Let $\textbf{SO}(2)$ act on $\mathbb{R}^2$ in the obvious way. Show that $\mathbb{R}^2$ is an irreducible representation under this action.

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The title is the question I am trying to answer. This is a homework problem so I do not want the answer given to me, I am just a bit confused about how to get started. I am taking a Lie groups paper but it is a reading course where I am the only student and the "lecturer" is overseas currently so I do not have anyone I can discuss the problem with.

I am not very familiar with representation theory apart from very basic definitions. I will outline my understanding of the problem and if anyone can tell me where I am going wrong (or right) it would be appreciated.

I want to show that $\mathbb{R}^2$ is an irreducible representation of $\textbf{SO}(2)$, so I show that there is a group homomorphism $\Pi:\textbf{SO}(2) \rightarrow GL(\mathbb{R}^2)$. For this map I think the obvious way is to not do anything, right? Since a rotation is already a linear transformation of $\mathbb{R}^2$. In order to show it is a homomorphism am I showing that $R_\theta(v_1+v_2) = R_\theta(v_1)+R_\theta(v_2)$? And is there more involved in showing that it is irreducible then just stating that the nontrivial subspaces of $\mathbb{R}^2$ are lines, none of which are invariant under rotations?

I don't think there is anything too mystical or difficult going on here, it's just my first time doing any exercises to do with representations and I am a bit confused about the whole idea.

Thanks