Linear Equations in Linear Algebra.7 LINEAR INDEPENDENCE
LINEAR INDEPENDENCE Definition: An indexed set of vectors {v,, v p } in n is said to be linearly independent if the vector equation x x x 2 2 p p v + v +... + v = 0 has only the trivial solution. The set {v,, v p } is said to be linearly dependent if there exist weights c,, c p, not all zero, such that c c c 2 2 p p v + v +... + v = 0 ----() Slide.7-2
LINEAR INDEPENDENCE Equation () is called a linear dependence relation among v,, v p when the weights are not all zero. An indexed set is linearly dependent if and only if it is not linearly independent. v = 2 3 4 v = 5 2 6 2 v = 3 0 Example : Let,, and. Slide.7-3
a. Determine if the set {v, v 2, v 3 } is linearly independent. b. If possible, find a linear dependence relation among v, v 2, and v 3. Solution: We must determine if there is a nontrivial solution of the following equation. 4 2 0 x 2 + x 5 + x = 0 2 3 3 6 0 0 Slide.7-4
LINEAR INDEPENDENCE Row operations on the associated augmented matrix show that 4 2 0 4 2 0 2 5 0 0 3 3 0 3 6 0 0 0 0 0 0 x and x 2 are basic variables, and x 3 is free. Each nonzero value of x 3 determines a nontrivial solution of (). Hence, v, v 2, v 3 are linearly dependent.. Slide.7-5
LINEAR INDEPENDENCE b. To find a linear dependence relation among v, v 2, and v 3, row reduce the augmented matrix and write the new system: 0 2 0 0 0 0 0 0 0 3 2 3 Thus, x 2x, x = x, and x 3 is free. = 3 2 3 2x = 0 Choose any nonzero value for x 3 say, x = 5. 3 Then x = 0 and x = 5. 2 x x + x = 0 0 = 0 Slide.7-6
LINEAR INDEPENDENCE Substitute these values into equation () and obtain the equation below. 0v 5v + 5v = 0 2 3 This is one (out of infinitely many) possible linear dependence relations among v, v 2, and v 3. Slide.7-7
LINEAR INDEPENDENCE OF MATRIX COLUMNS A = a L Suppose that we begin with a matrix instead of a set of vectors. The matrix equation A x = 0 x a + x a +... + x a 0 2 2 n n = can be written as. [ ] Each linear dependence relation among the columns of A corresponds to a nontrivial solution of. A x = 0 a n Thus, the columns of matrix A are linearly independent if and only if the equation A x = 0 has only the trivial solution. Slide.7-8
SETS OF ONE OR TWO VECTORS A set containing only one vector say, v is linearly independent if and only if v is not the zero vector. This is because the vector equation the trivial solution when. v 0 x v = 0 has only The zero vector is linearly dependent because has many nontrivial solutions. x 0 = 0 Slide.7-9
SETS OF ONE OR TWO VECTORS A set of two vectors {v, v 2 } is linearly dependent if at least one of the vectors is a multiple of the other. The set is linearly independent if and only if neither of the vectors is a multiple of the other. Slide.7-0
SETS OF TWO OR MORE VECTORS Theorem 7: Characterization of Linearly Dependent Sets S = {v,...,v } An indexed set p of two or more vectors is linearly dependent if and only if at least one of the vectors in S is a linear combination of the others. v 0 In fact, if S is linearly dependent and, then some v j (with j > ) is a linear combination of the preceding vectors, v,,. v j Slide.7-
SETS OF TWO OR MORE VECTORS Proof: If some v j in S equals a linear combination of the other vectors, then v j can be subtracted from both sides of the equation, producing a linear dependence relation with a nonzero weight ( ) on v j. v = c v + c v, then 2 2 3 3 0 = ( )v + c v + c v + 0v +... + 0v 2 2 3 3 4 p [For instance, if Thus S is linearly dependent. Conversely, suppose S is linearly dependent. If v is zero, then it is a (trivial) linear combination of the other vectors in S..] Slide.7-2
SETS OF TWO OR MORE VECTORS v 0 Otherwise,, and there exist weights c,, c p, not all zero, such that c c c 2 2 p p v + v +... + v = 0 Let j be the largest subscript for which. j = c v = 0 v 0. If, then, which is impossible because j. c 0 Slide.7-3
SETS OF TWO OR MORE VECTORS So j >, and c v +... + c v + 0v + 0v +... + 0v = 0 j j j j+ p c v = c v... c v j j j j c c j v = v... v. j + + j c c j j Slide.7-4
SETS OF TWO OR MORE VECTORS Theorem 7 does not say that every vector in a linearly dependent set is a linear combination of the preceding vectors. A vector in a linearly dependent set may fail to be a linear combination of the other vectors. 3 u = 0 v = 6 0 Example 2: Let and. Describe the set spanned by u and v, and explain why a vector w is in Span {u, v} if and only if {u, v, w} is linearly dependent. Slide.7-5
SETS OF TWO OR MORE VECTORS Solution: The vectors u and v are linearly independent because neither vector is a multiple of 3 the other, and so they span a plane in. Span {u, v} is the x x 2 -plane (with x = ). 3 0 If w is a linear combination of u and v, then {u, v, w} is linearly dependent, by Theorem 7. Conversely, suppose that {u, v, w} is linearly dependent. By theorem 7, some vector in {u, v, w} is a linear combination of the preceding vectors (since ). u 0 That vector must be w, since v is not a multiple of u. Slide.7-6
SETS OF TWO OR MORE VECTORS So w is in Span {u, v}. See the figures given below. Example 2 generalizes to any set {u, v, w} in u and v linearly independent. The set {u, v, w} will be linearly dependent if and only if w is in the plane spanned by u and v. 3 with Slide.7-7
SETS OF TWO OR MORE VECTORS Theorem 8: If a set contains more vectors than there are entries in each vector, then the set is linearly n dependent. That is, any set {v,, v p } in is linearly dependent if. p > n A = v L v p n p A x = 0 Proof: Let. Then A is, and the equation corresponds to a system of n equations in p unknowns. p > n If, there are more variables than equations, so there must be a free variable. Slide.7-8
SETS OF TWO OR MORE VECTORS A x = 0 Hence has a nontrivial solution, and the columns of A are linearly dependent. See the figure below for a matrix version of this theorem. Theorem 8 says nothing about the case in which the number of vectors in the set does not exceed the number of entries in each vector. Slide.7-9
SETS OF TWO OR MORE VECTORS S = {v,...,v } Theorem 9: If a set p in contains the zero vector, then the set is linearly dependent. n Proof: By renumbering the vectors, we may suppose v = 0. Then the equation that S in linearly dependent. v + 0v +... + 0v = 0 2 p shows Slide.7-20