Quick doubt about multiplying vectors by scalars I've just started studying linear algebra, and I'm having trouble understanding something that apparentely seems obvious to everyone else. My question is, why does multiplying a vector by a positive scalar does not change its direction? I've seen some visual representations of this and it does seem reasonable to me that that should be the case. However, I wonder whether there is a way to actually show it mathematically? In the 2d coordinate plane, for instance, I guess way one of showing that vector $A$ has the same direction of $cA,c>0$ would be to show that the line containing $A$ has the same slope as the one containing $cA$ (I'm not sure if this approach is correct though). However, how to proceed in the case of $n$ dimensions? I think I must be missing out something basic here, but I really can't seem to get my head around it. Any help will be highly appreciated. Thanks very much in advance. 
 A: It is because of similar triangles.
Consider all possible triangles
that the vector makes
with the
coordinate axes.
When all the lengths
are multiplied by $c$,
the angles of the
transformed triangles 
remain the same
because the lengths
are proportional.
A: For each $\alpha > 0$
$$\text{angle}(\mathbf{x}, \, \alpha \mathbf{x}) = \arccos \frac{(\mathbf{x}, \, \alpha \mathbf{x})}{\|\mathbf{x}\| \, \|\alpha \mathbf{x}\|} = \arccos\frac{\alpha (x,x)}{\alpha \|\mathbf{x}\| \, \| \mathbf{x} \|} = \text{angle}(\mathbf{x}, \, \mathbf{x}) = 0$$
A: One way you might think of defining "direction" in a space of general dimension $n\geq 1$ is the set of vectors $u$ with $\|u\|_{2}=1.$ In two dimensions, this is just the unit circle. If we accept this definition, then we can write any nonzero vector $v\in\mathbb{R}^{n}$ as $$v=\|v\|_{2}\left(\frac{v}{\|v\|_{2}}\right),$$ where the vector in parentheses is the "direction" of $v$ in the sense defined above, and $\|v\|_{2}$ is a positive scalar. But this makes it clear that if $a>0,$ then $$av=(a\|v\|_{2})\left(\frac{v}{\|v\|_{2}}\right)$$ has the same direction vector, but a different positive scalar multiplying it. In this sense, they have the same direction.
A: To what degree do the truth of the following claims seem clear to you?


*

*Adding a vector to itself results in a vector pointing in the same direction.

*Multiplying a vector by a positive integer results in a vector pointing in the same direction.

*Multiplying a vector by a positive rational number results in a vector pointing in the same direction.

*Multiplying a vector by a positive real number results in a vector pointing in the same direction.

