Homogeneous equations, Linear independence
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- Godfrey Jordan
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1 Homogeneous equations, Linear independence 1. Homogeneous equations: Ex 1: Consider system: B" #B# œ! B" #B3 œ! B B œ! # $ Matrix equation: Ô " #! Ô B " Ô! "! # B # œ! œ 0Þ Ð3Ñ Õ! " " ØÕB Ø Õ! Ø $ Homogeneous equation: At least one solution: Ex œ 0. x œ 0Þ Other solutions called nontrivial solutions. Theorem 1: A nontrivial solution of Ð$Ñ exists iff [if and only if] the system has at least one free variable in row echelon form. The same is true for any homogeneous system of equations. Proof: If there are no free variables, there is only one solution and that must be the trivial solution. Conversely, if there are free variables, then they can be non-zero, and there is a nontrivial solution. Ex 2: Reduce the system above: Ô " #! l! Ô " #! l! as before "! # l! Ä! " " l! Õ! " " l! Ø Õ!!! l! Ø
2 Ê B #B œ!à B B œ!à!œ!þ " # # $ Note that B $ œ free variable (non-pivot); hence general solution is B# œ B$ à B" œ #B# œ #B$ Þ Ô B " Ô #B $ Ô # x œ B# œ B $ œ B$ " ß ÕB Ø Õ B Ø Õ " Ø $ $ B $ arbitrary: straight line - Parametric vector form of solution. Theorem 2: A homogeneous system always has a nontrivial solution if the number of equations is less than the number of unknowns. Pf: If we perform a Gaussian elimination on the system, then the reduced augmented matrix has the form:
3 Ô " + "# + "$ á l!!! " + #% á l!!!! " + $& á l ã á l ã Õ ã l! Ø with the remaining rows zeroes on the left side. If the number of equations is less than the number of unknowns, then not every column can have a 1 in it, so there are free variables. By previous theorem, there are nontrivial solutions.
4 1. Inhomogeneous equations: [we should briefly mention the relationship between homogeneous and inhomogeneous equations:] Consider general system: Ex œ bþ (1)(1) Suppose p is a particular solution of (1), so Ep œ b. If x is any other solution of (1), we still have Ex œ b. Subtracting the two equations: So v2 œ x p Ex Ep œ 0 Ê EÐx pñ œ 0Þ satisfies the homogeneous equation. Generally: Theorem 1: M0 p is a particular solution of (1), then for any other solution x, we have that v2 œ x p solves the homogeneous equation (i.e., with b œ 0). Thus every solution x of (1) can be written x œ p v 2, where v 2 is a solution of the homogeneous equation. 2. Application: Network flows Traffic pattern at Drummond Square:
5 Quantities in cars/min. What are the flows on the inside streets? One equation for each node: B B B œ %! " $ % B B œ #!! " # B B B œ "!! # $ & B B œ '! % & Ô "! " "! l %! " "!!! l #!!! " 1! " l "!! Õ!!! " " l '! Ø
6 Ô "! " "! l %!! " " "! l "'!! " 1! " l "!! Õ!!! " " l '! Ø Ô "! " "! l %!! " " "! l "'!! " 1! " l "!! Õ!!! " " l '! Ø Ô "! " "! l %!! " " "! l "'!!! 0 " " l '! Õ!!! " " l '! Ø Ô "! " "! l %!! " " "! l "'!!!! " " l '! Õ!!! " " l '! Ø Ô "! "! " l "!!! " "! " l "!!!!! " " l '! Õ!!!!! l! Ø So: B œ"!! B B " $ & B œ"!! B B # $ & B œ '! B ß % & where BßB $ & are free.
7 Constraint: if for example all flows have to be positive; then we require B 3! for all 3. Therefore: This corresponds to a region in the BßB! $ & "!! Ÿ B B Ÿ "!! $ & B Ÿ '! & BßB $ & plane - can be plotted if desired. If they closed off road B$ and Bß & then we have B$ œb& œ!, so that B" œ 100ß B# œ " 00, B% œ '! note that then traffic flow becomes uniquely determined. Definition 1: A collection of vectors v" ßv# ßáßv8 is linearly independent if no vector in the collection is a linear combination of the others. Equivalently, Definition 2: A collection of vectors v" ßáßv8 is linearly independent if the only way we can have - v - v á - v œ 0 is if all of the - œ!þ Equivalence of the definitions: Def 1 Ê Def 2 " " # # If no vector is a linear combination of the others, then if -" v" -# v# á -8v8 œ 0 we will show that -ßáß- " 8 have also to be 0. Proof: Suppose not (for contradiction). Without loss of generality, assume -" Á! (proof works same way otherwise). Then we have: v" œ -# Î-" v# á -8 Î-" v8ß
8 contradicting that no vector is a combination of the others. Thus the - 3 all have to be 0 as desired. Note: If W# is a collection of vectors and W" is a subcollection of W2ß then If W # is linearly independent Ê no vector in W # is a linear combination of the others Ê no vector in W" is a linear combination of the others (since every vector in W is also in W ) " # Ê W is linearly independent. " Logically equivalent [contrapositive] If W " is linearly dependent (i.e., not independent) Ê W is linearly dependent # [ These are stated more formally in the book as theorems.] Theorem 2: P/> W œ v" ßáß v8 be a collection of vectors in. Then W is linearly dependent if and only if one of the vectors v 3 is a linear combination of the previous ones v ßáßv Þ " 3 " Proof: ( Ê ) If W is linearly dependent, then there is a set of constnats - 3 not all! such that Let - 5 zero, and -" v" á -8v8 œ 0. be the last non-zero coefficient. Then the rest of the coefficients are -" v" -# v# á -5 " v5 " -5v5 œ 0 Ä v5 œ -" Î-5 v" -# Î-5 v# á -5 " Î-5v5 " i.e. one of the vectors is a linear combination of the previous ones. ( É ) Obvious..
9 3. Checking for linear independence: Example 2: Consider the vectors " " " v" œ ß v# œ ß v$ œ " "! Are they linearly independent? Ê -" v" -# v# -$ v$ œ œ! " # $ Reduced matrix: - - œ! " # " " " l! " "! l! Ä " " " l!! # " l! Ä " " " l!! " "Î# l! Conclude: there are free variables. By theorem on homogeneous equations there is a nontrivial solution, so - need not be!. Thus not all - 3 must be! Ê not linearly independent. [note that if number of vectors is greater than the size of the vectors, this will always happen]. More generally thus: 3
10 Theorem 3: In 8, if we have more than 8 vectors, they cannot be linearly independent. From above we have: Algorithm: To check whether vectors are linearly independent, form a matrix with them as columns, and row reduce. (a) If reduced matrix has free variables (i.e., b a non-pivot column), then they are not independent. (b) If there are no free variables (i.e., there are no nonpivot columns), they are independent.
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