lecture_21 - MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 27...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 27 Linear Independence (contd) Example. Let V = M 22 be the real vector space corresponding to the 2 2 matrices. Consider the following three vectors: v 1 = 2 1 0 1 , v 2 = 1 2 1 0 , and v 3 =- 3- 2 1 . We determine whether these vectors are linearly independent. We consider the equation a 1 v 1 + a 2 v 2 + a 3 v 3 = = 2 a 1 + a 2 a 1 + 2 a 2- 3 a 3 a 2- 2 a 3 a 1 + a 3 = 0 0 0 0 . This corresponds to homogeneous linear system 2 a 1 + a 2 = 0 a 1 + 2 a 2- 3 a 3 = 0 a 2- 2 a 3 = 0 a 1 + a 3 = 0 The augmented matrix and its reduced row echelon form are 2 1 1 2- 3 0 1- 2 1 0 1 = 1 0 1 0 1- 2 0 0 0 0 This corresponds to the system a 1 + a 3 = 0 a 2- 2 a 3 = 0 = a 1 a 2 a 3 = - r 2 r r . Since there are infinitely many solutions, there is at least one nontrivial solution. In particular, when r = 1 we have the linear combination (- 1) 2 1 0 1 + (2) 1 2 1 0 + (1)- 3- 2 1 = 0 0 0 0 . This shows that the vectors are not linearly independent. More Criteria For Determining Linear Independence. Let ( V, + , ) be a real vector space, and consider a set S = { v 1 , v 2 , ..., v k } of k vectors in V . We focus on the case where V = R n . We can form an n k matrix where the vectors are its columns: A = a 11 a 12 a 1 k a 21 a 22 a 2 k ....
View Full Document

This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

Page1 / 3

lecture_21 - MA 265 LECTURE NOTES: WEDNESDAY, FEBRUARY 27...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online