Understanding the meaning of basis

Question

I am having a really hard time when I try to grasp vectors and its relation to the basis in which they re expressed. Let me explain. In an exercise, I am given three vectors and I am told that the image of the first of them, which is the vector v1 (1,2,3), equals 2 times v1. Then what I do is the product of a vector by a scalar and so I would get (2,4,6), However, the right answer provided is simply (2,0,0). This is the thing I can’t get. I know it is related to the fact that v1 is expressed in the standard basis but I have not been given a clear explanation on what is really going on with it. Can someone shed some light on this in the clearest way possible?

Answer 1

To emphasize the point in the second paragraph of gimusi’s answer, don’t think of $(1,2,3)$ as the vector $v_1$, but as its representation relative to the standard basis. If you’re used to visualizing vectors as little arrows, think of $v_1$ as the arrow itself. To describe this arrow, you need to pick a coordinate system. In the coordinate system that we consider “standard,” the head of this arrow has the coordinates $(1,2,3)$—from its tail you go a distance of $1$ in the $x$-direction, then a distance of $2$ in the $y$-direction, and then a distance of $3$ in the $z$-direction. That’s what those coordinates mean—they’re instructions for reaching the arrow’s head from its tail.

If you pick a different coordinate system, that is, a different basis, the arrow itself doesn’t change, but the instructions for reaching its head from its tail—its coordinates—do. Each vector in a basis defines a direction and a unit of length in that direction. If $v_1$ is the first element of a basis, then the first coordinate of a vector tells us how many multiple of $v_1$’s length to move in its direction as the first step in reaching the head of the vector.

It’s common to call the standard basis vectors $e_1$, $e_2$ and $e_3$. Going by the meaning of coordinates relative to a basis explained in the previous paragraph, then, relative to the standard basis the coordinates of $2v_1$ are $(2,4,6)$, meaning that to get to the head of $2v_1$ you need to go twice the length of $e_1$ in that direction, then $4$ times the length of $e_2$ in its direction, and finally $6$ times the length of $e_3$ in that direction. Alternatively, you can just go twice the length of $v_1$ in the direction of $v_1$. That puts you at the head of $2v_1$, so there’s no need to move any more, thus, the coordinates of $2v_1$ in any basis in which $v_1$ is the first vector are $(2,0,0)$.

Answer 2

With respect to the basis which contains $\vec v_1$ as first vector, the vector $v_1$ is expressed by $(1,0,0)$ and $2\vec v_1$ is exactly expressed by $(2,0,0)$.

The key point is to distinguish between vectors, which are invariable entities, and their representation with respect to a given basis which is expressed in term of components.

Notably, once we have fixed a basis $\vec v_1,\vec v_2,\vec v_3$ we say that $\vec w=(a,b,c)$ if

$$\vec w=a\vec v_1+b\vec v_2+c\vec v_3$$

since we identify $\vec w$ with its components $(a,b,c)$ with respect to the given basis.