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# Class Note for CSCI 124 at GW (3)

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Date Created: 02/07/15

Vector Spaces Linear Independence Notes for CSci 124224 Poorvi L Vora We reproduce Example 4 from the previous set of notes Example 4 Determine if 3 4 3 is a linear combination of 1 1 V1 7 1 V2 7 71 V3 7 1 1 1 1 O 1 1 0 a1 3 1 71 1 a2 4 1 1 0 a3 3 if it exists Gaussian elimination gives 1 1 0 a1 3 0 72 1 a2 1 0 0 0 a3 0 where the nal equation is of the form 0 O This means you have a consistent set of equations with several solutions One solution is 13 1 a2 0011 3 That is 1 1 0 3 3 1 0 71 1 1 4 2 1 1 0 3 Notice that three equations in three unknowns give several solutions This happens because one of the equations is redundant in the presence of the other two it provided no new information which is why it disappeared on gaussian elimination There were essentially two equations in three unknowns and hence several solutions This in turn happens because one of the vectors vi is redundant That is any one of the three vectors viEl1 lies in the span of the other two and hence Vector SpacesVoraGWU 2 a third vector does not provide an additional independent direction For example of the three vectors in equation 1 1 1 1 1 0 5 1 75 71 1 1 1 0 That is one can add some amounts of the vectors to cancel one another out and get the zero vector For example 1 1 0 0 1 1 517 717100 3 1 1 0 0 We can add any amounts of the zero vector equation 3 to equation 2 to get several solutions to our problem Suppose we add the zero vector 71 times we get equation 2 71x equation 3 1 1 0 1 1 1 1 0 3 3 1 0 71 1 1 n7 1 if 71 7 1 4 1 1 0 1 1 0 3 or n n 1 0 3 35 1 75 1 17n 1 4 1 1 0 3 Henceingenerala13a27a317nforallintegersn The above issue does not arise when the vectors V1 V2 V3 are linearly independent Loosely speaking a set of vectors Vi1 is linearly independent if the only way of getting the zero vector by combining them is to take a zero amount of each vector 1 Linear Independence De nition A set of vectors Vi1 C V is linearly independent if ELI aivi 0 ltgt a 0 fori 12 De nition A set of vectors Vi 21 C V that is not linearly independent is linearly dependent Example 5 The vectors 2 1 3 V1 2 V27 3 V3 5 3 4 7 are linearly dependent because V3 V1 1 V2 That is because V1 1 V2 7 V3 0 Hence 21 aiVl 0 does not imply that ai 0 fori 1 2 3 Vector SpacesVoraGWU 3 Example 6 The vectors we saw earlier 1 1 0 V1 1 7 V2 1 7 V3 1 1 1 O are also linearly dependent because 1 7 0 2 V1 2 V2 V3 7 and 21 aivi 0 does not imply that ai 0 fori 1 2 3 Example 7 On the other hand the vectors 1 0 0 V1 0 7V27 1 Vgi 0 0 0 1 are linearly independent because 21 aivi 0 implies that ai 0 fori 1 2 3 In general how would one determine if a set of vectors was linear independent Again linear equations 2 Determining the Linear Independence of a Set of Vectors v2 211 1 The question is are there values a1 a2 an not all zero such that a1v1 a2V2 1 any 0 2 Form a matrix A with vi1 as its columns A v1v2vn 3 Write a vector 3 containing the unknown coef cients ai a1 a2 an Vector SpacesVoraGWU 4 4 Solve the set of linear equations Aa 0 using gaussian elimination How many solutions does it have a We know that there is one solution for these equations it is a O b If there is exactly one solution and no more the vectors vi1 are linearly indepen dent c If there are several solutions the vectors Vi 1 are not linearly independent Example 8 Is the set of vectors Vllllw llwglllw ll 111 111 111 111 linearly independent In other words the question is are there values a1 a2 a3 and a4 not all zero such that a1V1 a2V2 asVs a4V4 0 l l l 1 a1 1 1 71 71 a2 0 111 11111 Now the above are four equations is four unknowns for which we already know one solution Of a1 a2 a3 a4 0 So the set of vectors is linearly independent only if there are no more solutions Gaussian elimination as always 1 1 1 1 a1 0 07272 a2 07272 0 a3 072 0727 14 Swap for a nonzero pivot Remember to swap rhs as well though in this case because the rhs is zero it doesn t matter 1 1 1 1 al 07272 0 a2 70 0 07272 a3 7072 07271147 391 1 1 1quota139 07272 0 a2 70 0 07272 a3 0 0 2727 L4 Vector SpacesVoraGWU 5 1 1 1 1 al 07272 0 a2 0 07272 13 0 0 074 L4 Because the nal equation contains a single unknown there is only one solution to this set of equations We know that an a2 a3 a4 0 is a solution now we know it is the only solution You can check that the gaussian elimination gives you that solution 7404 O or L4 0 Working backwards you get that all four unknowns are zero We could have determined the same by checking if the matrix corresponding to the equations was invertible If it was then there would be exactly one solution We didn t invert the matrix because this works only when there are n equations in n unknowns which is not always the case 3 Spans and Linear Equations Consider a set of n equations in m unknowns We have observed examples of the following facts 1 The set of equations AX b has a salutian that is at least one solution ifb E SpanAi1 where Al represents the ith column of A That is the equations have a solution if b can be expressed as a linear combination of Ai 2 The converse is also true the set if equations has no solution ifb is not in SpanAl1 3 If b E SpanAi1 the set of equations has at least one solution then a The set of equations has more than one solution if A3211 are not linearly independent b The converse is also true The set of equation has only one solution if A3211 are linearly independent 4 Bases of a Vector Space Consider the vector space R2 Consider any vector in it This vector can be written as a linear v11s1v2111 ml lqv yvz R2 Q Spanv1 vz combination of That is Vector SpacesVoraGWU 6 Further there is no vector in Spanv1 V2 that is not in R2 Hence R2 SpanV1V2 and we say that V1 vz is a basis for R2 A basis ofa vector space is a set oflinearly independent vectors whose span is the vector space De nition A set of vectors B C V is a basis for V if and only if B is linearly independent and SpanB V In general the vectors 1 0 0 0 0 1 0 0 v1 vz 7 quotvi V 7 0 0 0 1 form a basis for R known as the standard basis Other bases exist see Example 9 and Exercises 2 and 3 for examples 5 Summary of Methodology to Determine if a set A is a Basis for a Vector Space V 1 Determine if the set A is linearly independent If it is not it is not a basis 2 Determine if every vector in V can be expressed as a linear combination of the vectors in A If so you have shown that V Q SpanA 3 Determine if every vector in SpanA belongs to V If so you have shown that SpanA Q V In the problems we address this will be easy V will be one of R R2 R3 or R4 As the vectors in A will be either one two three or four dimensional it will be straightforward to say whether these vectors and any linear combinations of them belong to the R R2 R3 or R4 See Example 9 and Exercises 2 and 3 for further clari cation 4 If 2 and 3 above hold we have V SpanA 5 If 1 and 4 above hold A is a basis for V Vector SpacesVoraGWU 7 Example 9 1 1 1 1 V1 1 V2 1 V3 1 V4 71 1 7 71 7 71 7 1 1 71 1 71 is a basis for R4 We have already shown the above set is linearly independent see Example 8 To show that it is a basis we also need to show that R4 is its span For this we need to consider any vector in R4 NQERS and show that it is a linear combination of the vectors v1 vz V3 and V4 That is we need to show that there exist a1 a2 13014 such that w a1V1 a2V2 asVs a4V4 Z 2 or 1 1 1 1 w 1 7 1 7 7 71 7 m 04 1 a2 71 a3 71 a4 1 i y 111 1711 111 1711 121 Again the equations may be written as 1 1 1 1 7 a1 w 1 1 71 71 a2 7 m 1 71 71 1 a3 7 y 1 71 1 71 a4 2 which may be reduced 1 1 1 1 0 072 72 O 2 0 72 72 a3 yiw LO 72 0 7 704 ziwj 1 1 1 1 701 w 0 72 72 0 a2 7 yiw 0 0 72 72 a3 7 ziw 0 72 0 72 a4 27w Vector SpacesVoraGWU 8 1 1 1 1 04 w 0 72 72 0 a2 7 y 7 w 0 0 72 72 a3 7 m 7 w 0 0 2 72 a4 27w7y7w27y 1 1 1 1 a1 w 0 72 72 0 a2 7 y 7 w 0 0 72 72 a3 7 m 7 w 0 0 0 74 a4 2 7 y m 7 w a4 w7m1y7z7 a3 w7mlyz7 a2 211444711727 a1 w11yz Thus given any vector in R4 we have obtained its coef cients wrt Vi211 hence R4 Q Spanvi1 Further because each of the Vl39 is a four dimensional column linear combinations of the Vl39 cannot lie outside the fourdimensional space over the real numbers Hence SpanVi 211 Q R4 Hence Spanvi 11 R4 Example 8 shows that vi 11 are linearly independent hence viEl1 forms a basis for R4 6 The Dimension of a Vector Space De nition The number of vectors in a nite basis of a vector space is its dimensian Fact The number of vectors in a basis for R is n This gives us one additional fact connecting linear equations to vector spaces Consider a set of n equations in n unknowns represented by the matrix equation AX b Notice that A is square We now add to the observations of section 3 for the special case of the square matrix 1 The equations have exactly one solution if A is invertible or equivalently if all its columns are independent or equivalently if the columns form a basis for R 7 Exercises 1 Which of the following sets of vectors is linearly independent a 1 1 1 V1 1 7V2 1 7V3 1 1 2 0 b 1 2 0 V1 2 V2 2 V3 1 Vector SpacesVoraGWU c 1 1 1 V1 1 7V2 1 7V3 1 1 71 1 2 Show that 2 1 0 W1 7 0 7W2 7 0 W3 7 1 1 1 0 is a basis for R3 3 Show that 0 1 1 V1 1 7V2 0 7V3 1 1 1 0 is a basis for R3 4 Express the vectors in a d as linear combinations of the set of vectors 2 1 0 W1 0 7W2 0 7W3 1 to Vector SpacesVoraGWU 10 Solutions 1 Which of the following sets of vectors is linearly independent Answer Try to nd nonzero coef cients that combine the vectors to get 0 If you can nd the nonzero coef cients the vectors are nut linearly independent Otherwise they are a 1 1 1 V1 1 V2 7 1 V3 1 1 2 0 Answer Solve Aa 0 or 1 1 1 a1 0 1 1 1 a2 0 1 2 0 a3 0 1 1 1 a1 0 0 0 0 a2 0 0 1 71 a3 0 1 1 1 a1 0 0 1 71 a2 0 0 0 0 a3 0 Final equation of form 0 0 hence there is more than one solution to the equations hence the vectors are not linearly independent b 1 2 0 V1 2 V2 2 V3 1 1 72 2 Answer Solve Aa 0 or 1 2 0 a1 0 2 2 1 a2 0 1 72 2 a3 0 1 2 0 a1 0 0 72 1 a2 0 0 74 2 a3 0 Vector SpacesVoraGWU l l 1 20 a1 0 0721 a2 0 0 00 a3 0 Final equation of form 0 0 hence there is more than one solution to the equations hence the vectors are not linearly independent c 1 1 1 V1 7 1 7V2 7 1 7V3 7 1 1 71 1 Answer Solve Aa 0 or 1 1 1 a1 0 1 1 71 a2 0 1 71 1 a3 0 1 1 1 a1 0 0 0 72 a2 0 0 72 0 a3 0 1 1 1 a1 0 0 72 0 a2 0 0 0 72 a3 0 The system has exactly one solution a1 a2 a3 a4 0 Hence the vectors are linearly independent 2 Show that 2 1 0 W1 7 7W2 7 0 7W3 7 1 1 1 is a basis for R3 Answer We need to show that W1 w2 and W3 are linearly independent and that Spanw 91 R3 First we show linear independence Find a1 a2 a3 such that 210 04 0 001 120 110 13 0 Vector SpacesVoraGWU 12 210 04 0 110 120 001 13 0 210 a1 0 00 120 001 a3 0 This system has exactly one solution a1 a2 a3 0 hence the vectors are linearly independent Next we show that Spanwi 1 R3 First we observe that the vectors are three dimensional hence all combinations are also three dimensional and Spanwi 1 Q R3 Next we wish to show that R3 Q Spanwi 1 We consider any vector in R3 z y We see if it can be expressed as a linear combination of wt91 2 1 0 a1 m 0 1 a2 y 1 1 0 a3 2 Gaussian elimination gives N 1 1 1 1 1 m r 2 1 0 1 1 0 a2 0 0 1 13 y 2 1 0 04 m 0 0 a2 z 7 g 0 0 1 13 y The solution is 13 34012 22 7 m a1 z 7 2 Hence R3 Q Spanwi 1 Hence we have shown that the given vectors form a basis of R3 3 Show that 0 1 1 V1 1 7V2 0 7V3 1 1 1 0 is a basis for R3 Answer We need to show that v1 vz and V3 are linearly independent and that Spanvi1 R3 First we show linear independence Find a1 a2 a3 such that 011 04 0 101 120 110 a3 0 Vector SpacesVoraGWU 13 101m 0 011a2 0 110 13 0 10104 0 011 120 0171a3 0 101 a1 0 011 120 0072 13 0 Has exactly one solution a1 a2 a3 0 Hence the vectors are linearly independent Next we show that Spanvi1 R3 First we observe that the vectors are three dimensional hence all combinations are also three dimensional and SpanVi 1 Q R3 Next we wish to show that R3 Q Spanvi 1 We consider any vector in R3 73172 We see if it can be expressed as a linear combination of vi 1 0 1 1 a1 m 1 0 1 a2 y 1 1 0 a3 2 1 1 a1 y 0 1 1 a2 m 1 1 0 a3 2 1 0 1 04 y 0 1 1 a2 m 0 1 71 a3 2 7 y 1 0 1 a1 y 0 1 1 a2 m 0 0 72 a3 ziyim a3 412 Lg a1 Hence R3 Q Spanvi 1 Hence we have shown that the given vectors form a basis of R3 4 Express the vectors in a d as linear combinations of the set of vectors 2 1 0 W1 0 7W2 0 7W3 1 1 1 0 a 1 2 14 Vector SpacesVoraGWU Answer Solve the equation a32a25a172 1 Answer Solve the equation 201 1a24a172 a3 1 Answer Solve the equation 15 Vector SpacesVoraGWU 1a20a10 a3 0 Answer Solve the equation a30a22a171

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