Applied Linear Algebra
Applied Linear Algebra MATH 2210
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This 19 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 2210 at University of Connecticut taught by Sarah Glaz in Fall. Since its upload, it has received 23 views. For similar materials see /class/205804/math-2210-university-of-connecticut in Mathematics (M) at University of Connecticut.
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Date Created: 09/17/15
45 The Dimension of a Vector Space THEOREM 9 If a vector space V has a basis b1bn then any set in Vcontaining more than n vectors must be linearly dependent Proof Suppose u1up is a set of vectors in therep gt n Then the coordinate vectors u1 up are in Rquot Sincep gt n u1 up are linearly dependent and therefore u1upare linearly dependent I THEOREM 10 If a vector space V has a basis of n vectors then every basis of V must consist of n vectors Proof Suppose 1 is a basis for Vconsisting of exactly n vectors Now suppose g is any other basis for V By the de nition ofa basis we know that 1 and g are both linearly independent sets By Theorem 9 if 1 has more vectors than g then cannot be the case is a linearly dependent set which Again by Theorem 9 if g has more vectors than 1 then cannot be the case is a linearly dependent set which Therefore g has exactly n vectors also I DEFINITION If Vis spanned by a nite set then Vis said to be finitedimensional and the dimension of V written as dim V is the number of vectors in a basis for V The dimension ofthe zero vector space 0 is de ned to be 0 If Vis not spanned by a finite set then V is said to be infinitedimensional EXAMPLE The standard basis for P3 is So dim P3 In general dim Pquot n 1 EXAMPLE The standard basis for Rquot is e1en where e1 e are the columns of So for example dim R3 3 EXAMPLE Find a basis and the dimension ofthe subspace ab2c 2 2b 4 d W CH C abcdarerea bcd 3a3cd 61 1 20 1 1 2 0 2 2b 4 d 2 2 4 1 Solution Since CH 0 a b c d b c 51 0 1 1 1 3a3cd 3 0 3 1 W spanv1v2v3v4 where 1 1 2 0 2 2 4 1 V1 V2 V3 V4 0 1 1 1 3 0 3 1 0 Note that v3 is a linear combination ofvl and v2 so by the Spanning Set Theorem we may discard v3 0 v4 is not a linear combination ofvl and v2 So V1v2v4 is a basis for W 0 Also dim W EXAMPLE Dimensions of subspaces of R3 0dimensional subspace contains only the zero vector 0 1dimensional subspaces Spanv where v 0 is in R3 These subspaces are through the origin 2dimensional subspaces Spanu v where u and v are in R3 and are not multiples of each other These subspaces are through the origin 3dimensional subspaces Spanuvw where u v w are linearly independent vectors in R3 This subspace is R3 itself because the columns ofA u v w span R3 according to the IMT THEOREM 11 Let H be a subspace of a nitedimensional vector space V Any linearly independent set in H can be expanded if necessary to a basis for H Also H is nitedimensional and dim H dim V 1 EXAMPLE LetH span 0 1 ThenH is a subspace of R3 and dimH lt dim R3 0 0 1 1 1 1 0 We could expand the spanning set 0 1 to 0 1 0 to form 0 0 0 0 1 a basis for R3 THEOREM 12 THE BASIS THEOREM Let V be ap 7 dimensional vector space p 2 1 Any linearly independent set of exactlyp vectors in Vis automatically a basis for V Any set of exactlyp vectors that spans Vis automatically a basis for V EXAMPLE Show that L1 7 t1 t7 t2 is a basis for P2 Solution Letv1tv2 17tV3 1t7t2 and 3 Luz Corresponding coordinate vectors 0 1 1 v1 1 V2 1 V3 1 0 0 71 v2 is not a multiple of v1 v3 is not a linear combination of v1 and v2 v1 v2 v3 is linearly independent and therefore V1v2v3 is also linearly independent Since dim P2 3 V1v2v3 is a basis for P2 according to The Basis Theorem Dimensions of Col A and Nul A Recall our techniques to find basis sets for column spaces and null spaces 1234 2478 1 2 3 4 1 2 3 4 2 4 7 8 0 0 1 0 So is a basis for ColA and dim ColA 2 EXAMPLE SupposeA Find dim ColA and dim NulA Solution Now solve Ax 0 by rowreducing the corresponding augmented matrix Then we arrive at 12340 12040 24780 00100 x1 ZXZ 4X4 X3 0 and x1 2 4 X2 1 0 xz X4 X3 0 0 X4 1 72 74 1 0 So Is a baSIS for Nul A and 0 0 0 1 dim NuIA 2 Note dim Col A number of pivot columns ofA I I dim Nul A number of free variables ofA 41 Vector Spaces amp Su bspaces Many concepts concerning vectors in Rquot can be extended to other mathematical systems We can think of a vector space in general as a collection of objects that behave as vectors do in Rquot The objects of such a set are called vectors A vector space is a nonempty set Vof objects called vectors on which are defined two operations called addition and multiplication by scalars real numbers subject to the ten axioms below 9 P 01 G l 00 D The axioms must hold for all u v and w in Vand for all scalars c and d uvisinV u v v u uvw uvw There is a vector called the zero vector 0 in Vsuch that u 0 u For each u in V there is vector 7u in Vsatisfying u iu 0 cu is in V Cu v cu cv cdu cudu Cdu Cdu 10 IU u Vector Space Examples EXAMPLE Let sz2 ab Cd a b c dare real In this context note that the 0 vector is EXAMPLE Let n 2 0 be an integer and let Pquot the set of all polynomials of degree at most n 2 0 Members of Pquot have the form pt a0 a1t aztz antquot where a0a1 We Will just verify 3 out of the 10 axioms here Let pt a0 a1t antquot and qt b0 b1t bntquot Let Axiom 1 an are real numbers and t is a real variable The set Pquot is a vector space c be a scalar The polynomial p q is de ned as follows p qt ptqt Therefore P cIX Pt lt t which is also a of degree at most Axiom 4 0 00t 0tquot zero vector in Pquot p 0t pt0 a0 0a1 0t a0 a1t antquot pt and so p 0 p Axiom 6 CPXI CPU t which is in Pquot The other 7 axioms also hold so Pquot is a vector space Iquot So pqisin Pquot an 0tquot tquot Su bs paces Vector spaces may be formed from subsets of other vectors spaces These are called subspaces A subspace of a vector space Vis a subset H of Vthat has three properties a The zero vector of Vis in H b For each u and v are in H u v is in H In this case we say H is closed under vector addition c For each u in H and each scalar c cu is in H In this case we say H is closed under scalar multiplication If the subsetH satis es these three properties then H itself is a vector space a EXAMPLE LetH 0 a and b are real Show thatH is a subspace of R3 Solution Verify properties a b and c of the de nition of a subspace a The zero vector of R3 is inH let a and b b Adding two vectors in H always produces another vector whose second entry is and therefore the sum of two vectors in H is also in H H is closed under addition c Multiplying a vector in H by a scalar produces another vector in H H is closed under scalar multiplication Since properties a b and c hold Vis a subspace of R3 Note Vectors a0b inH look and act like the points ab in R2 X2 X1 EXAMPLE sH x1 x is real a subspace of x e does H satisfy properties a b and c X1 05 05 l 15 2 Graphical Depiction of H Solution All three properties must hold in order forH to be a subspace of R2 Property a is not true because Therefore H is not a subspace of R2 Another way to show that H is not a subspace of R2 Let 1 1 and so u v 3 which is in H 80 property b fails and so H is not a subspace of R2 05 05 1 15 2 Property a fails Property b fails A Shortcut for Determining Subspaces THEOREM 1 lfv1vp are in a vector space V then Spanv1 vp is a subspace of V Proof In order to verify this check properties a b and c of definition of a subspace a 0 is in Spanv1 vp since 0 7v17v2 vp b To show that Spanv1 vp closed under vector addition we choose two arbitrary vectors in Spanv1vp u a1v1a2v2 Clpr and V b1v1 bzvz bpvp Then u V a1V1a2v2 apvpb1v1 bzvz bpvp 7V17V17V2 7v2 7vp 7vp 511 b1V1a2 bzV2 0p prp So u v is in Spanv1 vp c To show that Spanv1 vp closed under scalar multiplication choose an arbitrary number c and an arbitrary vector in Spanv1 vp V b1v1 bzvz bpvp Then CV Cb1V1 bzvz bpvp v1 v2 vp 80 av is in Spanv1 vp Since properties a b and c hold Spanv1 vp is a subspace of V Recap 1 To show that H is a subspace of a vector space use Theorem 1 2 To show that a set is not a subspace of a vector space provide a speci c example showing that at least one ofthe axioms a b or c from the de nition of a subspace is violated EXAMPLE ls V a 2b 2a 7 3b a and b are real a subspace of R2 Why or why not Solution Write vectors in Vin column form liaiiillzilliillillil So V Spanv1v2 and therefore Vis a subspace of by Theorem 1 a 2b EXAMPLE ls H a 1 a and b are real a subspace of R3 Why or why not a Solution 0 is not in H since a b 0 or any other combination of values for a and b does not produce the zero vector So property fails to hold and therefore H is not a subspace of R3 2 EXAMPLE Is the setHof all matrices ofthe form 3 a b 3b a subspace ofMM a Explain Solution Since 2a b 7 2a 0 0 b 3ab 3b 3a 0 b 3b 2 0 0 1 ThereforeH Span 3 0 1 3 and soHIs a subspace ofMM 61 Inner Product Length amp Orthogonality Not all linear systems have solutions 1 2 3 EXAMPLE No solution to x1 exists Why 2 4 x2 2 1 Axis a point on the line spanned by 2 and b is not on the line So Ax b for all x x1 14 Using information we will learn in this chapter we will find that 2 0 so that A 1 4 2 8 39 Segment joining AR and b is perpendicular or orthogonal to the set of solutions to Ax b Need to develop fundamental ideas of length orthogonality and orthogonal projections The Inner Product ul V1 H2 V2 Inner product or dot product of u and v uquot Vn v1 T V2 u39vuvu1 2 un u1V1u2V2unvn Vn Note that V u v1u1vzu2 vnun u1v1u2vz unvn u V THEOREM 1 Let u v and w be vectors in Rquot and let c be any scalar Then a u v v u b uvw uwvw C CH V cu V u CV d uu20anduuOifandonlyifu0 Combining parts b and c one can show 01u1 cpup w 01u1 w cpup w Length of a Vector For v v2 the length or norm of v is the nonnegative scalar ilvil de ned by iviWJv v wv and ivi2vv For example ifv Z then ilvi Jaz b2 distance between 0 and v Picture For any scalar c ilcVil ici ilVil Distance in Rquot The distance between u and v in Rquot distuv lu 7 vl This agrees with the usual formulas for R2 and R3 Let u u1u2 and v v1vz Then u 7 V u17 v1u2 7 V2 and distuv lu 7 Vll 7V1u2 7x22 1117 v12 u2 7x222 Orthogonal Vectors 039i5tuyV2 lluivll2 quotV39UV u39uVV39uV uu7uv7vuvvlul2lvl272uv 2 distuv2 lul2lvl272uv Similarly distu7v2 lu H2 lvl2 2u v Since distu7v2 distuv2 u v Two vectors u and v are said to be orthogonal to each other if u v 0 Also note that if u and v are orthogonal then lu vl lu H2 lvl2 THEOREM 2 THE PYTHAGOREAN THEOREM Two vectors u and v are orthogonal if and only if lu vl lu ll2 lvl2 Orthogonal Complements If a vector 2 is orthogonal to every vector in a subspace W of Rquot then 2 is said to be orthogonal to W The set of vectors 2 that are orthogonal to W is called the orthogonal complement of W and is denoted by Wi read as W perp Row Null and Columns Spaces THEOREM 3 LetA be an m x n matrix Then the orthogonal complement ofthe row space ofA is the nullspace ofA and the orthogonal complement ofthe column space ofA is the nullspace ofAT RowAY NulA Colif NulAT Why See complete proof in the text Consider Ax 0 gtllt gtllt gtllt 0 gtllt gtllt 7 0 gtllt gtllt gtllt 0 r1 X r2 X 0 Note that Ax and so X Is orthogonal to the rowA since X Is orthogonal rm x 0 to r1rm EXAMPLE LetA 1 0 1 2 0 2 0 1 Basis for NuIA 1 0 and therefore NuIA is a plane in R3 0 1 1 Basis for RowA 0 and therefore RowA is a line in R3 71 Basis for CoIA i and therefore CoIA is a line in R2 72 Basis for NuIAT 1 and therefore NuIAT is a line in R2 Subspaces NuIA and RowA Subspaces NuIAT and CoIA 23 Characterizations of Invertible Matrices Theorem 8 The Invertible Matrix Theorem LetA be a square n x n matrix The the following statements are equivalent ie for a given A they are either all true or all false A is an invertible matrix on A is row equivalent to In A has n pivot positions 90 The equation Ax 0 has only the trivial solution F The columns ofA form a linearly independent set h The linear transformation x an is onetoone The equation Ax b has at least one solution for each b in Rquot 3quot The columns ofA span Rquot The linear transformation x an maps Rquot onto Rquot There is an n x n matrix C such that CA In k There is an n x n matrixD such thatAD In AT is an invertible matrix EXAMPLE Use the Invertible Matrix Theorem to determine ifA is invertible where 1730 A 74111 2 73 Solution 1730 173 0 A 74111 0711 2 73 0 016 3 pivots positions Circle correct conclusion MatrixA is is not invertible EXAMPLE Suppose H is a 5 x 5 matrix and suppose there is a vector v in R5 which is not a linear combination ofthe columns ofH What can you say about the number of solutions to Hx 0 Solution Since v in R5 is not a linear combination of the columns ofH the columns ofH do not R5 So by the lnvertible Matrix Theorem Hx 0 has lnvertible Linear Transformations For an invertible matrix A 14 le x for all x in Rquot and AA lx x for all x in Rquot Pictures A linear transformation T Rquot a Rquot is said to be invertible if there exists a function S Rquot a Rquot such that STx x for all x in Rquot and TSx x for all x in Rquot Theorem 9 Let T Rquot a Rquot be a lineartransformation and letA be the standard matrix for T Then Tis invertible if and only ifA is an invertible matrix In that case the linear transformation S given by Sx A lx is the unique function satisfying STx x for all x in Rquot and TSx x for all x in Rquot
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