I've come across two different definitions of vectors.

The first one is from linear algebra and it just defines vectors to be the elements of a vector space.

The other one is the one that we were taught in the physics class. There a vector was defined as anything whose coordinates transform in a particular way under certain transformations of space (the professor specifically gave example of the pairs of numbers $(x, y)\in\mathbb R^2$ and the transformation on $\mathbb R^2$ as rotation by angle $\theta$ so that $(x, y)\mapsto (x\cos\theta+y\sin\theta, -x\sin\theta + y\cos\theta)$).

Question: Is there a relation between the two? And can someone help making the second definition mathematically precise, and more general? (I don't know how to generally define "space" in "transformations of space".)

  • 1
    $\begingroup$ "vector" to physicists means "element of a tangent space on a manifold". (Or maybe an element of a cotangent space, or a field of such elements etc). $\endgroup$ Commented Jun 21, 2020 at 18:09
  • $\begingroup$ You may want to contemplate differing views of a complicated situation. The vector space article has historical background. $\endgroup$
    – Somos
    Commented Jun 21, 2020 at 18:12
  • $\begingroup$ Wikipedia mentions specific differences between mathematics and physics vector concepts. $\endgroup$
    – Somos
    Commented Jun 21, 2020 at 18:22
  • $\begingroup$ When you said triples, did you mean pairs? $\endgroup$ Commented Jun 21, 2020 at 18:29
  • $\begingroup$ It's worth noting that your physics definition is the definition of a contravariant or tangent vector given here $\endgroup$ Commented Jun 21, 2020 at 18:31

1 Answer 1


To oversimplify, mathematicians love abstraction and physicists are practical: they want to calculate stuff.

You are being subjected to both viewpoints. Physics often uses $R^3$, which satisfies the defining properties (axioms) of a vector space that were given in your math class. Thus, $R^3$ satisfies all the theorems about vector spaces that were proved there.

However, there is no harm in just understanding $R^n$ for now. It's much easier to understand the theory there.

BTW, the linear transformations from $R^3$ to itself (or the matrices that represent them) have a multiplication defined but, as a vector space, it's just $R^9$.


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