# Proving that a representation is irreducible

Let $$V_n = \mathbb{C}^n$$ the representations of $$SU(n)$$ given by matrix multiplication $$SU(n) \times V_n \rightarrow V_n, (A, v) \mapsto A \cdot v .\,$$ Show that $$V_n$$ is irreducible.

I tried to prove this by induction using the fact that the representation is completely reducible, because $$SU(n)$$ is a compact Lie group, but I'm not sure is the right way to do it.

Any suggestions? Thanks in advance!

## 1 Answer

Let $$v\in V_n\setminus\{0\}$$. For each $$w\in V_n$$ such that $$\|w\|=\|v\|$$, there is a $$M\in SU(n)$$ such that $$M.v=w$$. Therefore, if $$U\subset V_n$$ is a vector subspace such that

• $$v\in U$$;
• $$SU(n).U\subset U$$,

then $$U\supset\{w\in V_n\,|\,\|w\|=\|v\|\}$$. But then $$U=V_n$$.

• Sorry, I'm not sure anymore to have all understood... How can you affirm the existence of a matrix $M \in SU(n)$ such that $M \cdot v = w$? It follows maybe from the fact that for all vectors $w \in V_n$ such that $||w|| = ||v||$ we have that $||A \cdot v|| = ||w||$ for all $A \in SU(n)$? – userr777 Nov 3 '18 at 15:39
• No. Let $v_1=\frac v{\|v\|}$, and consider vectors $v_2,\ldots,v_n$ such that $(v_1,v_2,\ldots,v_n)$ is an orthonormal basis. Then the matrix $P$ whose columns are the $v_i$'s belongs to $SU(n)$ and $P.e_1=v_1$. Now, you construct a matrix $Q$ from $w$ by the same process. Then $Q.P^{-1}\in SU(n)$ and $Q(P^{-1}.v_1)=Q.e_1=w_1$. Therefore, $Q(P^{-1}.v)=w$. – José Carlos Santos Nov 3 '18 at 16:08
• Ok now it's very clear, thanks! – userr777 Nov 3 '18 at 17:04