6.2 - 6.2 Orthonormal Sets Quick Section 6.1 Review: u, v...

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Unformatted text preview: 6.2 Orthonormal Sets Quick Section 6.1 Review: u, v ∈ Rn , Theorem: u+v 2 u + v ≤ u + v (Triangle Inequality) 2 =u +v 2 ⇔ ⇔ u · v = 0 i.e. orthogonal angle between u and v is π 2 Corollary: u, v, w ∈ Rn d(u, w ) ≤ d(u, v ) + d(v, w) Definition: W a subspace of V = Rn W ⊥ = {v ∈ V : v · w = 0 for all w ∈ W } EX: u= 1 1 , v v= −1 1 , u·v =0 u EX: 1 u = 1, 0 1 v = 0, 0 1 u·v =√ uv 2 1 z y θ v x u 1 cosθ = √ 2 → θ= π 4 Definition: ui · uj = 0 for all i = j . EX: Subset {u1 , · · ·, up } ⊆ Rn is orthogonal if EX: {e1 , · · ·, en } ⊆ Rn is orthogonal 1 −1 2 u1 = 0, u2 = 4 , u3 = 1 1 1 −2 u1 · u2 = 0, u1 · u3 = 0, u2 · u3 = 0 Theorem: If {u1, · · ·, up} ⊆ Rn is orthogonal and each ui = 0, then {u1 , · · ·, up } is linearly independent. True because if c1 u1 + · · · + cp up = 0 then for any i, ui · (c1 u1 + · · · + cp up ) = ci ui · ui = 0 ⇒ ci = 0 Definition: B is an orthogonal basis for a subspace W of Rn if it is a basis which is orthogonal. 2 Theorem: subspace W of Rn , then for any y ∈ W , y = c1 u1 + · · · + cp up , Proof: ci = If {u1 , · · ·, up } is an orthogonal basis for a ui · y ui · ui ui · y ui · y = ui · (c1 u1 + · · · + cp up ) = ci ui · ui ⇒ ci = EX: B = {u1 , u2 , u3 } orthogonal basis for R3 u1 · y u ·u 1 1 u ·y 2 y ∈ R3 ⇒ [y ]B = u2 · u2 u3 · y u3 · u3 ui · ui EX: 3 −1 −1/2 1 1 , 2 , −2 B= orthogonal basis y = 1, 1 1 7/2 1 = {u1 , u2 , u3 } 3 1 1 · 1 · 1 1 u1 · y 5 c1 = == u1 · u1 11 3 3 1 · 1 1 1 3 −1 1 2 · 1 · 1 1 1 u2 · y 2 = = = c2 = u2 · u2 6 3 −1 −1 2 · 2 1 1 −1/2 1 −2 · 1 · 1 7/2 u3 · y 4 2 1 = c3 = = = = u3 · u3 1/4 + 4 + 49/4 66 33 −1/2 −1/2 −2 · −2 7/2 7/2 ∴ y= 1 2 u1 + u2 + u3 11 3 33 5 Definition: {u1 , · · ·, up } is orthonormal if ui · uj = 0 i = j (1 ≤ i, j ≤ p) ui · ui = 1, i = 1, · · ·, p EX: 3 −1 1 1 u1 = √ 1 , u2 = √ 2 , u3 = 11 6 1 1 −1/2 1 −2 33 2 7/2 {u1 , u2 , u3 } is orthonormal. 4 ∴ B = {u1 , u2 , · · ·, up } orthonormal basis of subspace W of Rn , y ∈ Rn ⇒ c1 . [y ]B = . , . cp ci = ui · y ← really simple Theorem: m × n matrix U has orthonormal columns if and only if U T U = In True because . U = [ u1 | · · · | un ] , U T = . ⇒ . uT n U T U = uT uj = [ui · uj ] = In i uT 1 Theorem: m × n U with orthonormal columns ⇒ for any x, y ∈ Rn , a. b. c. Ux = x (U x) · (U y ) = x · y Ux · Uy = 0 ⇔ x · y = 0 Why? Because (a) ||U x||2 = (U x)T (U x) = xT U T U x = xT x = ||x||2 (b) (U x) · (U y ) = (U x)T (U y ) = xT U T U y = xT y = x · y 5 ...
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This note was uploaded on 04/09/2011 for the course MATH 232 taught by Professor Russel during the Spring '10 term at Simon Fraser.

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