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In the book Representation Theory A First Course the authors write on page 6 that

Corollary 1.6 Any representation is a direct sum of irreducible representations.

This property is called complete reducibility or semisimplicity. [...] The additive group $\mathbb{R}$ does not have this property: the representation $a \mapsto \begin{bmatrix}1 & a\\0 & 1\end{bmatrix}$ leaves the $x$ axis fixed, but there is no complementary subspace.

Could someone enlighten me why there isn't a complementary subspace to this representation? It is because the image of the representation does not contain the zero matrix?

Epsilon Away
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  • It follows from the Jordan normal form theorem. Also a proof that it is indecomposable is given in this post. – Dietrich Burde Apr 09 '22 at 08:32
  • @DietrichBurde: Not really, it doesn't. It's just an elementary calculation, no deep theorems required. Now, if you wanted to classify all finite-dimensional representations of $\mathbf R$, then maybe Jordan's theorem would be relevant, but this is just a fixed one. – tomasz Apr 09 '22 at 09:38
  • @tomasz Jordan normal form is no deep theorem. It is just a tool available from linear algebra a long time before any representation theory. I don't see why we shouldn't use it (and it is indeed used in many duplicates here). – Dietrich Burde Apr 09 '22 at 10:08
  • OK, "deep" may not be the right phrasing. Perhaps I should have said "nontrivial" instead. Anyway, these matrices are very concrete. How would Jordan's theorem help with (or simplify) the argument? – tomasz Apr 10 '22 at 03:16

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