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# 6.1 - 6.1 Inner Product Length Orthogonality Deﬁnition...

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Unformatted text preview: 6.1 Inner Product, Length, Orthogonality Deﬁnition: Vector space V is set of objects (vectors) which we can add and multiply by real numbers satisfying v +u ∈ V; the following for all u, v, w ∈ V, c, d ∈ R : (u + v ) + w = u + (v + w ); u+v = 0, −u ∈ V , satisfying u + 0 = u, u + (−u) = 0, cu ∈ V, c(u + v ) = EX: V = {polynomials of degree ≤ 2 : p(t) = a + bt + ct2 } = Span{1, t, t2} cu + cv, (c + d)u = cu + du, (cd)u = c(du), 1 · u = u. (But we’ll almost always use Rn as our vector space.) v1 u1 . . Given u, v ∈ Rn , u = . , v = . , the dot product . . vn un or inner product of u, v is deﬁned as follows: v1 n . . = u1 v1 + · · · + un vn = ui vi u · v = [ u1 · · · un ] . i=1 vn Note: u · v = uT v 1 −2 EX: u = 2 , v = 1 3 0 u · v = 1 · (−2) + 2 · (1) + 3 · (0) = 0 1 Theorem u, v, w ∈ Rn, c ∈ R ⇒ a. b. c. d. u·v =v·u (u + v ) · w = u · w + v · w (cu) · v = c(u · v ) = u · (cv ) u · u ≥ 0, u · u = 0 ⇔ u = 0 Show d: u · u = u2 + · · · + u2 = 0 1 n ⇔ u1 = u2 = · · · = un = 0 Deﬁnition: The length or norm of the vector u is u2 + · · · + u2 n 1 u = (u · u)1/2 = EX: 1 u = 2 , 3 Property: In R2 : cu = u= √ √ 1 + 4 + 9 = 14 c2 u2 + · · · + c2 u2 = |c| u 1 n e2 u1 2 + u2 2 u2 e1 u1 2 In R3 : e3 u2 u2 1+ u2 2+ u2 1+ 3 e1 e2 u2 2 If u = 0, v = unit length: EX: u u p oints in the same direction but has 1 u u= 1 u u =1 v= u= −1 2 , u= √ 5 v= √ −1/ 5 √ 2/ 5 e2 u v e1 3 Cauchy-Schwarz Inequality u, v ∈ Rn ⇒ (u · v )2 ≤ u 2 v 2 Deﬁnition: and v by the formula u, v ∈ Rn , deﬁne the angle θ between u u·v = u v cosθ Note: |u · v | u v ≤ 1 so there is a 0 ≤ θ ≤ π satisfying this. Justiﬁcation in R2 or R3 : v θ u u−v By Cosine Law: u = 2 +v 2 2 = uT u − uT v − v T u + v T v = u 2 = (u − v )T (u − v ) +v 2 u−v − 2cosθ u · v − 2u · v u·v so cosθ = u v 4 EX: 1 u = 1 , 0 cosθ = 1 u·v =√ uv 2 1 v=0 0 ⇒ θ= π 4 Theorem Proof: Triangle Inequality u + v ≤ u + v u+v 2 ≤ ( u + v )2 2 ⇔ 2 u 2 + 2u · v + v ≤u 2 +v +2 u v which is true by Cauchy-Schwarz inequality. Deﬁnition: u, v ∈ Rn are orthogonal if u · v = 0 u·v u·v = 0 so θ = π 2 This means cosθ = Theorem u, v ∈ Rn are orthogonal ⇔ ⇔ =u 2 u·v =0 because u + v 2 u+v 2 =u 2 +v 2 + 2u · v + v v u v 2 v 0 This is Pythagorean Theorem. u+ 5 Deﬁnition: Let W be a subspace of V ∈ Rn , W ⊥ = {v ∈ V : v · w = 0 for all w ∈ W } W ⊥ is a subspace of V — Why? Important Examples: (RowA)⊥ = Nul(A) (ColA)⊥ = Nul(AT ) First one: x ∈ (RowA)⊥ ⇔ ai ·x = 0 for every row ai of A ⇔ Ax = 0 Deﬁnition: v is u, v ∈ Rn , the distance between u and d(u, v ) = u − v Theorem d(u, w ) ≤ d(u, v ) + d(v, w) Proof: By the triangle inequality, u − w = (u − v ) + (v − w ) ≤ u − v + v − w w v u 6 ...
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