Can anyone help me with this proof? I'm a bit confused about it.

I want to show that in a finite dimensional euclidean vector space $(V, \cdot)$ for two vectors $\vec{x}$ and $\vec{y}$ is true that

$|\vec{x}\cdot \vec{y}|=\|\vec{x}\| \|\vec{y}\| \implies \vec{y}=\lambda \vec{x}$

The problem is about the absolute value.

If $\vec{x}\cdot \vec{y}=+\|\vec{x}\| \|\vec{y}\| $

Than I must prove that $ \vec{y}=\lambda \vec{x}$

I know that $|\lambda|=\frac{\| \vec{y}\|}{\|\vec{x}\|}$

but, here is the thing, can I say that, since $\vec{x}\cdot \vec{y}$ is positive and $\cos\theta=\frac{\vec{x}\cdot \vec{y}}{\|\vec{x}\| \|\vec{y}\| }=+1$ and so $\theta = 0$ in this case than $\lambda=+\frac{\| \vec{y}\|}{\|\vec{x}\|}$ ?

To make a comparison I know that for vectors in $V_3$ this would be about the direction of the two vectors, and the sign of $\lambda$ is determined by the fact that $\vec{x}$ and $\vec{y}$ have the same or opposite direction, while $|\lambda|=\frac{\| \vec{y}\|}{\|\vec{x}\|}$ is always valid. So in this case $\theta=0$, which would mean they have the same direction and I would take $\lambda>0$ .

But here, in dimension $n$ is the concept of "direction" still valid?

Anyway in this case the proof is the following

$\|\frac{\vec{y}}{\|\vec{y}\|}-\frac{\vec{x}}{\|\vec{x}\|}\|^{2}=0$ which is true only if I consider $\lambda=+\frac{\| \vec{y}\|}{\|\vec{x}\|}$

The same problem is when the dot product is negative so

If $\vec{x}\cdot \vec{y}=-\|\vec{x}\| \|\vec{y}\| $

So I can only prove that $\|\frac{\vec{y}}{\|\vec{y}\|}+\frac{\vec{x}}{\|\vec{x}\|}\|^{2}=0$, therefore assuming that $\lambda=- \frac{\| \vec{y}\|}{\|\vec{x}\|}$

I'm not looking for alternative proofs of this fact but I really need to know if this makes sense. Does $\cos\theta=-1$ means that two vectors have "opposite directions" also in dimension $n$?

And in general is the concept of direction still valid in the same meaning of $V_3$?

  • $\begingroup$ Did you eventually mean to proof the Cauch-Schwarz inequality ??? $\endgroup$ – XPenguen Dec 18 '15 at 23:44
  • $\begingroup$ That's the particular case of parallel / linearly dependent vectors.. $\endgroup$ – Gianolepo Dec 19 '15 at 8:31
  • $\begingroup$ If the two vectors $x,y$ are linearly dependent then it it is $x=\lambda y \;$ and then .... $\endgroup$ – XPenguen Dec 19 '15 at 8:36

To show that $x\;$ and $y\;$ are linearly dependent you just need to show that. $\; y=\lambda x\;$ does not have a trivial solution $\lambda =0$
You can assume that both $x\;$ and $y\;$ aren´t $0$. Hence you get $| \lambda |=\frac{||y||}{||x||}>0$
Hence $x,y\;$ are linearly dependant. You just needed to show that $\; \lambda \;$ is not $\; 0$


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