In this definition:

The set of $N$ quantities $V_j$ is said to be the components of an $N$-dimensional vector $\mathbf{V}$ if and only if their values relative to the rotated coordinate axes are given by

$$V_i'= \sum_{i=1}^N a_{ij}V_j, \quad i=1,2,\dots,N$$

As before, $a_{ij}$ is the cosine of the angle between $x_i'$ and $x_j$. Often the upper limit $N$ and the corresponding range of $i$ will not be indicated. It is taken for granted that you know how many dimensions your space has.

a vector is characterized as a set of components $\{V_j\}$ such that relative to some other coordinate system these components are given by that formula.

This extends to covectors and tensors of all ranks.

The problem is: I believe this definition assumes we have $V_j$, we have $V_i'$ and we can judge whether or not the relation holds. If it holds, $V_j$ are the components of a vector.

But how on earth are $V_i'$ being defined if not by that equation?

What I learned in linear algebra was: given a vector space $V$ and a vector $v\in V$ if we have a basis $\{e_i\}$ we can write uniquely $v = \sum V_i e_i$. This gives the $V_i$.

If we have another basis $\{f_i\}$ we can also write uniquely $v = \sum V_i'f_i$ and this is equivalent to saying

$$v = \sum V_i e_i = \sum V_i a_{ij} f_j$$

and again by uniquenes we have $V_j' = \sum V_i a_{ij}$.

But wait a minute, this means that $V_j'$ is equal to that by definition. In other words, given $V_i$, there is no other option for $V_j'$. This would mean that given any set of components there could be no way to fail this condition.

But I'm obviously missing the point. If the quantities $V_i'$ were defined like this for all sets of components $V_i$ there would be no reason to check if the transformation law is obeyed.

In that context: given a set of numbers $V_i$ how the primed quantities $V_i'$ are defined in order to check that $V_i$ is a vector? Given components $V_i$, what physicists mean by $V_i'$?

  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/314736/2451 $\endgroup$
    – Qmechanic
    Feb 25, 2017 at 18:17
  • 1
    $\begingroup$ As you said, there is no way to say if $(2,6,99) $ are the components of a vector with respect to some basis (other that admit that trivially any components do in fact define a vector, which depends on the choice of basis used to define it.) The idea in the definition is supposed to apply to algebraic EXPRESSIONS that are defined in terms of components of things for which we (for whatever reason) KNOW how they transform. Then one has two ways to transform the expression: Plug in the transformed objects or transform the whole expression like a vector component. If the two agree, it's a vector $\endgroup$ Feb 25, 2017 at 18:47

1 Answer 1


Following up my comment, I would like to present an example. Consider a basis $(e_1, \dots, e_n )$ and also a basis $(f_1,\dots , f_n) $. Then there is a unique set of numbers such that $f_i=T^j_i e_j$. If using summation convention some vector $V=v^i e_i=w^jf_j$ then as you pointed out $$v^i e_i=w^j T^i_j e_i $$ and hence $v^i=T^i_j w^j $ by linear independence. This is the ttansformation rule for vector components. Now let's apply the definition above. Suppose $\{a^i \}$ and $\{b^i\} $ are components of vectors. Then by your definition $X^i:=a^i+3b^i $ are also the components of a vector (I know it's stupid but as long as you don't have derivatives of vector-FIELDS this is about the most general thing you can do). The proof: Transform constituents: $$a'^i+3b'^i=T^i_j a^j+3T^i_j b^j=T^i_j (a^j+3 b^j)=T^i_j X^j=X'^j $$ Hence transforming the individual constituents of the expression and transforming the whole thing as a vector component give the same answer.

Not-Example: $Y^i:=a^i\cdot a^i $ are not the components of a vector. (in your terminology; surely there is still the possibility to malliciously misunderstand this and counter "Any numbers define a vector, what are you talking about?") Proof: Transform constituents $$a'^i a'^i=\sum_{j,k} T^i_j a^j T^i_k a^k\neq T^i_j Y^j. $$

As I said, this only gets interesting if you start talking about vector fields and their derivatives. Exercise: Show that the cross product of two vectors is not a vector, if one is allowed to consider coordinate transformations including reflection (I don't see why your definition would restrict to rotations in the first place)


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