# Definition of Vector

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.

• Is there any other context that could be missing here? – Aaron Stevens Apr 6 at 22:42
• I hope it doesn’t really define a vector as something that transforms like a vector. That would be a circular definition. – G. Smith Apr 6 at 22:49
• $\vec{p}'$ is not a vector, based on how you've defined it, anyway. It appears to be a 2x2 matrix. – probably_someone Apr 6 at 22:50
• 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. – Josh Pilipovsky Apr 6 at 22:55
• @G.Smith it's not circular, no. for an important reason! – CR Drost Apr 6 at 22:56

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 $$a = b$$.