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How do I prove that $$ \left \|\vec{u} \right \|^2 = \left \|\vec{v} \right \|^2 = \langle\vec{u},\vec{v}\rangle $$ if and only if $\vec{u} = \vec{v}$.

I don't have any idea on how to prove this but I hope some one could help me.

LiziPizi
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3 Answers3

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Using bilinearity, \begin{align*} \langle u - v, u - v \rangle & = \langle u, u - v \rangle - \langle v, u - v \rangle\\ & = \langle u, u \rangle - \langle u,v\rangle - \langle v, u\rangle + \langle v, v \rangle\\ & = \|u\|^2 - \langle u, v \rangle - \overline{\langle u, v \rangle} + \| v\|^2\\ & = \|u\|^2 - \|u\|^2 - \overline{\|u\|^2} + \|u\|^2\\ & = 0 \end{align*} (using the given condition, and since the norm is a real number). But $\langle u - v, u - v \rangle = 0$ implies that $u - v = 0$, that is, $u = v$.

The converse is trivial.

M. Vinay
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  • but it is $u = 0$ or $u = v$, they wanted only $u = v$ – LiziPizi Jun 01 '16 at 13:38
  • It is not true that $\langle X, Y\rangle = 0 $ implies that $X = 0$ or $Y = 0$. – Eff Jun 01 '16 at 13:38
  • @m.vinay I don't understand where you took the $\langle u - v, u - v \rangle$ from? – LiziPizi Jun 01 '16 at 13:52
  • @LiziPizi Are you asking about my motivation? As far as proving is concerned, you don't have to justify your choice of the LHS you start from, provided all your steps are valid. But my motive was this: Trying to use bilinearity to combine the norm (which is itself an inner product) with the other inner product in the condition, and keeping in mind that the end result should be $u - v = 0$, try to come up with an expression that can be expanded to given terms equal to the norm and the inner product in the question. This is one such expression. – M. Vinay Jun 01 '16 at 13:55
  • @m.vinay the thing is that every time I start proving something I almost always start with my assumptions that in this case is $\left |\vec{u} \right |^2 = \left |\vec{v} \right |^2 = \langle\vec{u},\vec{v}\rangle$, so I want to know how you knew to start with $\langle u - v, u - v \rangle$? you probably told me in the last comment but I did not understand – LiziPizi Jun 01 '16 at 14:01
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    @LiziPizi When trying to prove a statement, you also should keep if mind what your last step should be. In fact sometimes it's easier to start at the end of the proof and work backwards. –  Jun 01 '16 at 14:03
  • @LiziPizi Good question. Sometimes you can start from the given LHS and neatly reduce it to the RHS (or vice versa). But that doesn't always work. Still, you can start with one and play around with it a bit till you hit upon something else in which everything "fits together". In this case, as I said, seeing the inner products immediately made me think of bilinearity, and I just looked for a way to bring that in. In fact (as you might have seen in my earlier invalid proof), I started off taking a different approach, and found this proof by attempting to fix the mistake in that one. – M. Vinay Jun 01 '16 at 14:07
  • @LiziPizi As a matter of fact, I did not start with $\langle u - v, u - v \rangle$. I started with $\langle u, u \rangle$, $\langle u, v \rangle$, and $\langle v, v \rangle$. When I put those together, I got $\langle u - v, u - v \rangle$. Then, I pretended to start with that (since, as I mentioned, you don't have to justify your choice of first step!). But yes, it's actually something I arrived at, by using the expected final result. Of course, when you write the actual proof you write the right way around (or the wrong way around, if you will!). – M. Vinay Jun 01 '16 at 14:10
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Hint: $\langle u,v \rangle = \cos(\theta) ||u|||||v||$ where $\theta$ is the angle between $u$ and $v$. Don't read the below if you want to try using this idea on your own!

Let $u,v$ be nonzero vectors such that $||u||^2 = ||v||^2 = \langle u,v \rangle$ and let $\theta$ be the angle between $u$ and $v$. Since we already know $||u|| = ||v||$, it suffices to show $\theta = 0$. We have $||u||^2 = \langle u,v \rangle = ||u|| ||v|| \cos(\theta)$, so $||u|| = ||v||\cos(\theta)$. Similarly, $||v||^2 =||u|| ||v|| \cos(\theta)$ gives $||v|| = ||u||\cos(\theta)$. Thus $\cos^2(\theta) = 1$, which implies $\cos(\theta) = \pm 1$. However since we have $||v|| = ||u||\cos(\theta)$, $\cos(theta)$ must be positive, so $\cos(\theta) = 1$. This implies $\theta = 0$. Since $u$ and $v$ have an angle of $0$ between them and $||u|| = ||v||$, it follows that $u=v$.

The other direction, i.e. that $u=v$ implies $||u||^2 = ||v||^2 = \langle u,v \rangle$, is trivial.

aras
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  • Why the downvote? If OP truly doesn't "have any idea" how to solve this, this approach might be more accessible than other solutions that use more powerful results, even if my solution is ugly and the other solutions are more elegant. – aras Jun 01 '16 at 13:39
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Observe that your assumptions imply that the Cauchy-Schwarz inequality $|\langle u,v \rangle| \leq \|u\| \|v\|$ is satisfied as an identity. It is known that this may happen if, and only if, $u = \lambda v$ for some $\lambda \in \mathbb{R}$.

Now, $\langle u,v \rangle = \lambda \langle v,v \rangle = \lambda \|v\|^2$ , and therefore $\lambda =1$.

Siminore
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