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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?

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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