In a book on General Relativity that I am reading, it defines a vector as an object or array of numbers that transforms like a vector (under rotations). I understand that under rotation $\theta$, a vector $\vec{p}_1 = (p_1, p_2)^\intercal$ transforms as $$ \vec{p}’ = R(\theta)\vec{p} = \begin{pmatrix} p_1\cos\theta + p_2\sin\theta\\ -p_1\sin\theta + p_2\cos\theta \end{pmatrix} $$ However, then he gives an example of an array of two numbers $\vec{p} = (ap_1, bp_2)^\intercal$, where $a\neq b$ as something that is NOT a vector, but this confuses me. How can you show this is not a vector from the action of the rotation matrix on it? Wouldn’t it just multiply as another other vector does under a rotation? There must be something simple here I’m missing.

  • $\begingroup$ Is there any other context that could be missing here? $\endgroup$ Commented Apr 6, 2019 at 22:42
  • $\begingroup$ I hope it doesn’t really define a vector as something that transforms like a vector. That would be a circular definition. $\endgroup$
    – G. Smith
    Commented Apr 6, 2019 at 22:49
  • $\begingroup$ $\vec{p}'$ is not a vector, based on how you've defined it, anyway. It appears to be a 2x2 matrix. $\endgroup$ Commented Apr 6, 2019 at 22:50
  • $\begingroup$ Sorry that was a typo. I think I understand what he means. I’m pretty sure he’s saying that under a rotation, the components of the “vector” should transform accordingly. So for example, $p_1\rightarrow \cos\theta p_1 + \sin\theta p_2$. Then, you plug this into the thing you want to check if it’s a vector and see if it transforms appropriately. $\endgroup$ Commented Apr 6, 2019 at 22:55
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    $\begingroup$ @G.Smith it's not circular, no. for an important reason! $\endgroup$
    – CR Drost
    Commented Apr 6, 2019 at 22:56

1 Answer 1


Let's not call the column with the same name as the vector $\vec{p}$. So we have two objects, $$\begin{align}\vec{p} &=(p_{1}, p_{2})^{T}\\ s(a, b) &= (a\,p_{1}, b\,p_{2})^{T},\end{align}$$ where the components of the vector $\vec{p}$ transform according to the equation you indicated and I assume $a$ and $b$ are scalars (so they don't change under a rotation; let's say they are just the temperature and pressure at the spatial point in question).

Now let's see how $s$ transforms, assuming its transformation is inherited from the transformations of the $p_{1}$ and $p_{2}$. We have $$s'(a, b) = \begin{pmatrix}a\left( p_{1} \cos(\theta) + p_{2} \sin(\theta)\right) \\ b\left(-\, p_{1} \sin(\theta) + p_{2} \cos(\theta)\right)\end{pmatrix}.$$ Now $s(a, b)$ deserves the name "vector" if it transforms as a vector, which would require $$s(a, b) \longrightarrow \begin{pmatrix}s_{1} \cos(\theta) + s_{2} \sin(\theta) \\ -s_{1} \sin(\theta) + s_{2} \cos(\theta)\end{pmatrix},$$ where $s_{1}$ and $s_{2}$ are the components of $s(a, b)$. You can see this is possible if and only if $a = b$.


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