A standard definition for vectors is that they are quantities with both magnitude and direction. This is often graphically represented as an arrow. The length of the arrow captures the magnitude of the vector, and the tip of the arrowhead represents its direction. While this definition is very intuitive and feels very natural, there are several subtleties that we should unpack.
An animation showing how the same vector gets represented differently due to the choice of different coordinate frames.
Whenever we talk about a direction, we implicitly assume that there is a standard reference frame or coordinate system with respect to which we are describing the direction of a vector. But what if two people choose two different coordinate systems? A vector’s direction is part of its definition and, in some sense, independent of the selected coordinate system to describe it. However, the explicit representation of its direction will most certainly depend on the coordinate system chosen. Let us illustrate this very fundamental but often puzzling aspect through an animation.
It still feels bizarre to talk about “direction” when no coordinate systems are chosen. We run into even more trouble when we try to define a zero vector, which upon adding to any vector $\mathbf{v}$ gives back the same vector $\mathbf{v}$. Clearly, the zero vector should have zero magnitude, but how to define its direction? How can we give an arrowhead to a single point? Our definition of vectors as arrows requires modification. We, therefore, seek a definition for vectors that captures all of the essences of the notion of vectors as arrows without inheriting any of its difficulties.
This leads us to abstract vectors.
We define abstract vectors as any mathematical objects which satisfy the following properties.
Let us denote vectors as $\mathbf{v}_1, \mathbf{v}_2,\dots, \mathbf{v}_n \in V$. We define a binary operation of vector addition, denoted by $+$ as follows.
We can work with both vectors and scalars. Let’s have $c_1,c_2,\dots,c_n \in \mathbb{R}$, so that