First of all I want to make clear that although I've already asked a related question here, my point in this new question is a little different. On the former question I've considered vector fields on a smooth manifold and here I'm considering just vectors.
In Physics vectors are almost always defined by their transformation properties. Quoting Griffiths:
Well, how about this: We have a barrel of fruit that contains $N_x$ pears, $N_y$ apples, and $N_z$ bananas. Is $\mathbf{N} = N_x\hat{\mathbf{x}}+N_y\hat{\mathbf{y}}+N_z\hat{\mathbf{z}}$ a vector? It has three components, and when you add another barrel with $M_x$ pears, $M_y$ apples, and $M_z$ bananas the result is $(N_x+M_x)$ pears, $(N_y+M_y)$ apples, $(N_z+M_z)$ bananas. So it does add like a vector. Yet it's obviously not a vector, in the physicist's sense of the word, because it doesn't really have a direction. What exactly is wrong with it?
The answer is that $\mathbf{N}$ does not transform properly when you change coordinates. The coordinate frame we use to describe positions in space is of course entirely arbitrary, but there is a specific geometrical transformation law for converting vector components from one frame to another. Suppose, for instance, the $\bar{x},\bar{y},\bar{z}$ system is rotate by an angle $\phi$, relative to $x,y,z$, about the common $x=\bar{x}$ axes. From Fig. 1.15,
$$A_y=A\cos \theta, A_z=A\sin\theta,$$
while
$$\bar{A}_y=\cos\phi A_y + \sin \phi A_z,$$
$$\bar{A}_z=-\sin\phi A_y + \cos\phi A_z.$$
More generally, for totation about an arbitrary axis in three dimensions, the transformation law takes the form:
$$\bar{A}_i=\sum_{j=1}^3R_{ij}A_j.$$
Now: Do the components of $\mathbf{N}$ transform in this way? Of course not - it doesn't matter what coordinates you use to represent position in space, there is still the same number of apples in the barrel. You can't convert a pear into a banana by choosing a different set of axes, but you can turn $A_x$ into $\bar{A}_y$. Formally, then, a vector is any set of three components that transforms in the same manner as a displacement when you change coordinates.
It is exactly this kind of definition I'm in trouble to understand. My point here is the following: as a mathematician would say, a vector is just an element of a vector space.
Let $V$ be a vector space over $\mathbb{K}$ and let $\{e_i\}$ be a basis. Then the mapping $f : \mathbb{K}^n\to V$ given by $f(a^1,\dots,a^n)=a^ie_i$ is an isomorphism by definition of basis.
This means that we can pick any numbers $a^1,\dots,a^n$ and they will give a unique vector no matter what those numbers are. If they represent numbers of pearls, bananas or apples, it doesn't matter. They are numbers.
Now, if we consider another basis $\{\bar{e}_i\}$ we are certain that exists numbers $a^i_j$ which are unique such that $e_j = a^i_j \bar{e}_i$.
In that setting if we have a vector $v = v^je_j$ then we have $v = v^ja^i_j \bar{e}_i$. In other words $v = \bar{v}^i\bar{e}_i$ with $\bar{v}^i = a^i_jv^j$. The transformation law is thus just a result from the theory of linear algebra!
Now, my whole doubt is: what is behind this physicists definition? They are trying to use a result of the theory to define vectors, but why this definition should make sense? As I've pointed out, because $f$ is isomorphism, by the definition of basis any set of numbers will form a vector and if we change the basis the new components will forcefully change as needed for the theory make sense.
EDIT: After thinking for a while I believe I have an idea of what's going on here. I believe we have two separate things: the mathematical idea of vector and the physical idea of a vectorial quantity.
I believe that is the source of the confusion since for a mathematician when we pick $(a^1,\dots,a^n)\in \mathbb{K}^n$ those are just arbitrary numbers while for a physicist if we pick $(a^1,\dots,a^n)$ each $a^i$ has a specific physical meaning as a measurable quantity. Is that the idea somehow?
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