### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# LINEAR ALGEBRA Math 121

UCI

GPA 3.73

### View Full Document

## 42

## 0

## Popular in Course

## Popular in Mathematics (M)

This 20 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 121 at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/201874/math-121-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

## Reviews for LINEAR ALGEBRA

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/12/15

Abstract These notes were taken in the Spring 2008 quarter at University of California at Irvine in Math 121A taught by AleX Sadovsky These notes served both as an opportunity for myself to become more comfortable with LaTeX as well as for studying Many of the proofs are original to me though some were taken directly from class The overall structure of the notes is the same as was in class although the rst section on vector spaces was taken almost directly from the course textbook Linear Algebra by F riedberg lnsel and Spence which is why I chose to omit many of the proofs The notes are rather condensed and l omitted almost all of the examples since those can be obtained by attending class or by formulating your own A familiarity with group theory is not needed but recommended when studying some of the proofs offered in these notes Any errors or inconsistencies are solely due to me Damek Davis 1 Vector Spaces De nition Vector Space A vector space V over a eld F on which two operations called addition and scalar multiplication respectively are de ned so that for each pair of elements X y E V and Oz 6 F there is a unique element X ay 6 V such that the following conditions hold 0 ForallXy VXyyX o For all Xyz EV X yz Xyz 0 There exists a vector 0 E V such that 0 X X 0 X for all X E V o For each X E V there eXists a y E V such that X y 0 y is denoted by 71 o For each X E V 1X X o For each pair of elements 043 6 F and each element X E V a X 04BX o For each element 04 E F and each pair of elements of Xy E V aX y 04X ay 0 For each pair of elements 043 6 F and each element X E V 04 BX 04X BX Observe that V is a group so the cancellation law automatically follows for vector spaces ie if V X V y then X y for all XyV E V lnverses of vectors are unique which follows from the group laws and the zero vector is also unique All of the trivial properties ie 0V 0 71V 7V 040 0 for all V E V and Oz 6 F follow naturally as well 0 De nition Subspaces A subspace W of a vector space V is a non empty subset which is a vector space in its own right If W is a subspace of V we say that W V Observations A non empty subset W of a vector space V is a subspace of V if and only if 0 0 E W o ley E V then Xy 6W 0 lfaEFX WthenaX W Lemma 11 The intersection of any collection of subspaces of a vector space V over a eld F is a subspace Proof Left as an exercise De nition If U W Q V are non empty then the sum of U and W denoted U W is de ned as UW uwu Uw W It is easily seen that if U W V then U W V 0 De nition Let V be a vector space over a eld F and let x1 xn E V A linear combi V L nation of x1 xn is any sum of the form Zaixi where 0416 F for i E 1 n i1 0 De nition Let V be a vector space over a eld F and let S Q V The span of S denoted span S is the set of all linear combination of the elements of S We de ne span 0 0 for convenience It is an immediate consequence that span S g V We say that S generates or spans V if span S V 0 De nition Linear Dependence A subset set S of a vector space V over a eld F is said to be linearly dependent if there exists a nite number of distinct vectors x1 xn E S and V L scalars 041 Ozn E F not all zero such that Zaixi 0 If S is not linearly dependent we say that S is linearly independent 111 0 Theorem 12 Let U Q W Q V If U is linearly dependent then W is linearly dependent lf W is linearly independent then U is linearly independent Proof Left as an exercise 0 Theorem 13 Let S be a linearly independent subset of a vector space V over a eld F and let V be any vector in V Then S U V is linearly dependent if and only if V E SpanS Proof See course textbook if needs clari cation 0 De nition Let V be a vector space over a eld F A basis of V is a linearly independent subset of V that generates V 0 Theorem 14 Let V be a vector space over a eld F and let X x1 xn be a subset of V Then X is a basis for V if and only if each V E V can be uniquely expressed as a linear combination of vectors in X Proof Left as an exercise 0 Theorem 15 If V is a vector space over a eld F that is generated by a nite subset S of V then there exists a subset of S that is a basis of V Hence V has a nite basis Proof See textbook 0 Theorem 16 Let V be a vector space that is generated by a set C containing exactly 71 vec tors and let L be a linearly independent subset of V containing exactly m vectors Then m g n and there exists a subset H of G containing exactly nim vectors such that LUH generates V Proof See textbook 0 Corollary 17 Let V be a vector space having a nite basis Then every basis for V contains the same number of vectors Proof Let X Y be two nite bases of cardinality m 71 respectively Since both sets t the criteria for L and G in the previous proof we see that m g n and n g m and hence m n 0 De nition A vector space V over a eld F is said to be nite dimensional if its basis is nite If it does not have a nite basis it is said to be in nite dimensional The dimension of V is the unique number of vectors contained in each basis of V and is denoted by dimV 0 Corollary 18 Let V be a vector space with dimension 71 over a eld F c Any nite generating set for V contains at least 71 vectors and a generating set for V that contains exactly 71 vectors is a basis for V c Any linearly independent subset of V that contains exactly 71 vectors is a basis for V 0 Every linearly independent subset of V can be extended to a basis of V Proof These are all direct consequences of the previous theorem 0 Theorem 19 Let W be a subspace of a nitedimensional vector space V Then W is nite dimensional and dimW dimV Moreover if dimW dimV then V W Proof See textbook 0 Corollary 110 If W is a subspace of a nitedimensional vector space V7 then any basis of W can be extended to a basis of V Proof The proof follows naturally from the previous theorems ltgt 2 Dual Spaces De nition A function f V a F is called a linear functional or a linear form on V if7 fax y afX fY for all 04 E F and Xy E V 0 De nition Let V denote the set of all linear forms on V On V de ne addition by7 f gX fX gX for all g 6 V 7 X E V and scalar multiplication by7 afX ozfX7 for all f E V and X E V and Oz 6 V Under these operations V is a vector space and is called the algebraic dual space of V 0 Notation fX is often written as X7 1 or X7 1 Also we have that7 lx y fl X7 fl ly7fl7 and X f gl X7 fl X79l for all g 6 V and Xy E V 0 Lemma 21 If V is nite dimensional and X X17 Xn is a basis of V7 and g 6 V satisfy the following condition fXk 9Xk for all Xk E V then f g Proof Let X 6 V7 then there eXists al 6 F such that X ogXi Hence7 i1 n n lel ZWlMJl Zailxhgl Xyl i1 21 Therefore 1 g 0 Lemma 22 If V is nite dimensional and X X17 7Xn is a basis then there eXists a unique subset X X 1 7Xn of V such that7 7 1 ifij lxXJli 07 my Proof By the previous lemma the subset is unique Theorem 23 X is albasis of V Proof Suppose that Z xix 0 then i1 7L ZailX17Xil 04739 0 i1 therefore the elements of X are linearly independent Let f E V and de ne fX 31 where the 316 F for i 1 n We see that f Z ilxwxil i1 71 then f Z ixg39xi E spanX by lemma 21 Hence the elements of X span V and is i1 therefore a basis 0 Terminology The set X is called the dual basis of X 0 Corollary 24 If V is nite dimensional then dimV dimV Proof It is immediately seen from theorem 23 0 De nition Annihilators Let V be a vector space over a eld F and let S C V The annihilator of S is the set SOf V Xf0forallx S Lemma 25 SO is a subspace of V Proof Left as an exercise Lemma 26 Let X5 X1 Xk be a basis of S then SO Xg k Proof Let f 6 X2 and X E S then X Zaixi and hence i1 k val Zailxi7f 0 i1 Therefore 1 6 SO so X2 Q SO Also 50 Q X2 by lemma 21 0 Lemma 27 Let S X1 Xk be a linearly independent set in an n dimensional vector space V then dimSO n 7 k Proof By the replacement theorem we can extend S to a basis for V Let Y X1 XkXk1 Xn Consider the dual basis for V YX17 7Xiwxis17 7xn We claim that X v x is a basis of SO They are linearly independent since they are a subset of a linearly independent set we need only show that they span SO Let X 6 SO then V L X E aixg for 04 E F Hence i1 04739Xjx 0forallj 12 k Thus 041 042 04k 0 and therefore 1 E spanx 1 and is therefore a basis for SO 0 3 Bilinear Forms Isomorphisms Direct Sums De nition Direct Sums Let U and V be a vector space over F furnish U x V with the addition 1117 V1 1127 V2 111 1127 V1 V2 and scalar multiplication by C111V1 6111 CV1 U x V is a vector space over F and is denoted by U 3 V Theorem 31 Let U and V be nite dimensional vector spaces over a eld F then dimU Q3 V dimU dimV Proof Let X be a basis of U and Y be a basis of V such that le n and lYl m We exhibit the basis X1707 7Xn707 Ovylt 707ym 0 De nition Let V U over F A subspace W U is a compliment ofV in U ifV W U 0 De nition Let U and V be a vector space over F A function w U V a F is called a bilinear form on U 3 V if 0 for all xed x E U w1y wx y E V o for all xed y E V w2x wxy E U Examples lnner product 0 Theorem 32 Let V be a nite dimensional vector space over a eld F and let X x1 xn be a basis of U and Y y1 ym be a basis of V If 04 ij E 1 n are xed scalars in F then there exists a unique bilinear form w on U 3 V such that wXi7yg39 041739 for all j e 1 7 71 Proof Let x E U y E V Then there exist unique scalars Blj Vii such that x Z ixi i1 and y 2211 Yij 1009 Y ME 519 2 ijz39 Z 1009 2 Wm Z Z Bi39YjwXi7 yj i1 i1 39 j1 i1 F1 1 m 2 BHjo 7391 M3 1 H 0 De nition Isomorphism Suppose U V are vector spaces over a eld F A bijection A U a V such that Ax ay Ax aAy for all x y E U and Oz 6 F is called an isomorphism from U to V 0 Lemma 33 Finite dimensional vector spaces U V over a eld F are isomorphic if and only if dim V dim U Proof Let X x1 xn and Y y1 ym be bases of V U respectively Suppose V U are isomorphic there isomorphism given by a function 1 V a U and dim V 7 dim U then without loss of generality since 1 is an isomorphism we can equally choose l and obtains the opposite scenario if dim V gt dim U we see that the vectors gtx1 gtxn are linearly dependent since 71 gt dim U Since they are linearly dependent there exists scalars 0416 F where i E 1 n not all zero such that 041 gtX1 an xn 0U but since 1 is one to one we must have that gtOV 0U and hence gt 1a1 gtX1 an xn 1 Q1X1 am mm aan 0v but the 041 s are not all zero7 this is a contradiction Therefore dim V dim U Suppose dim V dim U7 then de ne the linear function 1 V a U by gtxi y we only need to de ne functions on the basis vectors since 1 is a linear function We see immediately that j is an isomorphism ltgt Consequence of lemma 33 0 Every n dimensional vector space V over a eld F is isomorphic F 0 Convention We shall identity V with V 3 x and W with V 3 x Theorem 34 Let U7 V7 W be vector spaces over a eld F The following conditions are equivalent 1 U V Q3 W 2 V WO and VWU 3 For all z E U the exists a unique x E V and a unique y E W such that z x y Proof Left as an exercise 0 Corollary 35 If V U then there exists a W U such that U V W W is not uniquely determined 0 31 Multilinear Forms Let Vk be vector spaces over a eld F for k E 17 7m and let W V1 Vm De nition A function p W a F is called an m linear form if for all xk7yk E Vk and all 04 E F 1 Mn ay17X27 7Xm pX17 7Xm apy17 7Xm 2 PX17X2 043 27X37 7Xm pX17X27 7Xm 0419X17YZ7 7Xm 3 pX17 7Xmaym pX17 7Xm apX1wquot 7ym 4 Permutations Parity Recall Let A be a nite set a permutation of A is a bijection a A 7 A Let 07 be permutations of A then the following are immediate o 0 1 is a permutation of A o 7 o a is a permutation of A 0 Notation We will more commonly write 7390quot dropping o and call this the product of 739 and a Note in general 7390quot 7 0739 Also note the composition 039 0 00 0 039 o 0 are written 0203 respectively 0 De nition Let x E A o The set orbma anz n E Z is called the orbit of z under 0 Since orbma C A we have immediately that lorbm al o A permutation a of A is said to x z E A if oz m It follows that am x if and only orbz a The orbit consisting of only one element of A is said to be trivial o A permutation with at most one non trivial orbit is called a cycle 0 Two cycles whose nontrivial orbits are disjoint are said to be disjoint o A cycle whose nontrivial orbit has exactly two elements is called a transposition o The identity permutation on A is the permutation that xes all z E A 0 Observations o If y E orbma then 0y 6 orbma o lorbmal Al as stated earlier Lemma 41 There exists an integer k 2 1 such that 0k z and orbz a z am ak l Proof Since orbz a is nite there exists an n E N such that o m am an lm are distinct o a is one of zaz an l We want to show that a z z Suppose not then there exists a k E 1 n 7 1 such that 01 anz Apply a kakz m anik But zaz a 1 are all distinct and n 7 k E 1 n 7 1 hence we have reached a contradiction so a z m 0 Theorem 42 Every permutation of a nite set is the product of disjoint cycles Proof Let a be a permutation of a set A Let B1 B2 BT be the orbits of 0 De ne the function 77 for i E 1 r as follows 0m if m 6 Bi z otherwise Then it is easy to see that 0 7391 77 Since the equivalence classes are all pairwise disjoint we must have that the cycles are all disjoint in the product 0 Theorem 43 Every cycle can be written as the product of transpositions Proof If 11 an is a cycle on A then 11 an a1 anal MA 1102 0 Theorem 44 Every permutation can be written as the product of an even number of trans positions or an odd number of transpositions but not both Sketch of Proof Let 251 tn 6 R and be all distinct Construct the function ft17 in H t7 7 jgtk V 137316371 We see that f is non zero since the symbols 251 tn are all distinct For each permutation 0 on a set A 1 71 de ne Uft17 H 7771 fltt017 397t07139 We see that 0f f or if So the formal ratio riff i1 Uif is called the parity signum or signature and is denoted by 39710 0 De nition A permutation 0 is odd respectively even is 39710 71 respective 39710 o 39710739 397103971739 o 3971tra713p031t1071 71 O 5 kLinear Forms on a Single Vector Space De nitions Let V be a vector space over a eld F and for k E N kVV V q 11 times Such forms are called k linear forms on V 0 Given a k linear form w on V and a permutation 0 on 1 k we de ne 10 awx1 xk wxa1 xak for all x1 xk 6 kV o A k linear form w on V is said to be 1 symmetric if aw w 2 skew symmetric if aw sgnaw Examples 0 Scalar Product on R3 wx1x2 x1 x2 wx2 x1 We see that this bilinear form is symmetric 0 Cross product on R3 77dotted77 with another vector For a xed vector 2 E R3 vx1 xy x1 x x2 2 is a skew symmetric bilinear form 0 De nition A k linear form w on V is said to be alternating if wx1 xk 0 whenever at least two of the argument of xj s are equal Theorem 51 An alternating k linear form on V is skew symmetric Proof For k 2 we can simply observe that if w is alternating 0 wXy7Xy wX7X wy7x wX7y wy7y wy7xwx7y hence wyx 7wx y for all x y E V Now suppose w is an alternating k linear form on a vector space V over F It is suf cient to prove that aw 7w for a a transposition on the set 1 k Suppose a Jake jo kg 6 1 k De ne a bilinear form w0x y where x y E V by xing all arguments of w except the the two arguments with indices jo k0 where xy are located respectively Hence arbitrarily chosen and fixed 7 A A waX7y w7 3 y joth 1711105 koth place By this construction wa E BilU is alternating and hence it is skew symmetric It follows that aw w 0 Observations o The converse of the previous theorem is not true namely a skew symmetric bilinear form is not necessarily alternating This happens only over elds that have characteristic k 7 2 o For all k E N the set of k linear forms constitutes a vector space over F Now let V be a vector space of nite dimension n over a eld F Theorem 52 Two alternating n 7 linear forms on V are linearly dependent Proof The following proof uses a theorem that is proved later in the notes namely if w is a nonzero alternating n linear form on a vector space V then wx1 Xn 7 0 if and only if X1 xn are linearly independent Let X X1 Xn be a basis of V and let wq be any two nonzero alternating n linear forms on V By the previous argument we see that wx1 Xn 39yw 7 0 and qx1 Xn 39yq 7 0 De ne an alternating n linear form h by h 39qu 39ywq Hence we see that hX17 7Xn vqwX17 7X 7 quX17 7X 397qu i 397qu 0 Thus h E O and we see that w q and hence the vector space of all alternating n linear forms on a vector space V over a eld F is 1 dimensional 0 Note We have not proven that there exists a nonzero alternating n linear form It is infor mative to look at Halmos7 Finite Dimensional Vector Spaces section 31 for a proof of this fact This note and theorem 52 directly proves that the vector space of all alternating n linear forms on a vector space V is one dimensional which is extremely important to determinant theory 6 Linear Transformations Let U V be vector spaces over a eld F De nition A function A U a V is called a linear transformation if AX 043 AX aAy Notation 0 Drop parentheses when of the form AX to form the same product AX other wise keep parentheses o The set of all linear transformations from U to V is denoted U V o The set of all linear transformations from U to U is denoted U o A linear transformation A E V over F is said to be invertible if it is bijective A is invertible if and only if A is an automorphism of V De nition Given A E UV the set kerA X E U AX 0 is called the kernel or nullspace of A Observe that kerA U Proposition 61 A E U7 V is injective if and only if kerA Proof Left as an exercise 0 De nition We de ne RankA DimRangeA as the rank of A7 and NullityA dimkerA as the nullity of A RankNullity Theorem 62 lfU is nitedimensional7 then 71 dimU dimRangeA dimkerA Proof We shall prove UkerA RangeA De ne a linear transformation f UkerA a RomgeA7 by gtx kerA Ax f is injective since ker gt kerA f is obviously onto Hence by construction f is an isomorphism and hence we are done 0 Proposition 63 If U7 V are vector spaces over a eld F7 U7 V is a vector space over F Proof Simple observation Just notice that U7 V U7 V the set of all functions from U to V 0 Proposition 64 If A7 B E U7 V7 then the mapping A o B de ned A o Bx ABx and denoted AB7 is also in U Proof Left as an exercise Let V be a vector space F Theorem 65 If B70 6 V are such that AB CA I then A is invertible and B C A l Proof We will rst prove that A is a bijection 0 Suppose Ax Ay where xy 6 A7 then we see that CAx CAy ltgt x y Hence A is injective 0 Let y E V We see that ABy y7 hence A is surjective Since A is bijective A 1 exists7 hence since AB CA I we see that B C A l Theorem 66 If A7 B E V are invertible7 then AB 1 B lA l Proof Left as an exercise 0 Theorem 67 Suppose dim V n For A E V to be invertible it is sufficient that A be either injective or surjective Proof Let X X17 Xn be a basis of V 0 Suppose A is injective7 we will show that AX17 Axn form a basis of V 7 Linear independence Let 04 E F for i E 17 7n Suppose 041AX1 anAXn 07 by the linearity of A we can pull it out77 it yielding7 A041X1 anxn 0 But by previous proposition A is injective if and only if kerA 0 hence 041X104nxn 0 Finally since X1 xn form a basis of V we have that 041 an O 7 Spanning set Since AX17 Axn form are n linearly independent vectors in the n dimensional vector space V7 they must span V 0 Suppose A is surjective7 then in particular there exists 21 2 E V such that Azi Xi for i E 1 771 We will rst show that zi7 zn from a basis of V Suppose 3121 nzn 07 where 317 3 E F then ABlzl nzn A0 ltgt BlAzl nAzn 0 gt31X1 nxn Oandhence i Oforalli E 1 771 Hence since 21 2 are n linearly independent vectors in an n dimensional vector space V they must span V We need to show that kerA Suppose z E V such that Az 07 then since 217 Zn is a basis of V there exists 0417 704 E F such that 71 z E aizi Thus7 i1 AZ A Zn aizi i1 H M E Cu N Thus 041 an 0 since X forms a basis of V Therefore we see that z 0 and hence A is injective 7 Linear Transformations and Matrices Let U and V be nite dimensional vector spaces over a eld F and let X x1 xm Y y1 yn be the bases of U and V respectively If A E U V then there exists 0W 6 F such that n AXJ39 Zaijyi i1 wherej E 1 m andi E 1 n The doubly indexed set wigij is called the matrix of A with respect the the bases X Y and is denoted AXy Observations Let U V be nite dimensional vector spaces over a eld F and let A B E UV Let X x1 xmY y1 yn be the bases of U and V respectively De ne the matrices AXy 0 ley Blj We see that A BlXY 04 Bum If we now suppose that V U then we see that ABXJ39 i1 H M3 BijAXi H H M3 51 aijk k1 H M Bijakjxk l H 39ijxk Mama w H H Hence we can de ne the new matrix Aley Wig If U V W are vector spaces over a eld F with bases X Y Z respectively and A E V W B E U V then we can similarly de ne ABXZ O 71 Invariance Let V be a vector space over a eld F and let A E V De nition Invariance A subspace W V is said to be invariant under A if AW C W From this de nition it is easily derived that if W invariant under A then AW W O De nition Reducibility Suppose V is a vector space over a eld F and V U W The sum U 3 W is said to reduce on A E V if U and W are each A invariant Example Let F R and let P where is the said of all polynomials with coefficients in F Consider A such that Am t2zlttgt We see that B the set of all polynomials of even degree and C the set of all polynomials of odd degree are both invariant under A hence P B 90 0 Let V be a vector space over a eld F such that V U 3 W for some subspaces U and W Recall by theorem 23 that for all X E V there exists unique vectors y E U and z E W such that X y 2 De ne a linear transformation A E V such that AX y where y is the vector in the sum above De nition Projections The A E V de ned above is called a projection on U along W Proposition 71 If V U W and if A is a projection on U along W then 0 kerA W o For all y E UAy y Proposition 72 Suppose A E V and for all y E ImA Ay y then A is a projection of ImA along kerA Proof By theorem 34 it suffices to show that ImA kerA 0 V ImA kerA For suppose X E ImA kerA then X AX 0 hence X 0 For we know that V ImA kerA ltgt VkerA ImA so we can simply de ne an isomorphism from VkerA ImA the rst isomorphism theorem from group theory De ne j VkerA a ImA by gtX kerA AX o lnjectivity We see that ker gt kerA and hence j is injective o Surjectivity Let X E ImA then AX X and hence gtX kerA AX X Thus 1 is surjective 0 Linear transformation Let X kerA y kerA E VkerA Oz 6 F We see that gtX kerA ay kerA ay kerA gtltx 1mm my We Thus we see that j is an isomorphism and hence V ImA kerA Therefore V ImA kerA and A is the projection of ImA along kerA ltgt 8 Determinant Theory Let V be a vector space of nite dimension 71 over a eld F Let AltVn be the set of all alternating n linear forms on Vs Recall o AltVn is a vector space 0 dim AltV n 1 Consider A E V and y1 yn E V For all w E AltVn de ne Aw by Awy17 yn wAy17 AM The mapping Aw E AltVn so there exists a 6A 6 F such that Aw 6410 since dim AltV n 1 De nition For all A E V we call 6A the determinant of A and denote it by detA ltgt The following properties are immediately evident o detAB detAdetB for all AB E V This is easily seen since detBAw BAw BAw BdetAw detABw detAdetBw detAdetBw o 15251 1 where I is the identity transformation This is clearly true since w w 1w 0 lfA E V is invertible then 1 15251 detA 1A detA 1detA hence detA 7 0 since F has no zero divisors and detA 1 18114 De nition Let A E V with detA 0 respectively 7 0 is said to be singular respec tively nonsingular Thus invertible implies non singularity We will quickly see that converse is true with the aid of the following two theorems 0 Theorem 81 If w E AltV n and if y1 yn are linearly dependent in V then wy1w 73 0 Proof Since y1 yn are linearly dependent there exists k g n such that kil y 2 My i1 where oz 6 F for all i E 1 k 71 Therefore we see that kil wy17 7Yk7Yk17 yn wltY17quot39vzaiYi7Yk17 397Yn i1 kil Zaiw 17quot 7Yi7Yk17 7yn 7 W 171 where symbolizes that yi already appears among the preceding vectors for each i E i k7 1 Therefore We see that wy17 7yn 21104139100 1739 7Yi7Yk17quot39 73 0 17 0 Theorem 82 If w E AltV n with w 7 0 and if y1 yn are linearly independent in V then wy1 yn 7 0 Proof Suppose that there exists y1 yn E V such that the vectors are linearly independent and wy1 yn 0 We will show that under this supposition w applied to any set of n linearly independent vectors is 0 and hence w E 0 since wy1 yn 0 if y1 yn are linearly dependent To show this let x1 xn be any set of n linearly independent vectors De ne an invertible linear transformation A E V such that Ayl x for all i E 1 n Since detA 7 0 we have that 0 wy1w 7yn detAwy1yn Awy1 yn wAY1 7AYn wx1 xn Therefore we see that w E O which is a contradiction and hence wy1 yn 7 0 Proposition 83 A is invertible if and only if A is nonsingular Proof We already noted above that A being invertible implies that A is nonsingular Let w be a nonzero alternating n linear form Suppose A is nonsingular then detA 7 0 Since V is nite dimensional it suf ces to show that A is injective which is equivalent to showing that kerA Suppose A is not injective then there exists y1 7 0 E V such that Ayl 0 We can extend yl to a basis Y y1 yn for V Since y1 yn are linearly independent we see that detAwy1 yn 7 0 But detAwy1 yn Awy1 yn wAy1 Ayn w0Ay2 AM 0 Thus we have reached a contradiction hence A is injective and A is invertible 0 Theorem 64 Let V be a vector space over a eld F dim V n X x1 xn and A E V Let AX 0 The determinant of A in terms of the entries of its matrix is given by the following formula detA ngnaHltkgtk k1 7 where a 6 Sn Proof For all w E AltV7 n7 we see that7 detAwX1 7Kn wAX1 7AX 71 TL ME aiilxz m 7 Z ainnxm i11 151 TL TL 71 E ai11wxi17 E O izlxi27 7 E ainnxin i11 i21 151 TL 71 E aillamwxil7 7X i11 i11 n TL Z aikkwxi1 Xin ij ik then whole term 0 i1 in k1 H O akkwxa17 H 7X00 k4 H O akk59n 7lwX17 am Therefore we see that7 V L detA Z sgna 7 k 040kk 1 9 Ad joints Let U7 V be vector spaces over a eld F and let A E U7 V De nition To every 1 E V 7 make correspond a g E U such that AX fl X7 gl for all X E U This 9 is determined uniquely since if there was another such 9 it would have the same value at every basis vector of U and hence would be the same function The function A V a U that sends A f g is an element of V 7 U Hence7 AX fl X7147 This function A is called the adjoint of A 0 Question Suppose V is nitedimensional7 X X1 Xn is a basis of V7 and X 1 7x n is the dual basis The matrix of a linear transformation A E V in the basis X is known to be Alx ahj j Find the matrix of A in X How are detA and detA related 71 Proof We know that ij Zaijxi Let X 6 V7 then 171 X73 14th ijv xtl 7L l aijxn xtl i1 TL Z 041739 X13 Xil i1 akjlxkvxlsl akllxijil akzlxjv X2l aknlxgv XLLl 7L Zakilxjvxgl i1 n Xj72akix2 i1 Hence we see that A XQ ELI aimX since the linear functionals7 X9 are completely de termined by their values on the basis vectors in X Hence if we let 3 041 for all i k E 17 771 we see that A XQ ELI ikxg which implies that A X Blj 047074 We know that detA Z sgn0H wow 7 k1 And hence 15254 Z sgnaH 50W Z sgnaH Mm detA c7 k1 U k1 0 Observation A linear transformation A E V is invertible if and only if A is invertible 0 Question Let V U 63 W 7 and suppose A E LV is a projection on U along W Prove that A is the projection on W0 along U0 Proof We shall prove that RangeA W07 kerA U0 and A 2 A o A 2 A Let V E V and V E V Since A2 A we see that7 V7 AVl Therefore since V7 V were arbitrary we see that A 2 A

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.