# C$^{*}$-algebra acting irreducibly on the finite-dimensional space $\mathbb{C}^{n}$ must be $M_{n}(\mathbb{C})$

Suppose $A\subseteq M_{n}(\mathbb{C})$ is a C$^{*}$-algebra acting irreducibly on the finite-dimensional space $\mathbb{C}^{n}$. I.e., there are no non-trivial subspaces of $\mathbb{C}^{n}$ invariant under $A$, or, equivalently, $W_{v}:=\operatorname{span}\{av:a\in A\}=\mathbb{C}^{n}$ for any $v\in\mathbb{C}^{n}$.

Is there a straightforward way to see that $A=M_{n}(\mathbb{C})$?

• There's something not quite right here, since, the irreps of any finite-dimensional $C^\ast$-algebra $M_{n_1}(\mathbb{C}) \oplus \cdots \oplus M_{n_N}(\mathbb{C})$ are all of this form, i.e., each is given by projection onto a particular matrix block. – Branimir Ćaćić Jun 18 '18 at 16:03
• @Branimir I think I understand now. I have updated the question, requiring that $A$ be a subset of $M_{n}(\mathbb{C})$. – ervx Jun 18 '18 at 17:59

## 3 Answers

A unital C$^*$-algebra $A\subset M_n(\mathbb C)$ is of the form $M_{n_1}(\mathbb C)\oplus\cdots\oplus M_{n_k}(\mathbb C)$, with $n_1+\cdots+n_k=n$. If $k\ne1$, then $A$ has central projections, and thus it has invariant subspaces: for a central projection $p$, $$a(px)=apx=pax\in p\,\mathbb C^n,\ \ \ a\in A,\ x\in\mathbb C^n,$$ and then $A(p\mathbb C^n)\subset p\,\mathbb C^n$. So the only way that $A$ may fail to have nontrivial invariant subspaces is when $k=1$, i.e. $A=M_n(\mathbb C)$.

If the unit $1_A$ of $A$ is not $I_n$, then $1_A$ is a central projection in $M_n(\mathbb C)$ and thus $A$ has nontrivial invariant subspaces.

I like s.harp's solution a lot. I also found what seems to be a very elementary solution to the problem (using only basic linear algebra). This result is referred to as Burnside's Theorem. Lomonosov and Rosenthal provide a very elementary proof in there paper https://www.sciencedirect.com/science/article/pii/S0024379503007225 which I summarize below.

Let $V$ be a finite-dimensional vector space. The following is proved by induction on the dimension of $V$: If $A$ is an algebra of linear operators on the vector space $V$ acting irreducible on $V$, then $A$ is the set of all linear transformations on $V$.

If $\dim(V)=1$, this is obvious. So suppose that $\dim(V)=n> 1$ and that the result holds for vector spaces of dimensional less than $n$.

We claim that $A$ must contain a non-zero operator $u$ which is not invertible. To see this, note that since $A$ acts irreducibly on $V$, it contains an operator $b$, which is not a scalar multiple of the identity. If $b$ is not invertible, we take $u=b$; otherwise, take $u=\lambda b-b^{2}$, where $\lambda$ is any eigenvalue of $b$.

Since $u$ is non-zero and not invertible, the vector space $uV$ has positive dimension less than $n$. Moreover, since $A$ acts irreducibly on $V$, there is for each $v_{1},v_{2}\in V$ an $a\in A$ such that $a(uv_{1})=v_{2}$. Thus, $ua(uv_{1})=uv_{2}$. Hence, the algebra $uA=\{ua:a\in A\}$ acts irreducibly on the vector space $uV=\{uv:v\in V\}$. Therefore, by the inductive hypothesis, $uA$ contains all linear transformations of the vector space $uV$. Hence, there is an $a\in A$ with the property that $ua|_{uV}$ is a rank one operator. But then, $uau$ is a rank one operator on $V$. I.e., $A$ contains a rank one operator.

Given $x\in V$ and a linear functional $f\in V^{*}$, we let $x\otimes f$ denote the rank one operator on $V$ given by $(x\otimes f)(v)=f(v)x$. It is routine to check that for a linear operator $w$ on $V$, $$w(x\otimes f)=(wx\otimes f)$$ and $$(x\otimes f)w=(x\otimes (f\circ w)).$$ By the previous paragraph, $A$ contains a rank one operator $x\otimes f$. By the two equations above, it follows that for $a\in A$, $ax\otimes f\in A$. Since $A$ is irreducible, this implies that $\{y\otimes f:y\in V\}\subseteq A$. We claim that we also have $\{f\circ w:w\in A\}=V^{*}$. Indeed, given $g\in V^{*}$, choose $v_{0}\in V$ such that $f(v_{0})\not=0$ and define $w$ by $wv=\frac{g(v)}{f(v_{0})}v_{0}$. Then, $f\circ w=g$. Therefore, $\{y\otimes g:y\in V,g\in V^{*}\}\subseteq A$. I.e., $A$ contains all rank one operators on $V$. Thus, $A=\mathcal{L}(V)$, since every linear transformation is the sum of rank one operators.

• You're done the moment $A$ contains one rank $1$ operator, but I couldn't figure out how to extract one for the life of me. Very nice! – Branimir Ćaćić Jun 20 '18 at 16:45

A corollary of Wedderburn's theorem is that if $\pi:A\to\mathrm{End}(V)$ is an irreducible algebra homomorphism where $A$ is a unital $\Bbb C$-Algebra and $V$ is a $\Bbb C$ vectorspace, then $\pi$ is automatically surjective.

In our case this gives us a surjective $C^*$-morphism $\pi:\tilde A\to M_{n}(\Bbb C)$, which for dimensional reasons means that $\tilde A=M_{n}(\Bbb C)$ . Here $\tilde A$ is $A$ if $A$ is unital and $A\oplus\Bbb C\Bbb 1$ otherwise (viewed as a sub-algebra of $M_n(\Bbb C)$).

This corollary will give you the result if you verify that there exists no (non-unital) sub-algebra of $M_n(\Bbb C)$ so that if you adjoin the unit you get $M_n(\Bbb C)$. This follows from $M_n(\Bbb C)$ being simple, ie admitting no ideals.

You can find this corollary for example in Serge Lang's algebra book (Corollary 3.6, page 649 here), although the language used here is a bit different.