# For any positive integer $n$, is there an irreducible representation of degree $n$?

I want to prove a problem: Prove that $Z(M_n(\Bbb{C}))=\{\lambda I\mid \lambda\in \Bbb{C}\}$.

If $M\in Z(M_n(\Bbb{C}))$, let $\varphi:G\to GL_n(\Bbb{C})$ be an irreducible matrix representation of degee $n$, then we have the following commutative diagram. $$\Bbb{C}^n\stackrel{\varphi_g}{\longrightarrow}\Bbb{C}^n$$ $$M\downarrow~~~~~~~~~~~~\downarrow M$$ $$\Bbb{C}^n\stackrel{\varphi_g}{\longrightarrow}\Bbb{C}^n$$ Then by Schur's lemma, $M=\lambda I$ for some $\lambda\in \Bbb{C}$.

I use an assertion in the proof: for any positive integer $n$, there exists an irreducible representation of degree $n$. But I am not sure that the assertion is true.

• Is your $G$ the matrix algebra $M_n(\mathbb{C})$? You can check easily that $M_n(\mathbb{C})$ has an irreducible representation of degree $n$ given by the natural action of $n$-by-$n$-matrix on the $n$-dimensional vector space, essentially by what you wrote. Also I don't understand why your argument proves $Z(M_n(\mathbb{C}))=\mathbb{C}I_n$. – Aaron Nov 29 '15 at 14:44
• @Aaron Thanks. You mean that let $\varphi:GL_n(\Bbb{C})\to GL_n(\Bbb{C})$ be the identity mapping. Then $\varphi$ is a representation of degree $n$. But how did you know that $\varphi$ is irreducible. The statement of Schur's lemma which I had learned one is: Let $\varphi$ be an irreducible representation of $G$ for $V$. If a linear operator $T$ on $V$ satisfy $\varphi(g)T=T\varphi(g)$ for all $g\in G$, then $T=\lambda I$ with $\lambda \in \Bbb{C}$. – bfhaha Dec 2 '15 at 10:45