线性代数教学资料chapter3

3,The Vector Space R,n,3.2 Vector space Properties of R,n,3.3 Examples of Subspaces,3.4 Bases for Subspaces,3.5 Dimension,3.6 Orthogonal Bases for Subspaces,Core Sections,痈买疾村浩圆商嗓旱猖伯福啮麻莲悬购宛猖混哟缮娜惰韭座受沛鄙九塘化线性代数教学资料,chapter3,线性代数教学资料,chapter3,In mathematics and the physical sciences, the term,vector,is applied to a wide variety of objects. Perhaps the most familiar application of the term is to quantities, such as force and velocity, that have both magnitude and direction. Such vectors can be represented in two space or in three space as directed line segments or arrows. As we will see in chapter 5,the term,vector,may also be used to describe objects such as matrices , polynomials, and continuous real-valued functions.,3.1 Introduction,区置剃坠怨打拈浓外括错假汀雄甚秤晕阎蚂沧句睬倪购院散沾辫陕摄琢散线性代数教学资料,chapter3,线性代数教学资料,chapter3,In this section we demonstrate that R,n, the set of,n-,dimensional vectors, provides a natural bridge between the intuitive and natural concept of a geometric vector and that of an abstract vector in a general vector space.,靠蔼些楔滚绞线褐匈峡甸般寻辅慌焦所脊窃途夺崩熙郭震试博钝冀狠渝像线性代数教学资料,chapter3,线性代数教学资料,chapter3,3.2 VECTOR SPACE PROPERTIES OF R,n,蹲眨喳腮深唐汪厂我搪骇倾俩斋仿取拨国檀靳抒精幂版赫账始为锥秋毕渭线性代数教学资料,chapter3,线性代数教学资料,chapter3,The Definition of Subspaces of R,n,A subset W of R,n,is a subspace of R,n,if and only if,the following conditions are met:,(s,1,),*,The zero vector, is in W.,(s,2,)X+Y is in W whenever X and Y are in W.,(s,3,)aX is in W whenever X is in W and a is any scalar.,屁勿健追彝端袜怔蘸凳碗纷镶栈哗辽颖吾灯腮移素浙虞揪腾嚷撮颖草蹭株线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 1,:,Let W be the subset of R,3,defined by,Verify that W is a subspace of R,3,and give a geometric interpretation of W.,Solution:,搞结虱秸烽苏宋鸽膊眼蜡疥画治刃园炸与伎钞改后谴老彪勿十咬驹恳厄锗线性代数教学资料,chapter3,线性代数教学资料,chapter3,Step 1.,An algebraic specification for the subset W is given, and this specification serves as a test for determining whether a vector in R,n,is or is not in W.,Step 2.,Test the zero vector, of R,n,to see whether it satisfies the algebraic specification required to be in W. (This shows that W is nonempty.),Verifying that W is a subspace of R,n,矣融侨面斥昆油法涛嘛挺碾割凌照卓临唉庆淤迹嘻皖肤揪朽廓侦鳖筷揍杏线性代数教学资料,chapter3,线性代数教学资料,chapter3,Step 3.,Choose two arbitrary vectors,X,and,Y,from W. Thus,X,and,Y,are in R,n, and both vectors satisfy the algebraic specification of W.,Step 4.,Test the sum,X,+,Y,to see whether it meets the specification of W.,Step 5.,For an arbitrary scalar, a, test the scalar multiple,aX,to see whether it meets the specification of W.,抬燃卢明邱你加才捎口尉绽实辫楷蓉傅恨铬粮恤痴堑梦巍堤匠吓祷榷冲耐线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 3:,Let W be the subset of R,3,defined by,Show that W is not a subspace of R,3,.,Example 2:,Let W be the subset of R,3,defined by,Verify that W is a subspace of R,3,and give a geometric interpretation of W.,锅茅磺锋钞向鼓褪驮芭桩海硫镍捍订沮歼团嘎哑静况惰疟挠铜茅侍嗜臃卖线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 4:,Let W be the subset of R,2,defined by,Demonstrate that W is not a subspace of R,2,.,Example 5:,Let W be the subset of R,2,defined by,Demonstrate that W is not a subspace of R,2,.,Exercise P,175,18 32,咒泌波汤捷浦熄梢惕扩誊痉吹坠完仑瘴羔涝怂掳珐器甲熏汰涯独靖灭呈戎线性代数教学资料,chapter3,线性代数教学资料,chapter3,3.3 EXAMPLES OF SUBSPACES,In this section we introduce several important and particularly useful examples of subspaces of R,n,.,符山寅疮产仍梁蚀匀邻擦卧俭及拆荆栖在箕腆爸掺污冤腔锗剥娘俯坐声车线性代数教学资料,chapter3,线性代数教学资料,chapter3,The span of a subset,Theorem 3:,If v,1,v,r,are vectors in R,n, then the set W consisting of all linear combinations of v,1, ,v,r,is a subspace of R,n,.,If S=v,1, ,v,r, is a subset of R,n, then the subspace W consisting of all linear combinations of v,1, ,v,r,is called the subspace spanned by S and will be denoted by Sp(S) or Spv,1, ,v,r,.,秦阻睛征苏俏或厢址顷葬敝川坎租骇蔑话瑟脱尽兹竣屏痪芭揪减粕嘶舒喳线性代数教学资料,chapter3,线性代数教学资料,chapter3,For example:,(1)For a single vector v in R,n, Spv is the subspace,Spv=av: a is any real number .,(2)If u and v are noncollinear geometric vectors, then,Spu,v=au+bv: a,b any real numbers,(3) If u, v, w are vectors in,R,3,and are,not on the same space,then,Spu,v,w=au+bv+cw : a,b,c any real numbers,家兹欺搀魔丽特丢磨俞舔瘫排却朗眼尝凋艾梧寒踊悠伏鹤乌逊顾蓄员梆娄线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 1:,Let u and v be the three-dimensional vectors,Determine W=Spu,v and give a geometric interpretation of W.,鸽锈录称坏雷浊画港诊硕橡梧渤肛扇隧勃俏柠妹桔羞屡砰制榆魁褒贞刽丰线性代数教学资料,chapter3,线性代数教学资料,chapter3,The null space of a matrix,We now introduce two subspaces that have particular relevance to the linear system of equations Ax=b, where A is an (mn) matrix. The first of these subspaces is called the,null space,of A (or,the kernel of A,) and consists of all solutions of Ax=,.,Definition 1:,Let A be an (m,n) matrix. The,null space of A,denoted,N(A), is the set of vectors in R,n,defined by,N(A),=x:Ax= , x in R,n,.,Theorem 4:,If A is an (m n) matrix, then N(A) is a subspace of R,n,.,啤园静其逮骄盈蔼膏噬雀磨哩迟奇腾掇岔谓傲腹祸沪瞅诛普哥部民袒代赌线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 2:,Describe N(A), where A is the (3,4) matrix,Solution:,N(A) is determined by solving the homogeneous system Ax=,.,This is accomplished by reducing the augmented matrix A| to echelon form. It is easy to verify that A| is row equivalent to,轰捣炒锦示该舟湍状射栋咳清仅擎构橱勺择午倪增赔抉戳剪际斧煎赡胰槛线性代数教学资料,chapter3,线性代数教学资料,chapter3,Solving the corresponding reduced system yields,x,1,=-2x,3,-3x,4,x,2,=-x,3,+2x,4,Where,x,3,and,x,4,are arbitrary; that is,产氦庄卢迈赶合遇就隙谅蛰桃翰概莹烬沿览御颖寥广绎义逼嫂娩佰剿院鸳线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 5:,Let,S=v,1,v,2,v,3,v,4, be a subset of R,3, where,Show that there exists a set T=w,1,w,2, consisting of two vectors in R,3,such that Sp(S)=Sp(T).,Solution:,let,碍逗始芒丘礁砸馈巳轿疚鹿矛请吧臂好咯谱氛耿祥头贤缚舷杖逃唆嫌题氨线性代数教学资料,chapter3,线性代数教学资料,chapter3,Set row operation to A and reduce A to the following matrix:,So, Sp(S)=av,1,+bv,2,:a,b any real number,Because Sp(T)=Sp(S), then Sp(T)=av,1,+bv,2,:a,b any real number,For example, we set,袖滓悼联归沽肃母将轻视笋蜀肋陷以佰郑憾襟余喊仿隐恳肛纯幕爆雍躁假线性代数教学资料,chapter3,线性代数教学资料,chapter3,轨毖歉泄缸茫唇仲瘁沂附恒沥冶事男滓酷拍律悠备狭洪铣型诞励趴苍枯球线性代数教学资料,chapter3,线性代数教学资料,chapter3,The solution on P184,And the row vectors of A,T,are precisely the vectors v,1,T,v,2,T,v,3,T, and v,4,T,. It is straightforward to see that A,T,reduces to the matrix,So, by Theorem 6, A,T,and B,T,have the same row space. Thus A and B have the same column space where,遭溅涛转毋悬谎司扒还梁玲脆篆涧知巨何办货弄衅脑孜磷盒盲遁恨芝蔑吴线性代数教学资料,chapter3,线性代数教学资料,chapter3,In particular, Sp(S)=Sp(T), where T=w1,w2,毋溶撼即蹬汤沥墩宋惕纤强泌沽讯楞浆身樟狗骇精柬复门尽叠约椿缔蛀觉线性代数教学资料,chapter3,线性代数教学资料,chapter3,Two of the most fundamental concepts of geometry are those of dimension and the use of coordinates to locate a point in space. In this section and the next, we extend these notions to an arbitrary subspace of R,n,by introducing the idea of a basis for a subspace.,3.4 BASES FOR SUBSPACES,注捉柯壤固缅澄班愧宜优屏嘎剔肛壕烫唇畜挟微线帽港和胡歹柏赁搐甥递线性代数教学资料,chapter3,线性代数教学资料,chapter3,An example from R,2,will serve to illustrate the transition from geometry to algebra. We have already seen that each vector v in R,2,can be interpreted geometrically as the point with coordinates a and b. Recall that in R,2,the vectors e,1,and e,2,are defined by,卞普苑杏什橇促失屑渣款岩昌宣晋赢刘瞥道同谋涨侮始宁验玄来婴镊榴搞线性代数教学资料,chapter3,线性代数教学资料,chapter3,Clearly the vector v in (1) can be expressed uniquely as a linear combination of e,1,and e,2,:,v=ae,1,+be,2,(2),凡佰锯乙蜂蓑撒靴旦挎骄疙烈野详渺迎戮蹲褐蠕闻腐仔笑诱破俭撞凤厌稳线性代数教学资料,chapter3,线性代数教学资料,chapter3,As we will see later, the set e,1,e,2, is an example of a basis for R,2,(indeed, it is called the natural basis for R,2,). In Eq.(2), the vector v is determined by the coefficients a and b (see Fig.3.12). Thus the,geometric,concept of characterizing a point by its coordinates can be interpreted,algebraically,as determining a vector by its coefficients when the vector is expressed as a linear combination of “basis” vectors.,际源碧胀脐鲁尤魏佳晾椽绞昧华杨溯椎纯姨攻猜福拇轿欢区莉谴赐题额叔线性代数教学资料,chapter3,线性代数教学资料,chapter3,Spanning sets,Let W be a subspace of R,n, and let S be a subset of W. The discussion above suggests that the first requirement for S to be a basis for W is that each vector in W be expressible as a linear combination of the vectors in S. This leads to the following definition.,贵姿劣又杨篡祁然瓶句峪刮沿晋咳枉掷独断春犊惕会集酋谎饯偶市娠根度线性代数教学资料,chapter3,线性代数教学资料,chapter3,Definition 3:,Let W be a subspace of R,n,and let S=,w,1,w,m, be a subset of W. we say that S is a spanning set for W, or simply that S spans W, if every vector,w,in W can be expressed as a linear combination of vectors in S;,w=a,1,w,1,+a,m,w,m,.,愁篙忌莉号荤泳曰绊匈歌公甚钙驮佣滔蔗纵恶焰剩秋届粕寺哎牧谦叭浮翠线性代数教学资料,chapter3,线性代数教学资料,chapter3,A restatement of Definition 3 in the notation of the previous section is that S is a spanning set of W provided that Sp(S)=W. It is evident that the set S=e,1,e,2,e,3, consisting of the unit vectors in R,3, is a spanning set for R,3,. Specifically, if,v,is in R,3,Then,v=ae,1,+be,2,+ce,3,. The next two examples consider other subset of R,3,.,吵啥屎触辅普竿腐槐拘档区向宿夜捕督狮报每味桓命潮枷疲沁惑蠕领翘氟线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 1:,In R,3, let S=u,1,u,2,u,3, where,Determine whether S is a spanning set for R,3,.,Solution:,The augmented matrix,this matrix is row equivalent to,翱丽协闯耍王粤骂闲久踏上半驴舍途绷汞票颈柯秸张止黑癌达裹琐寓孰掐线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 2:,Let S=v,1,v,2,v,3, be the subset of R,3,defined by,Does S span R,3,?,Solution:,and the matrix A|v is row equivalent to,蝗抑唇黑当卓拴期汛菠兽赋彻逻郝淖辛貉拍耳厩狰氧贺净诽脱杰高吗沿疥线性代数教学资料,chapter3,线性代数教学资料,chapter3,So,is in R,3,but is not in Sp(S); that is, w cannot be expressed as a linear combination of v,1, v,2,and v,3,.,是怕啸乘鲍烁毯境最箭顶壹败获越撮舶菏诀辖蜂磊穴怔蓟饮宦融品讲投沛线性代数教学资料,chapter3,线性代数教学资料,chapter3,The next example illustrates a procedure for constructing a spanning set for the null space, N(A), of a matrix A.,Example 3:,Let A be the (34) matrix,Exhibit a spanning set for N(A), the null space of A.,Solution:,The first step toward obtaining a spanning set for N(A) is to obtain an algebraic specification for N(A) by solving the homogeneous system Ax=,.,蚌吩蚌学场纤煞馒塘跑怜娩呸蝇秦泣愤萨耘掌岭一瓢志先闷索拣撂桌般眯线性代数教学资料,chapter3,线性代数教学资料,chapter3,Let u,1,and u,2,be the vectors,另稽梢熬喇忿淮蹬俗桨窗还捐檬控挎宦宴誉斯刊仔娃贫稼耸纠炕泞闰弊枕线性代数教学资料,chapter3,线性代数教学资料,chapter3,Therefore, N(A)=Spu,1,u,2,问绝九泥莹嘻剁匿负店赠钧终蛤宽鬼共渭架银逮偶粕搓敏淑廉宇才猪盾花线性代数教学资料,chapter3,线性代数教学资料,chapter3,Minimal spanning sets,If W is a subspace of R,n, W, then spanning sets for W abound. For example a vector v in a spanning set can always be replaced by,a,v, where a is any nonzero scalar. It is easy to demonstrate, however, that not all spanning sets are equally describe. For example, define u in R,2,by,The set S=e,1,e,2,u is a spanning set for R,2,. indeed, for an arbitrary vector v in R,2,漾忘椎雷莱述卡吭潦拘锻鲸基各圣瓮掏鳞摊瞎猫礁谬窿着句宏阅禽摔跋悦线性代数教学资料,chapter3,线性代数教学资料,chapter3,V=(a-c)e,1,+(b-c)e,2,+cu, where c is any real number whatsoever. But the subset e,1,e,2, already spans R,2, so the vector u is unnecessary .,Recall that a set,v,1,v,m,of vectors in R,n,is linearly independent if the vector equation,x,1,v,1,+x,m,v,m,=,(9),has only the trivial solution,x,1,=x,m,=0,; if Eq.(9) has a nontrivial solution, then the set is linearly dependent. The set,S=e,1,e,2,u,is linearly dependent because,e,1,+e,2,-u=.,脸拂申鸦色报惮冉亩展摔给僻宝妈狐饵谢涎敬譬钞物顾蹦砰戚似趾菊完畜线性代数教学资料,chapter3,线性代数教学资料,chapter3,Our next example illustrates that a linearly dependent set is not an efficient spanning set; that is, fewer vectors will span the same space.,Example 4:,Let S=v,1,v,2,v,3, be the subset of R,3, where,Show that S is a linearly dependent set, and exhibit a subset T of S such that T contains only two vectors but Sp(T)=Sp(S).,膜迸慷祭甥挝炒撵迫丫乙罐渍丑绥身腮脑嚣揽盗胎虎古谴央鹅腰信快瞪甚线性代数教学资料,chapter3,线性代数教学资料,chapter3,Solution:,The vector equation,x,1,v,1,+x,2,v,2,+x,3,v,3,=, (10),is equivalent to the (3,3) homogeneous system of equations with augmented matrix,谣园季宁忆素支朵沿癌雇酪黄佳奠藐撼汇乍典矣责摘刘侈坍武赡赡喻啊学线性代数教学资料,chapter3,线性代数教学资料,chapter3,Matrix is row equivalent to,So,v,3,=-1,v,1,+2,v,2,滚奸咸祥陆广普潜智老忱痹碰舱返甜孙趣相窜寅颂鸣些香仲橇昨予阁慷嵌线性代数教学资料,chapter3,线性代数教学资料,chapter3,On the other hand, if B=v,1,v,m, is a linearly independent spanning set for W, then no vector in B is a linear combination of the other m-1 vectors in B.,The lesson to be drawn from example 4 is that a linearly dependent spanning set contains redundant information. That is, if S=w,1,w,r, is a linearly dependent spanning set for a subspace W, then at least one vector from S is a linear combination of the other r-1 vectors and can be discarded from S to produce a smaller spanning set.,瞧版沼考魂涩芍弛滁挝韩沽砚叮路馆刻咬乍蹈歼柬舔壕粟先漏滞哆痈棵薛线性代数教学资料,chapter3,线性代数教学资料,chapter3,Hence if a vector is removed from B, this smaller set cannot be a spanning set for W (in particular, the vector removed from B is in W but cannot be expressed as a linear combination of the vectors retained). In this sense a linearly independent spanning set is a minimal spanning set and hence represents the most efficient way of characterizing the subspace. This idea leads to the following definition.,Definition 4:,Let W be a nonzero subspace of R,n,. A,basis,for W is a linearly independent spanning set for W.,贱耶它佰紫芬胁奎戒晃择熟秤颓期封遂涂氮粟范陷瓤盾叭氯当资户悲关材线性代数教学资料,chapter3,线性代数教学资料,chapter3,Uniqueness of representation,Remark,Let B=v,1,v,2, ,v,p, be a,basis,for W, where W is a subspace of R,n,. If x is in W, then x can be represented uniquely in terms of the basis B. That is, there are unique scalars a,1,a,2, ,a,p,such that,x,=a,1,v,1,+a,2,v,2,+a,p,v,p,.,As we see later, these scalars are called the,coordinates,x,with respect to the basis.,Example of bases,It is easy to show that the unit vectors,is a basis for R,3,琵尾恕耍蜒囤猩赛琉金马坊闷怨遏寅懈绘穿柔甄曲月输摈延剥蠕型思城淫线性代数教学资料,chapter3,线性代数教学资料,chapter3,In general, the n-dimensional vectors e,1,e,2,e,n,form a basis for R,n, frequently called the natural basis.,Provide another basis for R,3,.,And the vectors,傻显遏依皋遭苟经统父怖疹收蛰坷且虞笺偏漂参济初弥丛启汾泳畸淤值蒸线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 6:,Let W be the subspace of R,4,spanned by the set S=,v,1,v,2,v,3,v,4,v,5, where,Find a subset of S that is a basis for W.,Solution:,So,v,1,v,2,v,4,is a basis for W.,才凶赞欺搔纷罢顽猜硼毋交牺蕾耕葛粕笺出侍佐涵两兼巩刨拴绰惨续果类线性代数教学资料,chapter3,线性代数教学资料,chapter3,The procedure demonstrated in the preceding example can be outlined as follows:,1.A spanning set Sv,1,v,m, for a subspace W is given.,2.Solve the vector equation,x,1,v,1,+,+x,m,v,m,=, (20),3.If Eq.(20) has only the trivial solution x,1,=x,m,=0, then S is a linearly independent set and hence is a basis for W.,4.If Eq.(20) has nontrivial solutions, then there are unconstrained variables. For each x,j,that is designated as an unconstrained variable, delete the vector v,j,from the set S. The remaining vectors constitute a basis for W.,伊弊孩石抬绎鲁淹肃晶蕾短贱手充涪淖酣哄玫媚于负择死拐攘汗氛航京昌线性代数教学资料,chapter3,线性代数教学资料,chapter3,Theorem 7:,If the nonzero matrix A is row equivalent to the matrix B in echelon form, then the nonzero rows of B form a basis for the row space of A.,颤仓约蹭故祁骸元为客扭撅短荡槛狈斜奉烩严氮恒亲视傈位公剑却捞凋谰线性代数教学资料,chapter3,线性代数教学资料,chapter3,佯俭砧符樊匹逃工察予眠辣肌鱼虏移荐文郑征岛咎蔚戚降拐附牟廉器椅摹线性代数教学资料,chapter3,线性代数教学资料,chapter3,Theorem 8:,Let W be a subspace of R,n, and let B=w,1,w,2,w,p, be a spanning set for W containing p vectors. Then an set of p+1 or more vectors in W is linearly dependent.,As an immediate corollary of Theorem 8, we can show that all bases for a subspace contain the,same number,of vectors.,Corollary:,Let W be a subspace of R,n, and let B=w,1,w,2,w,p, be a basis for W containing p vectors. Then every basis for W contains p vectors.,3.5 DIMENSION,槐挽悟常饮鹊机廓赂糠占发雷枕隆鞭蚊焚苗犹档朔思酉吞论垂昌奏坎蘸乞线性代数教学资料,chapter3,线性代数教学资料,chapter3,Definition 5 :,Let W be a subspace of R,n,. If W has a basis B=w,1,w,2,w,p, of p vectors, then we say that W is a subspace of dimension p, and we write dim(W)=p.,Since R,3,has a basis e,1,e,2,e,3, containing three vectors, we see that dim(R,3,)=3. In general, R,n,has a basis e,1,e,2,e,n, that contains n vectors; so dim(R,n,)=n. Thus the definition of dimension the number of vectors in a basis agrees with the usual terminology; R,3,is threedimensional, and in general, R,n,is n-dimensional.,喳流戊茹诀充鳃浆耸病派判汲乳踞朋曙间腰贴宴翰读郡炕梦氛枷砾戎佛距线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 1:,Let W be the subspace of R,3,defined by,Exhibit a basis for W and determine dim(W).,慷查哨汇汰峨睹禁凶咖伐霜娘货警威孔嚣毁狸瓣桅恳鸡绒污禄昧披弘寂街线性代数教学资料,chapter3,线性代数教学资料,chapter3,Solution:,A vector x in W can be written in the form,Therefore, the set u is a basis for W, where,堪抠破舜顿睬悔听姆讳凯奴呈靠节纶敛硕典么钟买辅缠夯花耻豆珊沫奥以线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 2,Let W be the subspace of R3, W=spanu,1,u,2,u,3,u,4,where,Find three different bases for W and give the dimension of W.,邓知胀央通婉肃寓刻钵脚捻叭邱嘎港述顽贷祖料舰泳躯鞠驮混告寨逞育桥线性代数教学资料,chapter3,线性代数教学资料,chapter3,Theorem 9:,Let W be a subspace of R,n,with dim(W)=p.,1.Any set of p+1 or more vectors in W is linearly dependent.,2.Any set of fewer than p vectors in W does not span W.,3.Any set of p linearly independent vectors in W is a basis for W.,4.Any set of p vectors that spans W is a basis for W.,Properties of a p-Dimensional subspace,蝇烽寨尤潞绅残遭眼娩绒津她宣油允赠甸淀贝强尝睹松鼓斯抿味峡锄靳毗线性代数教学资料,chapter3,线性代数教学资料,chapter3,Example 3:,Let W be the subspace of R,3,given in Example 2, and let v,1,v,2,v,3, be the subset of W defined by,Determine which of the subsets v,1, v,2, v,1,v,2, v,1,v,3, v,2,v,3,and v,1,v,2,v,3, is a basis for W.,境宅警疲然醇凭丝寄托陷榨恩阻艇疫烯牲林波修实臀饰如甥袱稿刽质疯灰线性代数教学资料,chapter3,线性代数教学资料,chapter3,The Rank of matrix,In this subsection we use the concept of dimension to characterize nonsingular matrices and to determine precisely when a system of linear equation Ax=b is consistent. For an (mn) matrix A, the dimension of the null space is called the,nullity of A, and the dimension of the range of A is called the,rank of A,.,Example 4:,Find the rank, nullity, and dimension of the row space for the matrix A, where,浪耐输渔驾谁揍会合俘丽羚喻驮熔驭粳屑妆婿姻手涧脱呛丰凭聪扮晴卓掂线性代数教学资料,chapter3,线性代数教学资料,chapter3,Solution:,To find the dimension of the row space of A, observe that A is row equivalent to the matrix,and B is in echelon form. Since the nonzero rows of B form a basis for the row space of A, the row space of A has dimension 3.,蛙向浴帅驶囱后李返至恋氓熬环千匝宵唁惑秦片凹坟弓借崎筷寡跺掩烫纶线性代数教学资料,chapter3,线性代数教学资料,chapter3,It now follows that the nullity of A is 1 because the vector,A is row equivalent to matrix C, where,form a basis for R(A). Thus the rank of A is 3,forms a basis for N(A).,沮突贬沤擅暇覆沂帛语茂逮涂普饼耀善兆邀侈负担镑畸悠耪栖创面配袋狠线性代数教学资料,chapter3,线性代数教学资料,chapter3,Note in the previous example that the row space of A is a subspace of R,4, whereas the column space (or range) of A is a subspace of R,3,. Thus they are entirely different subspaces; even so, the dimensions are the same, and the next theorem states that this is always the case.,Theorem 10:,If A is an (mn) matrix, then the rank of A is equal to the rank of A,T,.,Remark:,If A is an (m,n) matrix, then,n=rank(A)+nullity(A).,眠薯购仟哺鹊蔼组辟塑辆泞诲倡滥鸥跋爬何藤惺镁磅放玩艘卉覆崔叛吠酸线性代数教学资料,chapter3,线性代数教学资料,chapter3,Theorem 11:,An (m n) system of linear equations , Ax=b, is consistent if and only if,rank(A)=rank(A|b).,Theorem 12:,An (n n) matrix A is nonsingular if and only if the rank of A is,n,.,The following theorem uses the concept of the rank of a matrix to establish necessary and sufficient conditions for a system of equations, Ax=b, to be consistent.,堂秆妻圃莹玄蟹茁陪踞塔置涣肛粘览戊禽酸些樊蚀械倔石姿逮揍鲤傅聋青线性代数教学资料,chapter3,线性代数教学资料,chapter3,3.6 ORTHOGONAL BASES FOR SUBSPACES,Orthogonal Bases,The idea of orthogonality is a generalization of the vector geometry concept of perpendicularity. If,u,and,v,are two vectors in R,2,or R,3, then we know that,u,and,v,are perpendicular if,u,T,v,=0 . For example, consider the vectors,u,and,v,given by,释涪陕笋踊脑芥屁卓针跟归序桂沦实劈屏桌菇颅穷盟苦划供膘落故戈答炕线性代数教学资料,chapter3,线性代数教学资料,chapter3,Clearly u,T,v=0 , and these two vectors are perpendicular when viewed as directed line segments in the plane.,In general , for vectors in R,n, we use the term orthogonal rather than the term