# $\|x+y\|^2=\|x\|^2+\|y\|^2$ iff $x=cy$ with $c\gt0$?

$|| \vec{x} + \vec{y} ||^2 = ||\vec{x}||^2 + ||\vec{y}||^2$ iff $\vec{x} = c\vec{y}$ with $c>0$

Is this statement correct or not, if not not, please explain. I just started a self-study of Linear Algebra a week ago so my knowledge of it is not too big yet.

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As written, no. You've squared 2 out of the 3 terms. Is that intentional? – Aaron Oct 7 '12 at 10:48
Oh, no, not intentional sir, I edited it. – JohnPhteven Oct 7 '12 at 10:49

$||\vec x+\vec y||^2 = (\vec x+\vec y)(\vec x+\vec y) = \vec x\vec x+2\vec x\vec y+\vec y\vec y=||\vec x||^2+||\vec y||^2+2\vec x\vec y$. Thus equality holds iff $\vec x\vec y=0$, i.e. $\vec x$ and $\vec y$ are orthogonal.

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So in particular, we can't allow $\vec x=c\vec y$ with $c>0$. – Cameron Buie Oct 7 '12 at 11:16
only if it's $L_2$ norm – chaohuang Oct 7 '12 at 12:27

If $x = cy$, $c \geq 0$, you have $||x+y|| = ||x|| + ||y||$, not $||x+y||^2 = ||x||^2 + ||y||^2$. (BTW, if you have $c \leq 0$, you get $||x+y|| = \left|||x|| - ||y||\right|$)

This is because you then have (in the non-squared case, $c \geq 0$ again) $$||x+y|| = ||x + cx|| = ||(1+c)x|| = (1+c)||x|| = ||x|| + c||x|| = ||x|| + ||cx|| = ||x|| + ||y||.$$

In the squared case with $c \geq 0$ you end up with $(1+c)^2||x||^2$ on the left side, and $(1+c^2)||x||^2$ on the right side, which is in general not the same.

To see why that is the case, look at these things geometrically. $||x+y||$ is the length of the vector $x+y$. Now, how does the length of two vectors behave if you add them? In general, the resulting length will be smaller than the individual lengths. If the vectors point in the same direction (which is what $x=cy$, $c \geq 0$ means), the resulting length will be the sum of the individual lengths, though.

If you add the squares of the lengths instead, the vetors have to be orthogonal (i.e., the angle between them has to be 90 degree). It works then because for an orthogonal triangle you have $a^2 + b^2 = c^2$ if $a$,$b$ and the lengths of the orthogonal sides, and $c$ is the length of the third side. Note that the third side if interpreted as a vector is the same as the sum of the two other sides, if also interpreted as vectors.

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What you wrote is true if $c>0$. In general you have $\|(1+c)x\|=|1+c|\cdot\|x\|$ and $\|cx\|=|c|\cdot\|x\|$. – Martin Sleziak Oct 7 '12 at 12:09
@MartinSleziak Ups, yeah, I was a bit sloppy there. Will edit to fix. – fgp Oct 7 '12 at 12:22