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Let B = {b¡, following by filling in the blanks: b,} be a basis for a vector space V. You will be proving the ... If a subset {u,- 1,} is linearly dependent in V, then the set of coordinate vectors {{u, la[u,]a} is linearly dependent in R". ... You need only write the word for each blank on our quiz, but be organized so I can grade your work.

Let B = {b¡, following by filling in the blanks: b,} be a basis for a vector space V. You will be proving the ... If a subset {u,- 1,} is linearly dependent in V, then the set of coordinate vectors {{u, la[u,]a} is linearly dependent in R". ... You need only write the word for each blank on our quiz, but be organized so I can grade your work.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.4: Linear Transformations
Problem 24EQ
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Let B = {b¡, following by filling in the blanks: b,} be a basis for a vector space V. You will be proving the ... If a subset {u,, ,} is linearly dependent in V, then the set of coordinate vectors .... U {{u, la [u,]} is linearly dependent in R". You need only write the word for each blank on our quiz, but be organized so I can grade your work. (a) If set of vectors {u,, u,} is linearly dependent in V, then there exist scalars c,…, (where at least one c; is non-zero), (b) such that cu¡ +c,u, + +c,u,=0, where the zero vector is in (с) Вy the of the coordinate mapping: | qu, + ... +c,u,=[qu,]a+ [c,u, =G[u,] ++c, ...+ u
Transcribed Image Text:Let B = {b¡, following by filling in the blanks: b,} be a basis for a vector space V. You will be proving the ... If a subset {u,, ,} is linearly dependent in V, then the set of coordinate vectors .... U {{u, la [u,]} is linearly dependent in R". You need only write the word for each blank on our quiz, but be organized so I can grade your work. (a) If set of vectors {u,, u,} is linearly dependent in V, then there exist scalars c,…, (where at least one c; is non-zero), (b) such that cu¡ +c,u, + +c,u,=0, where the zero vector is in (с) Вy the of the coordinate mapping: | qu, + ... +c,u,=[qu,]a+ [c,u, =G[u,] ++c, ...+ u
(d) (Note: u, la is a vector in (e) Because the coordinate mapping is one-to-one, and since 0b, +Ob, + -+0__= 0,, (1) [0, la = the zero vector in (g) Thus c[u,]+ ·+c, u,= 0 , the zero vector in (this is obtained by "taking [ ]g of both sides of the equation in part (b)). These are the same c; scalars from part (a)! (h) Therefore, the set of vectors is
Transcribed Image Text:(d) (Note: u, la is a vector in (e) Because the coordinate mapping is one-to-one, and since 0b, +Ob, + -+0__= 0,, (1) [0, la = the zero vector in (g) Thus c[u,]+ ·+c, u,= 0 , the zero vector in (this is obtained by "taking [ ]g of both sides of the equation in part (b)). These are the same c; scalars from part (a)! (h) Therefore, the set of vectors is
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