Math Help - Linear Algebra problem

1. Linear Algebra problem

I was wondering how to go about a certain problem that confused me...

Here's the question:

Let x and y be nonzero vectors in the set of real n-vectors.

Prove that the length(x+y) is equal to the length(x) + length(y) if and only if y = cx for some c greater than 0.

2. Originally Posted by faure72
I was wondering how to go about a certain problem that confused me...

Here's the question:

Let x and y be nonzero vectors in the set of real n-vectors.

Prove that the length(x+y) is equal to the length(x) + length(y) if and only if y = cx for some c greater than 0.
Here is the idea.

This is the triangular inequality.
|x+y|<=|x|+|y|
Meaning the triangle you form out of these vectors follows the rule that the sum of two sides of a triangle is greater than the third side.
The exception (special case when we have equality) is when the vector, say "y" lies directly upon the other vector x. In that case we do not form a triangle and have equality.
Thus, we need that y lies (coincides) with x. Meaning they have the same direction. And two vectors has the same direction iff they are non-zero scalar multiples of each other. Thus, y=cx

3. Thanks! So I get the part about the triangular inequality. What did you mean, though, when you said "y lies directly upon the other vector x," though, when you were referring to the exception? Is it possible to show a picture of this?

4. Originally Posted by faure72
Thanks! So I get the part about the triangular inequality. What did you mean, though, when you said "y lies directly upon the other vector x," though, when you were referring to the exception? Is it possible to show a picture of this?
Heir.

5. Gee, I hate to do this without the use of TeX. Please try to follow the notation.
If x=cy with c>0 then the angle between x & y is 0: thus x.y=|x||y|. (|V| is the length of V and V.W is the dot product of V with W.)
Recall that |x+y|^2= |x|^2 + 2(x.y) + |y|^2.
So if x.y=|x|y| then |x+y|^2= |x|^2 + 2|x||y| + |y|^2=(|x|+|y|)^2.
Thus |x+y|=|x|+|y|.

If |x+y|=|x|+|y| then |x+y|^2= |x|^2 + 2|x||y| + |y|^2.
But |x+y|^2= |x|^2 + 2(x.y) + |y|^2 which implies (x.y)=|x||y| or x is positive multiple of y.