Action of $\text{GL}(V)$ on $V$ is irreducible

Question

There is some kind of natural action of $\text{GL}(V)$ on a vector space $V$ via $f.v:=f(v)$ but why is that action irreducible?

Assume we have an invariant subspace $U$ in $V$, $U\neq 0$. Then we want to get $U=V$, so let $v\in V$. As there is an $u\in U$, we need to find a linear invertible map $f:V\to V$ with $f(u)=v$. Then $f.u=f(u)=v$ is in $U$.

Is there such an $f$? If not: How does the proof go?

Tsemo Aristide · Answer 1 · 2017-10-23T21:25:39.700

1

Let $u\in U$ not zero, for every $v\in V$ there exists $f\in Gl(V)$ such that $f(u)=v$. To see this consider bases $(u,e_1,...,e_n)$ $(v,f_1,...,f_n)$ of $V$, write $f(u)=v, f(e_i)=f_i, 1\leq i\leq n$. This implies that $U=V$.

If the dimension of $V$ is infinite, use the fact that two bases have also the same cardinal ad the proof above mutatis mutandis.

Two infinite bases for a vector space have equal cardinality

edited Oct 23 '17 at 21:25

answered Oct 23 '17 at 21:20

Tsemo Aristide

89,587

Thanks! What is behind the 'mutatis mutandis'? – Oct 23 '17 at 21:41
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@R.A. "change what needs to be changed." I personally like "and mutatis mutandis, QED." – Kyle Miller Oct 24 '17 at 01:23

Action of $\text{GL}(V)$ on $V$ is irreducible

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