6.2review

# 6.2review - Some Review Oldest areas of mathematics...

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Unformatted text preview: Some Review Oldest areas of mathematics: geometry nonlinear dynamics (studying movement of stars/planets) greatest scientiﬁc discovery of all time: Kepler’s discovery of elliptic orbits of planets in 1609 motivated Newton to develop calculus some basic areas of mathematics: algebra calculus - studying dynamic continuous models linear algebra - study discrete events involving more than one variable multivariate calculus diﬀerential equations computational mathematics 1 Let’s take stock of where we are in linear algebra, largely by looking in detail at an example 1 3 −1 − 2 1 c1 Solve Ac = 1 2 −2 c2 = y = 1 EX: 7 c3 1 11 2 Can show c = 5 11 1 , 3 2 33 so Ac = 1 1 3 −1 − 1 2 2 1 5 11 1 −2 3 7 2 2 33 1 = y = 1 1 Some ways to solve for c: (1) Reduce A to RE form, solve for c by back substitution (this is Gaussian elimination) (2) Find A−1 , form c = A−1 y Q: What is the diﬀerence between these? A: Almost always, (2) is more work and less accurate on a computer (from ﬂoating point errors), so we seldom ﬁnd the inverse A−1 !! However, when we discuss orthogonal matrices, we’ll see a situation where the inverse is easy to use. 2 An interpretation of this linear system using vectors: 3 −1 −1 2 1, u2 = 2 , u3 = −2 a basis? Is u1 = 7 1 1 2 Yes ... Why? c1 c2 = y for any y . c3 2 u 33 3 ∴ , one can solve Ac = u1 u2 u3 5 1 In this particular case, y = 11 u1 + 3 u2 + ∴ by deﬁnition [y ]B = 5 11 1 3 2 33 Now we look at some geometrical interpretations of vectors: ||u1 ||2 = u1 · u1 = 3 · 3 + 1 · 1 + 1 · 1 = 11, ||u1 || = √ 11 √ 6 33 2 ||u2 ||2 = u2 · u2 = −1 · −1 + 2 · 2 + 1 · 1 = 6, ||u2 || = 1 7 1 ||u3 ||2 = u3 ·u3 = − 2 ·− 2 +−2·(−2)+ 7 · 2 = 2 33 , 2 ||u3 || = u1 · u2 = 3 · (−1) + 1 · (2) + 1 · (1) = 0 and can show u1 · u3 = u3 · u2 = 0. u1 u1 · u2 u2 cosθ = 0 0 =√ √ 2 18 π → θ= 2 3 u2 u1 ∴ orthogonal basis B u1 , u2 , u3 } since ui · uj = 0 for ={ 1 uy uy 5 all i = j , and for y = 1, c1 = u11··u1 = 11 , c2 = u22··u2 = 1 , 3 1 c3 = u3 ·y u3 ·u3 = 1 1/4+4+49/4 = 1 66/4 = 2 33 ∴ Note y= 1 2 u1 + u2 + u3 11 3 33 5 (u1 ) u1 · u1 u1 · u2 u1 · u3 AT A = (u2 )T u1 u2 u3 = u2 · u1 u2 · u2 u2 · u3 = (u3 )T u3 · u1 u3 · u2 u3 · u3 11 0 0 0 6 0 33 002 What if we make {u1 , · · ·, up } orthonormal, so 3 −1 1 1 u1 = √ 1 , u2 = √ 2 , u3 = 11 6 1 1 −1/2 1 −2 33 2 7/2 T and B = {u1 , u2 , · · ·, up } orthonormal basis of W ⊆ Rn ? 4 c1 . Then y ∈ Rn ⇒ [y ]B = . , . cp ci = ui ·y ← really simple Theorem: thonormal columns if and only if U T U = In . True because U T U = uT uj = [ui · uj ] = In i T u1 · u1 u1 · u2 u1 · u3 (u1 ) U T U = (u2 )T u1 u2 u3 = u2 · u1 u2 · u2 u2 · u3 = EX: u3 · u1 u3 · u2 u3 · u3 (u3 )T 5 u1 · y 100 11 0 1 0 = I =⇒ c = U −1 y = U T y = u2 · y = 1 3 2 001 33 u ·y 3 m × n matrix U = [ u1 | · · · | un ] has or- 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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