When is the norm of the sum of 2 vectors=sum of norms

1. Feb 4, 2015

cpsinkule

1. The problem statement, all variables and given/known data
When is |x+y|=|x|+|y| for arbitrary non-zero vectors x,y∈Rn ie, when does equality hold for the well known inequality |x+y|≤|x|+|y|

2. Relevant equations
|x|2=<x,x>=Σixi2
|x+y|≤|x|+|y|
3. The attempt at a solution
squaring both sides of the inequality we have
(|x+y|)2≤|x|2+|y|2+2|x||y|,
but (|x+y|)2=∑i(xi+yi)2=∑ixi2+∑iyi2+2∑ixiyi, then, by definition of norm we can cancel out all terms in the inequality but
2∑ixiyi=<x,y>≤2|x||y| so we see that equality can only hold when both sides of this inequality are equal, therefore we square both sides
<x,y>2=|x|2|y|2, by definition this expands to ∑ixiyi⋅∑ixiyi=
i(xiyi)2+∑i≠jxixjyiyj
but ∑i(xiyi)2=|x|2|y|2
so our condition for equality of |x+y|=|x|+|y| simplifies to ∑i≠jxixjyiyj=0 or
i≠jxixjyiyj=0
this is the answer I arrived at, however I don't feel confident in it for some reason. I found other solutions online for the problem in this exact book that state "when x and y are linearly dependent", however in the newest edition of the book, which I have, the problem actually states : "hint: the answer is not 'they are linearly dependent.'"

Edit: I've already found some errors in my work, including the fact that I did not expand (|x||y|)^2 properly

Last edited: Feb 4, 2015
2. Feb 4, 2015

cpsinkule

my ammended solution is ∑i≠jxiyj-xjyi=0

3. Feb 5, 2015

Dick

You are really squaring too much here and drawing wrong conclusions. Think about the dot product. You should get just x.y=|x||y| which you had about halfway through. Then remember that the dot product of two vectors is $|x||y|cos(\theta)$ where $\theta$ is the angle between the two vectors. So?

Last edited: Feb 5, 2015
4. Feb 5, 2015

cpsinkule

i see that, however the book explicitly states "the answer is NOT that they are linearly dependent" aka θ=0 which is why I am so confused, the book is "Calculus on Manifolds" by Spivak if you were wondering.

5. Feb 5, 2015

Staff: Mentor

"Linearly dependent" doesn't necessarily mean that θ=0, as you said. If the angle between two vectors were 180°, so that they point in opposite directions, the vectors would be linearly dependent, but it does not follow automatically that |x + y| = |x| + |y|.