Roberto s Notes o Liear Algebra Chapter 11: Vector spaces Sectio 2 Some examples of vector spaces What you eed to kow already: The te axioms eeded to idetify a vector space. What you ca lear here: Some examples of vector spaces related to, but also differet from the Euclidea oes see so far. The whole idea of vector spaces started with Euclidea vectors i mid, so it is ot surprisig that, as you discovered i the previous sectio, is a vector space for ay. But it should also ot be surprisig that there are ay other sets that are also vector spaces. We shall make their acquaitace gradually, startig, i this sectio, with sets that still cosist of Euclidea vectors. But before goig there, I wat to itroduce a strage fellow, a vector space that we shall meet ofte, despite, or i fact because of its simplicity. Techical fact The trivial vector space satisfies all vector space axioms ad is therefore truly a vector space. Defiitio A set cosistig of a sigle object, deoted by 0, with operatios defied by 0 0 0 ad c0 0 is called the trivial vector space. The proof of this fact is very simple, although it may create some headscratchig because of its simplicity. I ecourage you to check its truth by completig the correspodig Learig questio. So, i particular, the zero vector of ay, with the usual operatios, is a istace of the trivial vector space. But so am I, so log as I use the operatios: me me me c me me Are you callig me a trivial vector space? Yes, but it is ot a offesive term! If the idea is still cofusig, file it away for ow ad wait util we put it to good use. Let us look at a less trivial example. Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 1
Techical fact The set of vectors i whose compoets add up to 0, together with the usual operatios, forms a vector space. Techical fact The set of solutios of a homogeeous liear equatio i variables, with the same operatios as i, forms a vector space. Sice we are usig the usual operatios ad we kow that for them all algebraic ad special item axioms work, we oly eed to check the closure axioms. Also, by what we have see, it is eough to check the LC axiom. So, if u ad v are vectors i c ad d are two scalars, the: This implies that: such that ui 0 ad i1 i1 cu dv cu dv cu dv cu dv 1 1 2 2 u v c d cu dy cu dy i i i i i i1 i1 i1 i1 c u d y c0 d0 0 i i1 i1 i v 0 i, ad if Therefore the liear combiatio is also i the set ad the LC axiom works. Notice that the key fact we used i the proof is the right side of the equatio beig 0, so that we ca geeralize this fact. Oce agai, all we eed to check is the LC axiom. Let s say that a1x 1 a2x2... a x a x 0 is our homogeeous equatio, that u ad v are two solutios for it ad that c ad d are ay two scalars. To check that cu dv is also a solutio we oly eed to use it as x i the equatio ad follow usual algebra: a cu dv a cu dv a cu dv 1 1 1 2 2 2... a1cu 1 a1dv 1 a2cu2 a2dv2... acu adv ca u ca u... ca u 1 1 2 2 da1v 1 da2v2... dav...... c a u a u a u d a v a v a v 1 1 2 2 1 1 2 2 But sice both u ad v are solutios, the quatities i brackets are both 0, hece so is the etire expressio ad the liear combiatio is also a solutio. But we also kow that a homogeeous equatio i variables describes a hyperplae i that cotais the origi. Sice we are dealig with full-force geeralizatios, here is a characterizatio of all vector spaces that cosist of Euclidea vectors with the usual operatios. Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 2
Techical fact A subset of is a subspace if ad oly if it is a vector space with the usual operatios. If it is a vector space with the usual operatios, the closure axioms hold ad the set is closed uder liear combiatios. That meas that it is a subspace. If it is a subspace, the closure axioms must hold for the same operatios. Sice all other axioms hold because we are dealig with the same operatios, it is a vector space. What exactly did we prove? It looks like you just said the same thig i differet ways! I did! But remember what I said about lookig at the same thig from differet perspectives: I was just checkig that the object was the same, despite the differet poit of view ad termiology used. Also, remember that we shall soo apply these cocepts to more geeral vector spaces, so that the fie poits of logic must be carefully checked each time, eve if they look kid of obvious. So, back to hyperplaes as a special case of the last fact. Techical fact The set of poits that make up ay hyperplae of cotaiig the origi, with the usual operatios, forms a vector space. Example: 3x2y 0 This is a hyperplae of for ay 2 (geeralizatio!) ad cotais the origi ad therefore it is a vector space. Notice that the origi must be i the hyperplae, or we do ot get a vector space with the usual operatios. That will chage soo! Example: x 3y 2z 4 3 This plae is a hyperplae of, but it does ot form a vector space with the usual operatios, as it is ot closed uder additio or scalar multiplicatio. To see this all we eed is oe couterexample, so here it is. The poit 0, 2,1 is o the plae, sice 032 21 6 2 4. However, its double, 0, 4, 2, as a vector, is ot o the plae, sice 0 3 4 2 2 12 4 8 4. Oe couterexample is eough. But a hyperplae that does ot cotai the origi still looks like oe that does, so why is it ot a vector space? Good poit: otice that I said that it does ot form a vector space with the usual operatios! Remember that a vector space cosists of both a set of objects ad two operatios. We ca fix thigs by chagig operatios, but i that case we eed to check the rest of the axioms. This is best doe by usig the vector equatio of a hyperplae. I 3 will state ad show you this i, sice the geeral case oly requires the use of ellipsis (ot ellipses!) What???!!! Do t believe me? Pay attetio to the proof! Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 3
Techical fact The vectors o the hyperplae of 3 defied by x p ha kb form a vector space with the followig operatios: u v u v p 2 2 c u cu 1 c p I the followig, let us assume that u p h1a k1b, v p h a k b ad that c ad d are scalars. The the closure axioms work because: C1) h k h k u v p a b p a b p 1 1 2 2 h h k k p a b 1 2 1 2 C2) 1 c u c p h a k b c p p h a k b 1 1 1 1 As for the algebraic axioms, A1 follows from usual commutativity of vector additio ad A2 works because: u v w u v w p u v w p p u v p w p u v w I leave to you the task to check the other algebraic axioms. The vector p plays the role of the 0 vector, sice: u p u p p u Also, the egative of u will be 2p u, sice: u 2p u u 2p u p p ad this is the zero vector, by our last verificatio. I also leave to you the task of checkig S3. This is crazy! What does it all mea? Aother excellet questio! If you reflect o what these weird operatios do, you will realize that they take the origial hyperplae, move it eough so as to icorporate the origi (by subtractig p), perform the usual operatios o the ew hyperplae that cotais the origi, ad fially move everythig back to the origial hyperplae! Of course we ca cotiue to play this game i all kids of directios, but I will ot take it too far for ow, except to show you oe further geeralizatio that will come i hady later. Techical fact The set of solutios of ay homogeeous system Ax 0 forms a vector space. The set of solutios of a cosistet ohomogeeous system Ax c ca be obtaied by pickig oe solutio ad addig to it every vector that is a solutio of the associated homogeeous system Ax 0. I bet you wat me to check this too! Yes, sice it will help you build familiarity with this strage ew cocept of differet operatios. But let me help you with that by showig you the details of aother, eve weirder example. But beware: we shall get far from the familiar territory of algebra oly to realize that we are actually still there! Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 4
Techical fact The set L of positive real umbers with the followig two operatios forms a vector space: Additio: For ay two umbers x ad y i L, x y xy, where the product o the right is the usual oe. Scalar multiplicatio: For ay umber x i c L ad ay scalar c, c x x, where the power o the right side is the usual oe. So, we are cosiderig the set of positive real umbers, but by additio we ow really mea multiplicatio ad by scalar multiplicatio we really mea expoetiatio. Let s see if the axioms all hold. To help you alog I will use bold to refer to oe of the weird operatios above ad to idetify a positive umber whe viewed as a vector i this weird space. Axiom 1: The additio of two positive real umbers is ideed a positive real umber, sice they are beig multiplied together. Axioms 2, 3: Commutativity ad associativity of additio hold sice it holds for the usual product of umbers. Axiom 4: Here we may seem to ru ito trouble, sice the usual umber 0 is ot eve i our set (remember that we are dealig with positive real umbers). But the axiom just requires the existece of oe vector with the property of 0, it does ot have to be the umber 0. I fact, the positive real umber 1 plays the role of the zero vector here. Look: x1 1x 1x x x Axiom 5: Axiom 6: Axiom 7: Axiom 8: Ad this is what we wat, is t it? Agai, we do ot have egative umbers, but we do ot eed them either. All we eed is a positive umber that fulfills the axiom for the weird operatios we have defied. Ad it so happes that reciprocals do the job icely: for ay positive umber x if y is its reciprocal as a usual umber, the 1 1 x y x x 1 x x Ad did t we just see that the umber 1 plays the role of the 0 vector? So we are fie here too. (I realize that the iterplay of usual umbers ad vectors i this vector space of umbers is cofusig the first time aroud, so you may wat to look at this proof agai). For ay positive umber x ad ay scalar c we have that c cx x, which is a positive umber whatever c is, so this closure axiom holds. Same otatio as before, try to follow the logic: c cd x x cd x dc x d c x d c d x Words should ot be eeded ay more: 1 x 1 x Axiom 9: c x y xy c x c y c c xc y Axiom 10: Oh, I am sure you ca check this oe yourself, so I will ot dey you such pleasure. As you puzzle over this really strage example, it may help you to cosider the fact that you have see these strage operatios before. Does the word logarithm rig ay bell? Time for a pause, some more practice ad maybe a pleasurable drik before movig o to meetig other vector spaces that are more familiar, while lookig less as usual vectors. x Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 5
Summary Yes, Virgiia, there are vector spaces that are ot made up of Euclidea vectors, but may of them still are! Commo errors to avoid Avoid ot practicig! This is a ew area of math ad if you do t play ad struggle with it, it will ever become familiar! Learig questios for Sectio LA 11-2 The Learig questios for this sectio are more streamlied tha usual, focusig o exercises to icrease your familiarity with the vector space cocept. Regular programmig will resume shortly. questios: For each the sets described i questios 1-6, determie whether the set, with the usual operatio, is or is ot a vector space. To determie that it is, show that all axioms are satisfied; to determie that it is ot, idetify at least oe axiom that fails. 1. The trivial vector space. 2 2. The set of all poits o the uit circle of with usual operatios. 3. The set of 3D vectors x y z ax by cz 0. v that satisfy a equatio of the type 4. The set of 3D vectors x y z ax by cz 1. 5. The set of all geometric vectors x y i. v that satisfy a equatio of the type 6. The set of solutios of ay homogeeous system Ax 0 2 2 that satisfy the equatio y x Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 6
Determie whether each of the sets described i questios 7-12 forms a vector space with respect to the idicated operatios. To determie that it is, show that all axioms are satisfied; to determie that it is ot, idetify at least oe axiom that fails. 7. The set of four dimesioal Euclidea vectors, with usual scalar multiplicatio ad dot product as additio. 8. The set of usual 3-dimesioal vectors, with the usual additio, but with scalar k k k k x, y, z x, y, z multiplicatio defied by 9. The set of usual 3-dimesioal vectors, with the usual additio, but with scalar k k k k x, y, z e x, e y, e z multiplicatio defied by 10. The set of three dimesioal Euclidea vectors with usual additio ad scalar k k k multiplicatio give by k a b c a b c. 11. The set of usual 3-dimesioal vectors, with the usual scalar multiplicatio, but with additio defied by x, y, z u, v, w x u, y v, z w. 12. The set of usual 3-dimesioal vectors, with the usual scalar multiplicatio, but x u y v z w with additio defied by x, y, z u, v, w,, 2 2 2. 13. Prove that the poits o a hyperplae of ot cotaiig the origi do form a vector space by usig suitably defie operatios. 14. Cosider the set cosistig of all real multiples of the variable x, all real multiples of the variable y ad a additioal object deoted by. That is, our set cosists of cx, dy,, where c ad d represet ay real umbers. Also, cosider the operatios of additio ad scalar multiplicatio defied by: Additio Scalar multiplicatio ax bx a b x cy dy c d y ax cy cy ax : ax ax cy cy k ax ka x : k cy kc y k Prove that this set, with these operatios, satisfies eight of the te axioms of a vector space. That is, prove that eight of the te axioms are true ad show why the remaiig axioms do ot work. Of course, you will have to idetify which axioms do ot work! Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 7
What questios do you have for your istructor? Liear Algebra Chapter 11: Vector spaces Sectio 2: Some examples of vector spaces Page 8