Linear Independence. e 1 = (1,0,0,...0) e 2 = 0,1,0,...0). e n = (0,0,0,...1) a 1 e 1 = (a 1,0,0,...0) a 2 e 2 = 0,a 2,0,...0)

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1 Linear Independence --17 Definition Let V be a vector space over a field F, and let S V The set S is linearly independent if v 1,,v n S,,, F, and v v n = implies = = = An equation like the one above is called a linear relationship among the v i ; if at least one of the coefficients a i is nonzero, it is ontrivial linear relationship Thus, a set of vectors is independent if there is no nontrivial linear relationship among finitely many of the vectors A set of vectors which is not linearly independent is linearly dependent (I ll usually say independent and dependent for short) Thus, a set of vectors S is dependent if there are vectors v 1,,v n S and numbers,, F, not all of which are, such that v v n = Example If F is a field, the standard basis vectors are e 1 = (1,,,) e =,1,,) e n = (,,,1) Show that they form an independent set in F n Write e 1 +a e + + e n = I have to show all the a s are Now e 1 = (,,,) a e =,a,,) e n = (,,, ) So e 1 +a e + + e n = (,a, ) Since by assumption e 1 +a e + + e n =, I get (,a, ) = (,,) Hence, = a = = =, and the set is independent Example Show that any set containing the zero vector is dependent If S, then 1 = is ontrivial linear relationship in S 1

2 Example Show that the vectors (, 1,4) and ( 6,3, 1) are dependent in R 3 I have to find numbers a and b, not both, such that a (, 1,4)+b ( 6,3, 1) = (,,) In this case, you can probably juggle numbers in your head to see that 3 (, 1,4)+1 ( 6,3, 1)= (,,) This shows that the vectors are dependent There are infinitely many pairs of numbers a and b that work In examples to follow, I ll show how to find numbers systematically in cases where the arithmetic isn t so easy Example Suppose u, v, w, and x are vectorsin a vector space Provethat the set {u v,v w,w x,x u} is dependent Notice that in the four vectors in {u v,v w,w x,x u}, each of u, v, w, and x occurs once with a plus sign and once with a minus sign So (u v)+(v w)+(w x)+(x u) = This is a dependence relation, so the set is dependent If you can t see an easy linear combinationof a set of vectorsthat equals, you may haveto determine independence or dependence by solving a system of equations Example Consider the following sets of vectors in R 3 If the set is independent, prove it If the set is dependent, find ontrivial linear combination of the vectors which is equal to (a) {(,, 3),(1,1,1),(1,7,)} (b) {(1,, 1),(4,1,3),( 1,1, 11)} (a) Write a linear combination of the vectors and set it equal to : a 3 +b c 1 7 = I have to determine whether this implies that a = b = c = Note: When you convert vectors given in parenthesis form to matrix form, you turn the vectors into column vectors as above This is consistent with the way I ve set up systems of linear equations Thus, (,, 3) became 3 The vector equation above is equivalent to the matrix equation a b = 3 1 c

3 Row reduce to solve: Note: Row operations won t change the last column of zeros, so you don t actually need to write it when you do the row reduction I ll put it in to avoid confusion The last matrix gives the equations Therefore, the vectors are independent (b) Write a This gives the matrix equation Row reduce to solve: This gives the equations 1 1 a =, b =, c = +b c a b c 1 1 = 11 = a+c =, b 3c = Thus, a = c and b = 3c I can get ontrivial solution by setting c to any nonzero number I ll use c = 1 This gives a = and b = 3 So ( ) = This is a linear dependence relation, and the vectors are dependent The same approach works for vectors in F n where F is a field other than the real numbers Example Consider the set of vectors {(4,1,),(3,3,),(,1,1)} in Z 3 5 If the set is independent, prove it If the set is dependent, find ontrivial linear combination of the vectors which is equal to Write This gives the matrix equation a 4 1 +b 3 3 +c 1 = a b = 1 c 3

4 Row reduce to solve the system: This gives the equations r 1 4r 1 1 r r +4r 1 r 3 r 3 +4r r 3 r 3 +3r a+3c =, b+c = r 1 r 1 +3r Thus, a = c and b = 4c Set c = 1 This gives a = and b = 4 Hence, the set is dependent, and = 1 Example Consider the following set of vectors in Z 4 3: {(1,,1,),(1,,,1),(,1,,1)} If the set is independent, prove it If the set is dependent, find ontrivial inear combination of the vectors equal to Write This gives the matrix equation Row reduce to solve the system: This gives the equations a +b +c = a b = 1 c a+c =, b+c = Hence, a = c and b = c Set c = 1 Then a = and b = 1 Therefore, the set is dependent, and =

5 To summarize, to determine whether vectors v 1, v,, v m in a vector space V are independent, I try to solve v 1 +a v + +a m v m = If the only solution is = a = = a m =, then the vectors are independent; otherwise, they are dependent It s important to understand this general setup, and not just memorize the special case of vectors in F n, as shown in the last few examples Remember that vectors don t have to look like things like ( 3,5,7,) ( numbers in slots ) Consider the next example, for instance Example R[x] is a vector space over the reals Show that the set {1,x,x,} is independent Suppose That is, a + x+a x + + x n = a + x+a x + + x n = + x+ x + + x n Two polynomials are equal if and only if their corresponding coefficients are equal Hence, a = a = = = Therefore, {1,x,x,} is independent In some cases, you can tell by inspection that a set is dependent I noted earlier that a set containing the zero vector must be dependent Here s another easy case Proposition If n > m, a set of n vectors in F m is dependent Proof Suppose v 1,v,v n are n vectors in F m, and n > m Write v 1 +a v + + v n = This gives the matrix equation v 1 v v n a = To solve, I d row reduce the matrix v 1 v v n Note that this matrix has m rows and n+1 columns, and n > m The row-reduced echelon form can have at most one leading coefficient in each row, so there are at most m leading coefficients These correspond to the main variables in the solution Since there are n variables and n > m, there must be some parameter variables By setting any parameters variables equal to nonzero numbers, I get ontrivial solution for, a, This implies that v 1,v,v n is dependent Example Is the following set of vectors in R independent or dependent? { (1, 3),(5, π),( } 4,7) 5

6 Any set of three (or more) vectors in R is dependent Proposition Let {v 1,v,v n } be vectors in F n, where F is a field {v 1,v,v n } is independent if and only if the matrix constructed using the vectors as columns is invertible: v 1 v v n Proof Suppose the set is independent Consider the system v 1 v Multiplying out the left side, this gives v n a = v 1 +a v + v n = By independence, = a = = = Thus, the system above has only as a solution An earlier theorem on invertibility shows that this means the matrix of v s is invertible Conversely, suppose the following matrix is invertible: Let A = Write this as a matrix equation and solve it: v 1 v v 1 v v n v 1 +a v + v n = a v n a A A 1 A a a = = = A 1 = This gives = a = = = Hence, the v s are independent 6

7 Note that this proposition requires that you have n vectors in F n the number of vectors must match the dimension of the space The result can also be stated in contrapositive form: The set of vectors is dependent if and only if the matrix having the vectors as columns is not invertible I ll use this form in the next example Example Consider the following set of vectors in R 3 : x 8, 7 x 1, For what values of x is the set dependent? I have 3 vectors in R 3, so the previous result applies Construct the matrix having the vectors as columns: x A = x The set is dependent when A is not invertible, and A is not invertible when its determinant is equal to Now deta = 3(x 1)(x 8) Thus, deta = for x = 1 and x = 8 For those values of x, the original set is dependent The next proposition says that a independent set can be thought of as a set without redundancy, in the sense that you can t build any one of the vectors out of the others Proposition Let V be an F-vector space, and let S V S is dependent if and only if some v S can be expressed as a linear combination of other vectors in S ( Other means vectors other than v itself) Proof Suppose v S can be written as a linear combination of other vectors in S: v = v v n Here v 1,,v n S (where v i v for all i) and,, F Then = v v n 1 v This is ontrivial linear relation among elements of S Hence, S is dependent Conversely, suppose S is dependent Then there are elements,a, F (not all ) and v 1,v,v n S such that v 1 +a v + + v n = Since not all the a s are, at least one is nonzero There s no harm in assuming that (If another a was nonzero instead, just relabel the a s and v s so and start again) Since, its inverse a 1 1 is defined So v 1 +a v + + v n = v 1 = a v v n a 1 1 v 1 = a 1 1 ( a v v n ) v 1 = a 1 1 a v a 1 1 v n 7

8 Thus, I ve expressed v 1 as a linear combination of other vectors in S Definition Let {f 1,f,f n } be a set of functions R R which are differentiable n 1 times The Wronskian is f 1 f f n f 1 f f n W(f 1,f,f n ) = f 1 f f n f n (n 1) f (n 1) f n (n 1) Thus, the rows of the determinant consist of successive derivatives of the original functions Theorem Let S = {f 1,f,f n } be a set of functions R R which are differentiable n 1 times If W(f 1,f,f n ) at some point x = c, then S is independent Thus, if you can find some value of x at which the Wronskian in nonzero, the functions are independent The converse is false: You can find functions which are independent on an interval in R, but whose Wronskian is identically on the interval The converse does hold with additional conditions: For example, if the functions in equation are solutions to a linear differential equation Proof Let f 1 (x)+a f (x)+ + f n (x) = I have to show all the a s are This equation is an identity in x, so I may differentiate it repeatedly to get n equations: I can write this in matrix form: f 1 (x)+a f (x)+ + f n (x) = f 1 (x)+a f (x)+ +f n (x) = f 1(x)+a f (x)+ +a n f n(x) = f (n 1) 1 (x)+a f (n 1) (x)+ + f (n 1) n (x) = f 1 (x) f (x) f n (x) f 1(x) f (x) f n(x) f 1 (x) f (x) f n (x) (x) f (n 1) (x) f n (n 1) (x) f (n 1) n a = Plug in x = c: Let f 1 (c) f (c) f n (c) f 1 (c) f (c) f n (c) f 1 (c) f (c) f n(c) (c) f (n 1) (c) f n (n 1) (c) f (n 1) n A = a f 1 (c) f (c) f n (c) f 1 (c) f (c) f n (c) f 1(c) f (c) f n(c) f (n 1) n = (c) f (n 1) (c) f n (n 1) (c) 8

9 The determinant of this matrix is the Wronskian W(f 1,f,f n )(c), which by assumption is nonzero Since the determinant is nonzero, the matrix is invertible So a A A 1 A a a = = A 1 = Since = a = = =, the functions are independent Example Demonstrate that the set of functions {x,x 3,x 5 } is independent Compute the Wronskian: x x 3 x 5 W(x,x 3,x 5 ) = 1 3x 5x 4 6x x 3 = 16x6 I can find values of x for which the Wronskian is nonzero: for example, if x = 1, then W(x,x 3,x 5 ) = 16 Hence, {x,x 3,x 5 } is independent c 17 by Bruce Ikenaga 9