University:
University of Michigan
Course:
MATH 217 | Linear Algebra
Academic year:

2024
Views:

494

Pages:

9
Author:

Koen D

Linear Dependence and Independence Last time, we talked about the span of a list of vectors: The span of ~v1 , ~v2 , . . . , ~vk is those vectors which can be written as a linear combination a1~v1 + · · · + ak~vk . Last time, we talked about the span of a list of vectors: The span of ~v1 , ~v2 , . . . , ~vk is those vectors which can be written as a linear combination a1~v1 + · · · + ak~vk . Sometimes, a vector can be written in more than one way as a linear combination of other vectors. For example, h the i h plane i 1 −1 0 x + y + z = 0 in R3 is spanned by the vectors h −1 i 0 . 1 h 1 i The vector 2 can be written as , 0 1 −1 and −3 h 1 −1 0 i +3 h 0 1 −1 i =2 h 1 −1 0 and many other ways. i +4 h 0 1 −1 i h −1 i h 1 i h + 0 = 3 −1 +5 1 0 0 1 −1 i +2 h −1 i 0 1 h 1 −1 0 i +3 h 0 1 −1 i =2 h 1 −1 0 i +4 h 0 1 −1 i h −1 i h 1 i h + 0 = 3 −1 +5 1 0 0 1 −1 i +2 h −1 i The basic reason this happened is that we have a linear relation h 1 i h 0 i h −1 i −1 + 1 + 0 = ~0 0 −1 1 between the vectors. So we can add multiples of this relation to any expression for ~v and get other expressions for the vector ~v . 0 1 We say that a collection of vectors ~v1 , ~v2 , . . . , ~vk are linearly dependent if there are coefficients c1 , c2 , . . . , ck , not all of which are zero, such that c1~v1 + · · · + ck~vk = ~0. The opposite of “linearly dependent” is “linearly independent”. Let V be the span of ~v1 , . . . , ~vk . If ~v1 , . . . , ~vk are linearly dependent, then any vector w ~ in V will be able to be expressed in many different ways as a linear combination of the ~v ’s. Indeed, if we have a linear relation c1~v1 + · · · + ck~vk = ~0 and if w ~ = a1~v1 + · · · + ak~vk then we also have w ~ = (a1~v1 + · · · + ak~vk ) + t(c1~v1 + · · · + ck~vk ) = (a1 + tc1 )~v1 + · · · + (ak + tck )~vk for any scalar t. On the other hand, the converse is also true: If ~v1 , . . . , ~vk are linearly independent, then there is only one way to write any given vector w ~ as a linear combination of the ~v ’s. Indeed, if there were two ways w ~ = a1~v1 + · · · + ak~vk = b1~v1 + · · · + bk~vk then we would have a linear relation (a1 − b1 )~v1 + · · · + (ak − bk )~vk = ~0. How can we tell if a list of vectors is linearly dependent? We solve linear equations! A linear relation c1~v1 + · · · + ck~vk = ~0 is the same thing as having a vector in the kernel:   cc12 | | ··· | ~ v1 ~ v2 ··· ~ vk  .  = ~ 0. . . | | ··· | ck So ~v1 , ~v2 , . . . , ~vk are linearly dependent if and only if has a nonzero kernel. | | ··· | ~ v1 ~ v2 ··· ~ vk | | ··· | h1i h1i For example, let’s see if 1 , 2 , 1 3 need to compute the kernel of  1 1  1 2  1 3 We run row reduction:  1 1  1 2  1 3 We see that the vector the linear dependency 3 −4 1 1 h1i 1 1 2 1 are linearly dependent. We  3  2 . 1  1  0  0   2  1 h −4 i h3i 0 1 0 4   −1 . 0 is in the kernel. This corresponds to + h1i 2 3 + h3i 2 1 = ~0.