MATH1131 Mthemtics 1A Alger UNSW Sydney Semester 1, 017 Mike Mssierer Mike is pronounced like Mich mike@unsweduu Plese emil me if you hve ny questions or comments Office hours TBA (week ), or emil me to ook time Office: Red Centre 4105 MATH1131 consists of Lectures: lger, clculus Tutorils: lger, clculus / clssroom, online Clss tests: lger (week 6, 1), clculus (week 5, 9) Mple: online lessons, online quizzes (week 5, 7), l test (week 10) Finl exm (exm period) Get the course notes! (UNSW ookshop, Moodle) 1
Moodle Course pck (red the info ook!), Lecturers notes nd slides Announcements, office hours Clssroom tutoril nd homework prolems Links to online tutorils nd ssessments Forum Further support Mthemtics Drop-in Centre (Red Centre 3064) All dministrtive mtters: Student Services Office (Mrkie Lugton) fymthsstts@unsweduu, 9385 7011, Red Centre 307 Course overview Chpter 1: Introduction to Vectors (Lectures 1-4) Chpter : Vector Geometry (Lectures 5-8) Chpter 3: Complex Numers (Lectures 9-15) Chpter 4: Liner Equtions nd Mtrices (Lectures 16-0) Chpter 5: Mtrices (Lectures 1-3)
Chpter 1: Introduction to Vectors Lecture 1: Vector quntities nd R n Definition Geometriclly, vector is direction nd length (or mgnitude) Algericlly, vector is n ordered set of coordintes These two definitions re equivlent ut offer different viewpoints Consider the vectors nd : 0 the zero vector + ( ) = 3 ( ) 4 = 1 = = ( ) 4 1 ( ) 4 = ( ) 8 + = = ( ) +4 3+1 ( ) 4 3 = = ( ) 6 4 ( ) Note tht the position of the vectors is irrelevnt We cn write nd s row vectors or s column vectors In MATH1131, we choose the column vector representtion We cn scle nd y some numer (clled sclr) We cn lso dd nd sutrct nd Geometriclly, vector v is scled y scling its mgnitude v Algericlly, scling, ddition, nd sutrction hppens coordinte-wise 3
Definition Loosely, vector spce over R is set of vectors tht cn e dded together nd tht is closed under ddition nd rel sclr multipliction Consider the set V consisting of ll vectors in R nd R 3 ( ) 0 1 This is not vector spce; for instnce, we cnnot dd nd 1 0 { ( ) Consider V = R x } + = R y : x,y > 0 : V This is not vector spce over R under usul + nd : since V is not closed under sclr multipliction The line R, the plne R, nd the spce R 3 re ech vector spces: R R R R 3 the rel line R the rel plne R the rel spce R 3 More generlly, R n is vector spce The complex numers C lso form vector spce (equivlent to R ) 4
Definition Formlly, vector spce V over R is set on which ddition + nd sclr multipliction re given so tht, for ll u,v,w V nd λ,µ R, Closure under Addition u+v V Associtive Lw of Addition (u+v)+w = u+(v+w) Commuttive Lw of Addition u+v = v+u Existence of Zero Some element 0 V stisfiesx+0 = x for ll x V Existence of Negtive Some element ( v) V stisfies v+( v) = 0 Closure under Sclr Multipliction λv V Associtive Lw of Sclr Multipliction λ(µv) = (λµ)v Multipliction y identity 1v = v Sclr Distriutive Lw (λ+µ)v = λv+µv Vector Distriutive Lw λ(u+v) = λu+λv Commuttive Lw of Addition Associtive Lw of Addition + + +(+c) + c +c + = + +(+c) = (+)+c 5
Simplify 3(+) 3(+) = 3( + ) +( ) (Definition of Sutrction) = (3() + 3) +( ) (Vector Distriutive Lw) = ( (3 )+3 ) +( ) (Associtive Lw of Sclr Multipliction) = ( 6+(+1))+( ) = ( 6+(+1) ) +( ) (Sclr Distriutive Lw) = ( 6+(+) ) +( ) (Multipliction y Identity) = (6+)+ ( +( ) ) (Associtive Lw of Addition) = (6+)+0 (Existence of Negtive) = 6+ (Existence of Zero) In prctice, we just write 3(+) = 6+3 = 6+ Exercise Simplify (4 5) 3( 4) Definition R n is vector spce with entry-wise ddition nd sclr multipliction: x1 + y1 = x 1 +y 1 nd λ x1 = λx 1 x n y n x n +y n λx n Show tht R n stisfies the Commuttive Lw of Addition: u+v = v+u Addition is commuttive in R: u i +v i = v i +u i Thus, u+v = u1 + v1 = u 1 +v 1 = v 1 +u 1 = v+u u n v n u n +v n v n +u n x n 6
( ) 1 Let = nd = ( ) 1 3+ = 3 ( ) 1 4 = 4 ( ) 4 Clculte 3+ nd 4 ( 4 + ( 4 ) ( ) ( ) ( ) ( ) 3 8 3+8 11 = + = = 6 6 4 ) ( ) ( ) ( ) ( ) 4 4 4 4 0 = = = 8 8 () 9 Exercise 0 Let = nd = 4 Clculte + nd 4 3 3 7