Can I move vectors?

Question

The next stupid question is here I guess. But this question has been appearing in my mind for very long time.

When we have a vector $(a, b)$ for example, we can imagine a vector drawn from the center $(0, 0)$ towards $(a, b)$, right? Now, what if I need to use the same vector, but drawn at some other point $(x, y)$? In terms of linear algebra a vector is matrix of the size $(1, n)$ or $(n, 1)$ (not necessarily, just thinking of vectors in $\mathbb{R}^n$) with elements that are magnitudes we multiply other vectors by. But what about geometrical meaning?

Does a vector remain identical if I just take it and draw in an arbitrary place? Can I use values of its elements in calculations regardless of where I draw it?

Can someone tell me please why this confusion might appear in my mind? What concepts do I probably miss? Thanks in advance!

Answer 1

Yes, you can move vectors.

Vector is fully defined by it's components in some basis. Not by "components and point of origin".

It's just a matter of how mathematicians defined what is "vector". It would be possible to define vector as "components plus origin point". It just would be not a very convenient definition. People would constantly have to deal with "components-only" part of these entities.

Does a vector remain identical if I just take it and draw in an arbitrary place?

Yes.

Can I use values of its elements in calculations regardless of where I draw it?

I think so. But to be sure I need an example of calculations or problem.

Can someone tell me please why this confusion might appear in my mind? What concepts do I probably miss? Thanks in advance!

Source of confusion seems to be obvious. Vector is an arrow, arrow has an origin.

But it's very convenient to ignore the "origin" in definition of vector.

Example from physics. Velocity is a vector. Two objects do not have to be located in the same point to have equal velocities.

Answer 2

Work in $\mathbb R^2$ for convenience. The extension to $\mathbb R^n$ is immediate. I assume you know how the vector space operations are defined.

A vector in $\mathbb R^2$ is just an ordered pair $\vec u=(a,b).$ You can "imagine" that it is represented by an arrow from the origin; that is, the directed line segment from $(0,0)$ to $(a,b)$. This is called the $ordinary$ representation of $\vec u.$

Now suppose you have two vectors, $\vec u_0=(x_0,y_0)$ and $\vec u_1=(x_1,y_1)$ such that $\vec u_1-\vec u_0=\vec v.$ Then, the ordinary representation of $\vec v$ is a directed line segment from the origin. But if you want to move $\vec v$ so that it begins at $\vec u_0$ and ends at $\vec u_1$, you get the $geometric$ representation of $\vec v$. It is an arrow from $\vec u_0$ to $\vec u_1$ with the same magnitude and parallel to the vector $\vec v$ in ordinary position (Drawing the picture will help here.)

Now, the set of points corresponding to the geometric representation of $\vec v$ is the line segment from $\vec u_0$ to $\vec u_1$. That is, the segment connecting the points $(x_0,y_0)$ and $(x_1,y_1)$, and this is given by the equation $(x,y)=(x_0,y_0)+t(x_1-x_0,y_1-y_0);\ 0\le t\le 1.$ It is therefore, a set of vectors (in ordinary position, one for each $t$), sweeping out the segment from $(x_0,y_0)$ to $(x_1,y_1).$