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  "body": "\n## Definition\n\nLet $G$ be a group, let $F$ be a field, and let $\\rho \\colon G \\to GL(V)$ be a [[representation|WYGRXCG2RN]] of $G$ on a [[vector space|000AYPA9AJ]] $V$ over $F$.\nA subspace $W \\subseteq V$ is $G$-invariant if $\\rho(g)(W) \\subseteq W$ for all $g \\in G$.\nThe representation $\\rho$ is *irreducible* if $V \\ne 0$ and the only $G$-invariant subspaces of $V$ are $0$ and $V$.\n\n## Remarks\n\nExample: let $G = \\Z/4\\Z$ act on $\\R^2$, with the class of $1$ acting as the rotation $R$ by $90$ degrees.\nA $G$-invariant line would be spanned by an eigenvector of $R$, but $R$ has characteristic polynomial $x^2 + 1$ with no real root.\nSo the only invariant subspaces are $0$ and $\\R^2$, and the representation is irreducible.\nNon-example: let $G = \\Z$ act on $\\C^2$, with $n$ acting by the matrix $\\begin{pmatrix} 1 \u0026 n \\\\ 0 \u0026 1 \\end{pmatrix}$.\nThe line spanned by $(1, 0)$ is invariant, so the representation is not irreducible.\nIt is also not a direct sum of irreducible representations, because this line has no invariant complement.\nOver $\\C$, the same rotation representation of $\\Z/4\\Z$ on $\\C^2$ is not irreducible: it splits into the eigenlines for $i$ and $-i$.\nSo irreducibility depends on the field.\nAn irreducible representation is also called simple, since it is the same as a simple module over the [[group algebra|MNP2HBWB5Q]].\n",
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