CHAPTER I: Vector Spaces

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1 CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig they have previously see. () Not every liear algebra course is the same, so it s possible some people ever saw this material. (3) Eve if you have already see it, ad still remember it, it s very importat, so twice is ice. I mathematics, we study may differet objects. We usually begi with defiitios of the objects to be studied ad the maybe itroduce a few of their basic characteristics. Very quickly however, the questio turs to What ca we do with these? I other words, give two of these thigs, how ca I combie them or chage them ad produce ew objects. That is where operatios come i. I some cases, operatios work o sigle objects (such as differetiatio) ad i others they require a pair of objects (such as additio or multiplicatio). I liear algebra, we have aother situatio. We have sets of objects that have two operatios defied o them: oe that pairs the objects up (additio) ad aother that pairs objects with other umbers, called scalars (scalar multiplicatio). If the operatios satisfy a certai list of axioms, the set is called a vector space. The term vector space is a very good oe, but also ca be misleadig. Normal Euclidea space R is ideed a vector space, ad its elemets ofte called vectors, but i more geeral vector spaces, we have to be very careful to uderstad which objects are the vectors ad which are the scalars. This is ot difficult, but just requires attetio. So what are the axioms that the operatios o a set of vectors must satisfy? I m glad you asked. Defiitio Let V be a set closed uder two operatios, usually deoted by (vector) additio ad scalar multiplicatio. The set is called a vector space if the followig axioms are satisfied: (1) x y= y x for all x, y V. () ( x y) z= x ( y z) for all x, y, z V. (3) there exists a elemet of V, usually deoted by 0, such that x 0= x for all x V. (This elemet is called the zero vector or the idetity.) (4) for every x V there exists a elemet, usually deoted by x V, such that x ( x) = 0. The elemet x is called the iverse of x. (5) α( x y) = αx αy for all x, y V ad scalar α. (6) ( α β )x= αx βx for all x V ad scalars α, β. (7) ( αβ ) x= α( βx) for all x V ad scalars α, β. (8) 1x= x for all x V.

2 Note: The set of scalars will usually be assumed to be the real umbers uless otherwise stated. To be clear, we will ofte idicate the set of scalars by sayig V is a vector space over R or V is a vector space over C. (For those with some exposure to moder algebra: I geeral, the set of scalars ca come from ay field F. I this case we say that V is a vector space over F. If the set of scalars is oly a rig, the the set of vectors is called a module.) Example 1.1. R R is ordiary Euclidea space; i.e. the set of all two dimesioal vectors with real etries. Vector additio is defied compoet-wise, as is multiplicatio by real umbers. It is a straightforward procedure to verify the eight vector space axioms ad see that R is a vector space over R. Example Similarly, R (the set of all -dimesioal vectors with real etries) is a vector space over R for ay atural umber. I particular, if = 1 we see that i fact R is a vector space over itself. Example m R This is the set of m matrices with real etries. You might recall that uder ordiary matrix additio (which is also compoet-wise) ad scalar multiplicatio, all of the above properties are satisfied. Example Let P deote the set of all polyomials of degree less tha. We will add polyomials ad multiply by real umbers as usual. Notice that the sum of two polyomials of degree less tha also has a degree less tha, so this set is closed uder additio. Similarly, multiplyig a polyomial by a scalar ca oly reduce its degree. The zero polyomial is the idetity. For p P, its iverse is p. Example Let Cab [, ] deote the set of all cotiuous real-valued fuctios that are defied o [ ab, ]. Our two operatios are agai additio ad scalar multiplicatio. Sice the sum of two cotiuous fuctios is still cotiuous (ad the same hold for scalar multiples), this set is closed uder these operatios. The zero fuctio is cotiuous ad is therefore the idetity. For f Cab [, ], its iverse is f. So far, all of the examples have used ordiary operatios ad objects, so I have t bothered to verify the axioms i great detail. Let s get a few more iterestig examples. Example Let S be the set of all ordered pairs of real umbers. (Note that I m ot callig this R. That symbol is reserved for Euclidea space with ormal additio ad scalar multiplicatio.) Defie scalar multiplicatio as ormal

3 α( x, x ) = ( αx, αx ), 1 1 but defie vector additio by ( x, x ) ( y, y ) = ( x y,0) (Aother ote: I m usig the symbol to distiguish it from ormal additio.) Let s check the axioms ad see if this is a vector space. ( x, x ) ( y, y ) = ( x y,0) = ( y x,0) = ( y, y ) ( x, x ). A1: (Oe more ote: To verify that was commutative, I was able to use the fact that ordiary additio is already commutative. We will use this techique agai for associativity o A.) A: ( ) ( x, x ) ( y, y ) ( z, z ) = ( x, x ) ( y z,0) = ( x ( y z ),0) = (( x y ) z,0) = ( x y,0) ( z, z ) ( x x y y ) = (, ) (, ) ( z, z ) A3: This axiom fails. There is o zero elemet. Let ( x1, x) S with x 0. Addig ay elemet to ( x1, x ) will yield a elemet whose secod coordiate is 0. Note: It would ot have bee eough to just show that (0,0) failed to be the zero elemet. It is possible i abstract vector spaces for some uexpected elemet to play the role of 0. Sice oe axiom fails, S is ot a vector space. We do ot eed to check the others. Example Let R deote the set of positive real umbers. Defie additio by x y= xy, ad scalar multiplicatio by α x= x α. A1: x y= xy= yx= y x. A: x ( y z) = x yz= x( yz) = ( xy) z= ( x y) z. A3: (Here is a example of a couterituitive zero elemet. Just because the umber 0 does t work, does t mea A3 fails.) x 1= x1= x. A4: (Note that the target product here is the zero elemet, ot the umber 0.) Let x R. Sice x is ot 0, x R also. The we have x x = x x = 1. Axioms 5 ad 6 are a good exercise i workig with otatio. Usig the operatios of

4 this example, Axiom 5 states α ( x y) = α x α y ad Axiom 6 states ( α β ) x= α x β x. That is what we eed to verify. A5: α ( x y) = α ( xy) = ( xy) α = x α y α = x α y α = α x α y. α β α β α β A6: ( α β ) x= x = x x = x x = α x β x. A7: ( αβ ) x= x αβ = x βα = ( x β ) α = α ( x β ) = α ( β x). 1 A8: 1 x= x = x. So R uder these operatios is a vector space. Exercise Let Z be the set of itegers. Defie additio i the usual way but defie scalar multiplicatio by α k = α k. The dot represets ormal multiplicatio ad the double brackets represet the greatest iteger fuctio. Is this a vector space over R? If yes, verify all 8 axioms. If o, fid oe axiom that fails. Exercise Let S be the set of all ordered pairs of real umbers. Defie ( x1, x ) ( y1, y) = ( x1 y1 1, x y 1) ad α ( x1, x) = ( α αx1 1, α αx 1). Is this a vector space over R? If yes, verify all 8 axioms. If o, fid oe axiom that fails. Exercise Let R be the set of real umbers with scalar additio defied as usual, but additio defied by x y= max{ x, y}. Is this a vector space over R? If yes, verify all 8 axioms. If o, fid oe axiom that fails. The eight axioms required of vector spaces are ot the oly properties that are true. How is a axiom differet from a property or characteristic? Axioms are properties that caot be deduced from other properties, so they have to be verified up frot. All other properties about vector spaces ca be deduced from these eight. For istace Example I every vector space V, 0x= 0 for all x V. (Here I am usig the symbol 0 i two differet ways i oe equatio. The first 0 is the scalar 0. The 0 o the right had side is the zero elemet, which as we have see may or may ot be the umber 0. Cofused yet?) First of all, ote that x 0x= 1x 0 x= (1 0) x= 1x= x. Addig x to the far left ad far right of that equatio gives us 0x= 0. Example I every vector space V, α 0= 0 for all scalar α. (Note: Both 0 s are the zero elemet.) First ote that α0 α0 = α(0 0) = α0. Now add ( α0), the iverse of α 0, to both sides.

5 ( ) α ( α α ) α0 α0 ( α0) = α0 ( α0) 0 0 ( 0) = 0 Exercise Prove that 1x= x for all x V. α0 0= 0 α0= 0 Exercise Prove that if α x= 0, the either α = 0 or x= 0. Exercise Prove that x y= 0 implies y= x (i.e. iverses are uique).