔ (i) W 6= φ,

(ii) au + bv ∈ W for u, v ∈ W, a, b ∈ R.

(10-properties를 검사하는 것을 대신한다)

idea of proof. (⇒) By proposition 3, there is 0 ∈ W

∴ W 6= φ

Let u, v ∈ W, a, b ∈ R,

then au ∈ W, bv ∈ W by proposition 6 (∵ the same operation 이기에) and au + bv ∈ W by proposition 1

Hence W < Rⁿ

(⇐) (1) Take a = 1, b = 1 then

au + bv ∈ W by assumption

⇒ u + v ∈ W (2) u + (v + w) (in W )

= u + (v + w) (in Rⁿ) 마찬가지로 Rⁿ에도 속하고

= (u + v) + w by proposition 2

= (u + v) + w (in W ) 같은 operation을 사용하기에 Proposition 2에 의해서 바꾼 벡터합 역시 W 에 속한다 (3) 0이 W 에 속함을 보여야 한다.

Since W 6= φ, pick u ∈ W

then take a = 1, b = −1, v = u (임의의 a, b를 적절하게 잡아서 증명) u + (−1)u ∈ W by assumption

⇒ 0 ∈ W (au + bv ∈ W !) (4) Let u ∈ W

then take a = −1, b = 0, then −u ∈ W (5) ∼ (10) : hold!

Example

(1) W₀ = {0} ∈ Rⁿ 0 = (0, · · · , 0)

(1) 0 ∈ W₀ ⇒ W₀ 6= φ

(2) for u, v ∈ W₀, a, b ∈ R then u = 0, v = 0

⇒ au + bv = a · 0 + b · 0 = 0 ∈ W₀ by (1), (2), W₀ < Rⁿ

(2) W ={(a, −a, b) ∈ R³|a, b ∈ R}

(1) Take a = b = 0

(0, 0, 0) ∈ W by definition of W

∴ W 6= φ

(2) Let u = (a, −a, b) ∈ W, v = (c, −c, d) ∈ W, (a, b, c, d ∈ R) for α, β ∈ R,

αu + βv = α(a, −a, b) + β(c, −c, d)

= (αa + βc, −(αa + βc), αb + βd)

∈ W ∵ (αa + βc, αb + βd ∈ R)

∴ W < R³

(3) Let A = {u1, · · · , u_k} ⊂ Rⁿ

V ={v ∈ Rⁿ | v : linear combination of A}

={v ∈ Rⁿ | v = a₁u₁ + · · · + a_ku_k, a_i ∈ R}

then V < Rⁿ sub vector space → subspace

idea of proof. (1) Take a₁ = 1, a₂ = 2, a₃ = 3, · · · , a_k = k ∈ R(아무거나) u₁ + 2u₂ + · · · + ku_k ∈ V by definition of V

∴ V 6= φ (2) α, β ∈ R,

v = a₁u₁ + · · · + a_ku_k, w = b₁u₁ + · · · + b_ku_k, u, w ∈ V αv + βw = (αa₁ + βb₁)u₁ + · · · + (αa_k + βb_k)u_k

∈ V by definition of V

∴ V < Rⁿ

(4)X ={(a, b) ∈ R²|0 ≤ a, b ≤ 1}, V < Rⁿ

X: not subspace

(1, 1) + (1, 1) = (2, 2) /∈ X

Theorem 1.3.3.

W₁, W₂ < Rⁿ ⇒ W₁ T W₂ < Rⁿ

idea of proof. (i) W₁, W₂ < Rⁿ

⇒ W₁ 6= φ, W₂ 6= φ

⇒ 0 ∈ W₁, 0 ∈ W₂

⇒ 0 ∈ W₁ T W₂

⇒ W₁ T W₂ 6= φ (ii) Let u, v ∈ W₁ T W₂

Let a, b ∈ R

Show au + bv ∈ W₁ T W₂ u, v ∈ W₁ T W₂

⇒ u, v ∈ W₁ and u, v ∈ W₂

⇒ au + bv ∈ W₁ and au + bv ∈ W₂

⇒ au + bv ∈ W₁ T W₂ by (i), (ii) using theorem,

W₁ T W₂ < Rⁿ

• W₁ S W₂ : not subspace

Let W₁, W₂ : 1-dim subspaces in R²

W₂ W₁

직선이 평면상에서 원점을 지나야한다.

W₁ + W₂하면 평면을 만들어낼 수 있다

Let W₁ ={(a, a)|a ∈ R}, W² ={(b, 0)|b ∈ R}, (1, 1) ∈ W₁, (1, 0) ∈ W₂ (1, 1) + (1, 0) = (2, 1) /∈ W₁ S W₂

Theorem 1.3.4.

W₁, W₂ < Rⁿ ⇒ W₁ + W₂ < Rⁿ

W₁ + W₂ :={u₁ + u₂|u₁ ∈ W₁, u₂ ∈ W₂}

idea of proof. (i) Since 0 ∈ W₁, 0 ∈ W₂, so 0 ∈ W₁ + W₂

∴ W¹ + W₂ 6= φ

(ii) Let u, v ∈ W₁ + W₂ a, b ∈ R for u ∈ W₁ + W₂

write u = u₁ + u₂, u₁ ∈ W₁, u₂ ∈ W₂ by 합의 정의 v = v₁ + v₂, v₁ ∈ W₁, v₂ ∈ W₂

Then au + bv = (au₁ + bv₁) + (au₂ + bv₂) ∈ W₁ + W₂ Hence W₁ + W₂ < Rⁿ