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Homogeneous equations, Linear independence
1. Homogeneous equations:
Ex 1: Consider system:
B #B œ! B #B œ! B B œ!
" # "
$
3
Matrix equation:
Ô ×Ô × Ô ×
Õ ØÕ Ø Õ Ø
" #! B!
"! # B!
! " " B!
œ œ Þ Ð Ñ
"
$
Homogeneous equation:
E x œ 0.
At least one solution:
x œ 0 Þ
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 × Ô × Õ Ø Õ Ø
" #! l! " #! l! "! # l!! " " l! ! " " l!!!! l!
Ä
as before
Ê
B (^) " #B (^) # œ !à B (^) # B (^) $œ !à! œ !Þ
Note that B (^) $œfree variable (non-pivot); hence general solution is
B (^) # œ B à$ B (^) " œ #B (^) # œ #B Þ$
x œ œ œ B ß
B #B #
B B "
B B "
Ô × Ô × Ô ×
Õ Ø Õ Ø Õ Ø
" $
$
$ $
$
Parametric vector form of solution.
B$ arbitrary: straight line -
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:
1. Inhomogeneous equations:
[we should briefly mention the relationship between homogeneous and inhomogeneous equations:]
Consider general system:
E x œ bÞ (1)(1)
Suppose p is a particular solution of (1), so E p œ b. If x is any other solution of (1), we still have E x œ b. Subtracting the two equations:
E x E p œ 0 Ê EÐ x p Ñ œ 0 Þ
So v (^) 2 œ x p satisfies the homogeneous equation. Generally:
Theorem 1: M0 p is a particular solution of (1) , then for any other solution x , we have that v (^) 2 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:
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 ×
Ö Ù
Ö Ù
Õ Ø
"! " "! l %! " "!!! l #!! ! "! " l "!! !!! " " l '!
Constraint: if for example all flows have to be positive; then we require B 3! for all. Therefore: 3
B ß B$ &!
"!! Ÿ B $ B &Ÿ "!!
B &Ÿ '!
This corresponds to a region in the B ß B$ &plane - can be plotted if desired.
If they closed off road B (^) $ and B ß& then we have B (^) $ œ B (^) &œ !, so that
B (^) " œ 100 ß B (^) # œ " 00,B (^) %œ '!
note that then traffic flow becomes uniquely determined.
Definition 1: A collection of vectors v (^) " ß v (^) # ß á ß v 8 is linearly independent if no vector in the collection is a linear combination of the others.
Equivalently,
Definition 2: A collection of vectors v (^) " ß á ß v 8 is linearly independent if the only way we can have - (^) " v " (^) - (^) # v # (^) á - 8 v 8 (^) œ 0 is if all of the - œ !Þ 3
Equivalence of the definitions: Def 1 Ê Def 2
If no vector is a linear combination of the others, then if
- (^) " v " (^) - (^) # v # (^) á - 8 v 8 œ 0
we will show that - ß á ß -" 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 - Î- 8 " v 8 ß
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 W ß 2 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 Ö v (^) " ß á ß v 8 × 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 (^) " ß á ß v 3"Þ
Proof: ( Ê ) If W is linearly dependent, then there is a set of constnats - 3 not all ! such that
- (^) " v " (^) á - 8 v 8 œ 0.
Let - 5 be the last non-zero coefficient. Then the rest of the coefficients are zero, and
- (^) " v " (^) - (^) # v # (^) á - (^) 5" v 5" (^) - 5 v 5 œ 0
Ä v (^) 5 œ - Î-" 5 v (^) " - Î-# 5 v (^) # á - (^) 5" Î- 5 v 5"
i.e. one of the vectors is a linear combination of the previous ones. ( É) Obvious.
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.
