Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$ || \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.

share|cite|improve this question
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
up vote 5 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.