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I'm taking a linear algebra course and we have just began learning about vectors.

Is the vector $(5,0)$ just a line on the $x$ axis or is it a point? In addition, does it have a direction or magnitude? I believe for direction it would be positive but I do not see the magnitude.

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    $\begingroup$ A vector is neither a line nor a point. It is more like an arrow, that points in a certain direction and has a magnitude. The coordinates $(5,0)$ define the end of the arrow, whereas the origin $(0,0)$ is the starting point. Your vector points in the direction of the positive x-axis and its magnitude is $5$. $\endgroup$
    – M. Wind
    Apr 11 '15 at 2:43
  • $\begingroup$ @M.Wind does it have to start at the origin? Can we shift it? $\endgroup$
    – Chilanie
    Apr 11 '15 at 3:03
  • $\begingroup$ Starting at the origin is convention when no other starting point is given. $\endgroup$
    – futurebird
    Apr 11 '15 at 3:26
  • $\begingroup$ Consider it an abstract object that holds information about a magnitude pointing in some direction. Geometrically, we can view these as arrows. But it doesn't belong to a certain point. Vectors are not fixed in space, though you can certainly use the vector $\vec{v}=(5,0)$ to point from the origin to point $(5,0)$. You could also use it to point from $(1,2)$ to $(6,2)$ or in general from $(a,b)$ to $(a+5,b)$. If you've studied physics, you can easily think of forces, velocities, and displacements in terms of vectors, among many other quantities. $\endgroup$
    – zahbaz
    Apr 11 '15 at 3:30
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    $\begingroup$ @Chilanie In principle you could shift it. But it is better not to do so. A good analogy is a standard map of a country. It shows a city A at coordinates $(1,2)$ and a city B at coordinates $(6,2)$. The vector $(5,0)$ is the difference between these pairs of coordinates. It tells you that to get from city A to city B you have to travel $5$ units (kilometers) in the positive x-direction (east). $\endgroup$
    – M. Wind
    Apr 11 '15 at 16:49
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$u = (5,0) \in \mathbb{R}^2$ can be interpreted as coordinates of the arrow from the origin $0=(0,0)$ to point $P =(5,0)$.

Its magnitude is $\lVert u \rVert = \sqrt{5^2 + 0^2} = 5$ using the euclidean norm.

Its direction is $u/\lVert u \rVert = (5,0)/5 = (1,0)$.

Note however that $u$ might be interpreted as something else, e.g. the function $f(x) = 5 = 5\cdot 1 + 0\cdot x$ and one could still say it has lenght $5$ and direction $g(x) = 1$.

Example of modeling a line:

The endpoints of the vectors $u(t) = t n + u_0$ are on a line through the end point of $u_0$, where $t$ is some real number and $n$ some vector which is representing the direction and $u_0$ is a vector.

One can assign $v = \vec{PQ}$ the vector from point $P$ to point $Q$, this structure is called an affine space.

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  • $\begingroup$ Does it have to start at the origin? Can it start at (1,0)? $\endgroup$
    – Chilanie
    Apr 11 '15 at 3:08
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    $\begingroup$ You can have vectors between any point $P$ and any point $Q$. $\endgroup$
    – mvw
    Apr 11 '15 at 3:11
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    $\begingroup$ It is not at all uncommon for you to get to choose where it starts from, and there can be very useful meaning in doing so. For example, if you wish to add two vectors $u+v$, you can do so by first drawing $u$ and then drawing the vector $v$ starting from the arrowhead of $u$. The vector $(u+v)$ then is the arrow which starts at the tail of $u$ and ends at the arrowhead of $v$. In this example, if $u$ started at the origin and $u$ is nonzero, then $v$ will not have been drawn starting from the origin. If that bothers you, $u+v$ can still be seen as the diagonal of the parallelogram. $\endgroup$
    – JMoravitz
    Apr 11 '15 at 3:12
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    $\begingroup$ You should get some text on analytic geometry or computer graphics to have some examples of this use of vectors. The abstract view is presented in books on linear algebra. $\endgroup$
    – mvw
    Apr 11 '15 at 3:14
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One of the goals of a first course in linear algebra is to abstract the geometric notions of objects you have become familiar with to purer mathematical objects. This means that interpreting the underlying meaning of a particular structure is, well, open-ended. This is especially true when one object can take on multiple interpretations in different contexts.

So, if you are working with vectors as directed line segments (as you might in a physics application), then $(5,0)$ might be interpreted as a rightward-pointing vector of magnitude $5$ on the $xy$-plane, under the usual coordinate system.

If you're interested in image processing or computer graphics, then perhaps $(5,0)$ represents a coordinate on an image, which you might think of as a simple point.

Formally, $(5,0)$ is an element of $\mathbb{R}^2$, which simply means that $(5,0)$ is a pair of real numbers (where order matters; that is, $(5,0)\ne (0,5)$). $\mathbb{R}^2$ is a vector space, which is a way of saying that it is a set of pairs $(x,y)$ obeying certain properties of addition and scaling. Abstractly, vectors are simply elements of vector spaces.

The interpretation determines whether geometric notions such as direction or magnitude should be considered. If, for example, $(5,0)$ is the solution to a system of linear equations, such as \begin{align} x+2y&=5 \\ 2x+y&=10, \end{align} then magnitude might not be a sensible concept to attribute to it.

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    $\begingroup$ "vectors are simply elements of vector spaces": +1 to pointing this out. It is the properties that are important, not the flavor/context. This becomes an important thing to note when eventually working with $l_p$ or $L^p$ spaces $\endgroup$
    – JMoravitz
    Apr 11 '15 at 2:49
  • $\begingroup$ This is very informative. thanks for the pointers. $\endgroup$
    – Chilanie
    Apr 11 '15 at 3:03

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