such that span{v₁,..., Vn} =1. Let {V₁, Vn} be a linearly dependent set. Prove that there exists some j with 1 ≤j≤nspan{V₁,...,j,..., Vn}, where vj denotes removing v; fromthe set {v₁,..., Vn}. Note: You can show that two sets A and B are equal by showing thatr€A+r€ B.

Suppose that be a linearly dependentset.Then there exists some such that where are scalars…

Answered: 1. Let {v₁,..., Vn} be a linearly…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

such that span{v₁,..., Vn} =
1. Let {V₁, Vn} be a linearly dependent set. Prove that there exists some j with 1 ≤j≤n
span{V₁,...,j,..., Vn}, where vj denotes removing v; from
the set {v₁,..., Vn}. Note: You can show that two sets A and B are equal by showing that
r€A+r€ B.

Expert Solution

Step 1: "Introduction to the solution"

Suppose that $open curly brackets v subscript 1 comma v subscript 2 comma........ comma v subscript n close curly brackets$ be a linearly dependentset.

Then there exists some $j element of open curly brackets 1 comma 2 comma..... comma n close curly brackets$ such that $v subscript j equals c subscript 1 v subscript 1 plus........ plus c subscript j minus 1 end subscript v subscript j minus 1 end subscript plus c subscript j plus 1 end subscript v subscript j plus 1 end subscript plus....... plus c subscript n v subscript n........ left parenthesis 1 right parenthesis$

where $c subscript i apostrophe s$ are scalars for $i equals 1 comma 2 comma..... comma n.$

We have to show that span $open curly brackets v subscript 1 comma...... comma v subscript n close curly brackets equals$ span $open curly brackets v subscript 1 comma.... comma stack v subscript j with hat on top comma... v subscript n close curly brackets comma$ where $stack v subscript j with hat on top$ denotes removing $v subscript j$ from the set

$open curly brackets v subscript 1 comma....... comma v subscript n close curly brackets.$