How is the square of a transformed vector not a vector?

[BHL 20]

So I learned that if a vector (a₁, a₂, a₃) is transformed to a different set of coordinates, and the components of the resulting vector are squared like so: (a₁'², a₂'², a₃'²), this result is not itself a vector. The proof for this simply shows that each component a_i'² does not transform to a_i² when brought to the original coordinates.

I'm having a lot of trouble understanding this. Say the vector (1,0,0) is transformed by a 90^O rotation about the z-axis. It becomes (0,-1,0). The "square" of this is (0,1,0). Sure this doesn't transform back to (1,0,0) but if this triplet is considered in isolation, without any reference to the original vector, there doesn't seem to be any reason not to consider it a vector.

 

[fresh_42]

(a₁'², a₂'², a₃'²) is of course a vector. It is simply not a result of a linear transformation of (a₁', a₂', a₃')

 

[suremarc]

I'm having a lot of trouble understanding this. Say the vector (1,0,0) is transformed by a 90^O rotation about the z-axis. It becomes (0,-1,0). The "square" of this is (0,1,0). Sure this doesn't transform back to (1,0,0) but if this triplet is considered in isolation, without any reference to the original vector, there doesn't seem to be any reason not to consider it a vector.

It is correct to say that the map f: (a_0,\ldots,a_n)\rightarrow(a'_0,\ldots,a'_n) maps vectors to vectors. The author's point appears to be that, given a linear map S, S f S^{-1}\neq f in general. This is true precisely when f is non-injective.

Where did you learn this? It may be that there is some context missing here.

 

[WWGD]

So I learned that if a vector (a₁, a₂, a₃) is transformed to a different set of coordinates, and the components of the resulting vector are squared like so: (a₁'², a₂'², a₃'²), this result is not itself a vector. The proof for this simply shows that each component a_i'² does not transform to a_i² when brought to the original coordinates.

I'm having a lot of trouble understanding this. Say the vector (1,0,0) is transformed by a 90^O rotation about the z-axis. It becomes (0,-1,0). The "square" of this is (0,1,0). Sure this doesn't transform back to (1,0,0) but if this triplet is considered in isolation, without any reference to the original vector, there doesn't seem to be any reason not to consider it a vector.

Or maybe you mean that somehow the transformation is not tensorial in some sense?

 

[BHL 20]

Where did you learn this? It may be that there is some context missing here.

It's from a vectors and tensors course. It was an example used to illustrate the definition of a vector.

 

[HallsofIvy]

And, what, exactly, is the definition of "vector"? It is NOT simply "three numbers"! The same vector
in different coordinate systems may have different "components". In order to be a vector, there must be a specific homogeneous relation between the components in one coordinate system and the components in another.

(The point of "homogeneous" is that a vector that has all components 0 in one coordinate system has all components 0 in any
coordinate system (the "0 vector"). From that, it follows that if U= V in one coordinate system, then U- V= 0 in that coordinate system so U- V= 0 in any coordinate system so U= V in any coordinate system. The result of that is that if an equation involving only vectors (and their extension, tensors) is true in one coordinate system, then it is true in any coordinate system.)

 

[Fredrik]

When books like this use the term "vector", they're not referring to an element of a vector space, or even a triple of numbers. They're referring to a function that associates a triple of real numbers with each ordered basis for $\mathbb R^3$ that can be obtained by applying some rotation R to the components of the standard ordered basis.

The authors never say that for some reason. I don't know if they think it's so obvious that it's not worth mentioning, or if they just don't have a clue what they're doing.

The function is such that the relationship between two triples associated with two different ordered bases is given by the tensor transformation law.

Suppose that the relationship between new (=primed) and old (=unprimed) basis vectors is given by
$$e_i' =Re_i = (Re_i)_j e_j =R_{ji} e_j.$$ Let x be a triple associated with the "unprimed" ordered basis, and let x' be the corresponding triple associated with the "primed" ordered basis. We have
$$x_j e_j =x_ie_i =x_i'e_i' =x_i' R_{ji} e_j$$ Since a basis is linearly independent, this implies that
$$x_j= R_{ji}x_i' = (Rx)'_j$$ and therefore
$$x=Rx'$$ and
$$x'=R^{-1}x.$$ The term "contravariant" is used because of the appearence of $R^{-1}$ instead of $R$ in this equation.

The claim in post #1 is saying that we shouldn't expect $(x_i')^2=R_{ji} (x_j)^2$ to hold just because $x_i'=R_{ji}x_j$ holds.

 

[FactChecker]

In the context of tensors, the term "vector" usually does not just refer to a mathematical triple (x, y, z) of numbers. A "vector" must be defined in such a way that it converts to other coordinate systems in the correct way. That definition of "vector" makes it an entity that is independent of the choice of coordinate systems. (It is an equivalence class of triples with their associated coordinate system). That allows it to represent a physical entity that exists even if no one is around to specify a coordinate system to measure it in. Your (a₁'², a₂'², a₃'²) will not undergo coordinate system transformations correctly. .