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# 5.3 - Dr Doom Lin Alg MATH1850/2050 Section 5.3 Page 1 of 5...

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Unformatted text preview: Dr. Doom Lin Alg MATH1850/2050 Section 5.3 Page 1 of 5 Section 5.3. Linear Independence. Deﬁnition: Linear Independence. If S 2 {V1, V2, . . . ,V,} is a nonempty set of vectors, then the vector equation k1V1+k2V2+H~+k7~Vr=0 has at least one solution, namely [\$120, [\$220, ..., 19,20 If this is the only solution, then S is called a linearly independent set. If there are other solutions, then S is called a linear dependent set. Example: A linearly dependent set. v1 = (1,0,—1), v2 = (—1,2,3), and v3 = (1,1,0) form a linearly dependent set because 3V1+V2—2V3=0 SO k1237k2=17k3=—2 is a nontrivial solution to the vector equation. Example: A linearly dependent set. The polynomials p1 = 1 + I, p2 = 1 + x + x2, and p3 = x2 form a linearly dependent set because Pi—P2+P3=0 SO [€121,1{722—17k321 is a nontrivial solution to the vector equation. The polynomials p1, p2, and p3 are vectors in P2 the vector space of polynomials of degree less than or equal to 2. Example: A linearly independent set. The standard basis vectors in R”: e1, e2, . . . , en form a linearly independent set, because the only solution to [€181 + [€282 + ' ' ' + knen : 0 is [C120, ICQZO, ..., [67,20 Dr. Doom Lin Alg MATH1850/2050 Section 5.3 Page 2 of 5 Example: A linearly independent set. The polynomials p0 = 1, p1 = I, p2 = x2, . . . , p7, = x” form a linearly independent set of vectors in P” (the vector space of polynomials of degree 3 71), because the only solution to [Capo + klpl + [€2P2 + ' ' ' + knpn = 0 is ICOZO, [C120, ICQZO, ..., [67,20 Example: Determining Linear Independence / Dependence. Deter- mine Whether the vectors V1 2 (1,0, —1), V2 2 (—1,2,3), and V3 2 (1,1,0) form a linearly dependent or linearly independent set. LET Mi \Ji ‘\"‘1 \h +1433); = 9 Elm 15' (mites: 1“ 4‘, -‘c L, = 0 1m €m\é\$; 1“; ‘\ k3 = 0 «51B emig: —k‘ 4&‘5kz = O A '\< 0*» Lieu: we cm use nﬁémmﬂs C (new. fwmx/tis mum) \—\\ \—\I \A\: o 2 | -l ‘5 o o ‘L n Zﬁ‘l. «I o .J = Q. (M118. ws me momma“) So guy-Hg3 A16 LN. moemsmv < \A\= 0 v) No» 11:ka stsLun‘ons exist). Dr. Doom Lin Alg MATH1850/2050 Section 5.3 Page 3 of 5 Example: Determining Linear Independence / Dependence. Deter- mine Whether the P2 vectors p1 = 1 — a: + 332, p2 = 1 — 3:2, and p3 = 1 + a: form a linearly dependent or linearly independent set. LET lL\Q\‘\’\‘1?~"\‘\$‘f% =0 7. ccusl‘gez ocsrixcfwrs as 36-54, x, x 0N Ban-r sfots. FOL x."—\: lC\ ‘\ k1. +l‘3 = 0' Coo. x = ~\q +‘42, = 0 \$01 XL: k\ —“-‘L = O t l t 2 n \ I l t l l \ \ 0 -\ O l o l 7, __——9- o \ 'L —a o \ ‘L o \ —\ 0 23:23-94 0 -'L -\ [3291*le o o 3 233% o o l 0 so (leklzllfo (5 THE euw Sowu'm). 50 942,95 ate Lin. IUBEPEM- Theorem 5.3.1. A set S with two or more vectors is: (a) Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. (b) Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S. Example. We saw that V1 2 (1,0,—1), V2 2 (—1,2,3), and V3 2 (1,1,0) form a linearly dependent set. Thus, we can write at least one of these vectors as a linear combination of the others. In particular, LET \ﬁx!‘+‘¢1y1+l‘3\b;0 14. k. ‘4‘ I —\ \ \ -—| l \ —\ I ° 0 1 I —; o ’L l ‘—3 O \ V7, 0 "l 3 O O 'L \ o o o 0 L61 \ha-J; , Tue» k1; "3‘94, 1“: “72:1: roll. i=9. Ger Ami Ti“!ka SOUL! ‘4‘:"‘5) Qua-L, \‘gz'l. Dr. Doom Lin Alg MATHl850/2050 Section 5.3 Page 4 of 5 U MIT Example. We saw that the standard basis vectors in R": e1,e2,...,en form a linearly independent set, so we cannot write any of these vectors as a linear combination of the others. For instance, in R3: e1 2 (1, 0,0), e2 2 (07170)) 83 = (0,0, 9" S‘ Tb, £3 7' \Li‘ﬁn '\-\‘1'§1 (0’0, 1): ‘c‘(\ooa°)+‘L-L(Qa \;O) k, = 0 k1 = O gymm \S UACOUSS‘EU‘, DO éUCH kn“; was“ Theorem 5.3.2. (a) A ﬁnite set of vectors that contains the zero vector is linearly dependent (b) A set with exactly two vectors is linearly independent if and only if neither vector is a scalar multiple of the other. Examples: 1' 81 = {V17V270} + O'!1+ = 9 La'ruis is A in) 3mm mien Comiuxﬁon , 2. 82 = {(3, —1,0,2), (2, —2/3,0,4/3)} mg :11 ._-, _% \L V\ 97. SP 32 is 559. 3' S3={(2717_2)7(17_273)} “gig 5“ # kgl ‘0'— AN kem’ in ‘12 50 \$3 is u‘u. rum». Dr. Doom Lin Alg MATH1850/2050 Section 5.3 Page 5 of 5 Geometric Interpretation of Linear Independence. 0 ln R2 or R3, a set of two vectors is linearly independent if and only if the vectors do not lie on the same line when they are placed with their initial points at the origin (see Figure 5.3.1 in the text on page 244). 0 ln R3, a set of three vectors is linearly independent if and only if the vectors do not lie in the same plane when they are placed with their initial points at the origin (see Figure 5.3.2 in text on page 245). If the three vectors lie in the same plane, then at least one can be written as a linear combination of the others. Theorem 5.3.3. Let S = {V1,V2, . . . ,VT} be a set of vectors in R". If r > n, then S is linearly dependent. For example, a set with 3 or more vectors in R2 must be linearly dependent, and a set with 4 or more vectors in R3 must be linearly dependent. For R3, say we have the four vectors: V1 = (71117711277113), V2 = (11217712277123), V3 = (71317713277133), V4 = (041,042,043) That is, n = 3 and r = 4, and r > n so these vectors should form a linearly dependent set. To see this, we can show that k1 = 0, k2 = 0, k3 = 0, k4 = 0 is not the only solution to the vector equation: kiVi + [€2V2 + [€3V3 + [€4V4 = 0 This vector equation can be written as a homogeneous system of linear equa- tions with the ki as the unknowns: 7111191 —— 7121192 —— 7131193 —— 7141/91 = 0 7112191 —— 7122192 —— 7132193 —— 7142/91 — 0 Ui3k1 —- U237€2 —— v33k?) —— 7143701 — 0 There are more unknowns than equations, so there are nontrivial solutions. See Theorem 1.2.1. ...
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5.3 - Dr Doom Lin Alg MATH1850/2050 Section 5.3 Page 1 of 5...

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