My Math Forum If (ui +ei) is LD, prove: sum{|ui|^2} >= 1

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 March 15th, 2019, 01:50 PM #1 Newbie   Joined: Mar 2019 From: brazil Posts: 1 Thanks: 0 If (ui +ei) is LD, prove: sum{|ui|^2} >= 1 Let u1, u2, ..., un be vectors in R^n and e1, e2, ..., en be the canonical basis. If (ui + ei) is a LD set, the exercise is to prove the inequality: sum{|ui|^2} >=1 I am trying to suppose, by contradiction, that the sum is less than 1, and then get that the vectors (ui+ei) are LI. I know this exercise is introductory to Linear Algebra, so it doesn't require deep theorems like "Frobenius norm" or matrices knowledge.
 March 16th, 2019, 01:46 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 What is LD? Thanks from topsquark
 March 16th, 2019, 04:03 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 LD presumably means linearly dependent.

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