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

I was trying to prove that $proj_v(u)=\frac{v·u}{||v||^2}$ v, and I was getting close, but then a friend spoiled the fun of completing the proof by giving me what he called a "hint". Blurting out a key part in a proof isn't a "hint". Needless to say, after this, I immediately knew what to do to complete the proof, but as I looked back at my original approach, I can't help but wonder what was wrong with it. I tried it, but with it I got answers like the zero vector or even an infinite number of answers, and I don't know why.

What I did was I defined the projection of u onto v to be the vector x such that (u- xv=0 (because of the orthogonality). If I break these vectors down into their component forms, with the components of x $\langle$ x$_1$, x$_2\rangle$ being treated as variables, (and the components of u and v as constants) then we get an infinite number of answers, since we have 2 variables. Why does this approach to the problem give me an infinite number of answers, when there is clearly one unique solution? Thank you.

share|cite|improve this question
You wrote that $(u-x).v$ has "clearly" only one solution. Are you sure about it? Try to draw a picture and find missing condition. (I hope this classifies as a hint.) – Martin Sleziak Jul 19 '11 at 7:01
That was a great hint; it told me what was wrong, without giving me too much information. Thank you. – Hautdesert Jul 20 '11 at 6:54
up vote 2 down vote accepted

Nice try, but as everyone said, if you're trying to define a projection, at least define it to be in the span of the vector $V$, i.e. $x$ should be co-linear to it.

What people are trying to explain is that it is not clear that $x$ is colinear to $v$ in your definition, and there are indeed counter-examples. For instance, if you suppose that $u$ and $v$ are already orthogonal, your choice of $x$ is simply any multiple of $u$, because $$ (u-\lambda u) \cdot v = (1-\lambda)(u \cdot v) = 0. $$ What you would want to do is the exact same thing, but suppose further more that $x$ is co-linear to $v$, hence $x = \lambda v$, and solve for $\lambda$. Your idea was pretty good ; you were just missing one point.

That does not classify as a hint though. I would've skipped the "What you would want to do" part if I just gave a hint.

Since you already know the answer from your spoiler, here it is in my opinion : Let $\lambda v$ be the projection of $u$ on $v$ defined as a solution of $(u-\lambda v)\cdot v = 0$. Then $u \cdot v - \lambda v \cdot v = 0$, hence $\lambda = \frac{u \cdot v}{v \cdot v}$, which gives you $$ \text{proj}_v(u) = \frac{u \cdot v}{v \cdot v} v. $$

share|cite|improve this answer
Ah, of course! I forgot that x MUST be a scalar multiple of v. Thanks. – Hautdesert Jul 20 '11 at 7:05

The idea was right, but we need a further condition on $x$, namely that $x$ is "along" $v$. I don't want to spoil the fun by translating the "along" into symbols! But draw a picture, and things should become clear.

If a translation proves necessary, please send a message.

share|cite|improve this answer

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.