Converse of Schur's First Lemma

Suppose $G$ is a closed subgroup of $SU(d)$, $d>1$, and let $\rho$ be a $d$-dimensional special unitary representation of $G$. Suppose that if a matrix $A$ commutes with all of $\rho(G)$ for all $g\in G$, then $A=cI$. Does it then hold that $\rho$ is an irreducible representation of $G$?

Schur's first lemma states that if $\rho$ is irreducible, then if a matrix $A$ commutes with all of $\rho(G)$ for all $g\in G$, then $A=cI$. I'm asking if the converse of this statement holds. If it doesn't hold in general, does it hold in any special cases? I'm particularly interested in the case where $d=3$ and $G$ is either finite or continuous and closed.

Suppose $\rho:SU(d)\to Gl(V^d)$ is reducible. Then we have a decomposition $V^d = U^m \oplus W^n$ (with $m+n=d$) where both $U^m$ and $W^n$ have positive dimension. (Representations which preserve $\langle \cdot , \cdot \rangle$ are automatically completely reducible by using orthogonal complements).
Choose a basis $\{u_i\}$ of $U$ and a basis $\{w_j\}$ of $W$, so the set $\{u_i, w_j\}$ is a basis of $V^d$. In this basis, every $\rho(g)$ will have block diagonal form where the top block is $m\times m$ and the bottom block is $n\times n$.
Now, consider the diagonal matrix $D$ where the first $m$ diagonal entries are $1$ while the last $n$ entries are $2$. By inspection, this matrix commutes with the block form we just mentioned above. In particular, we have exhibited a matrix not of the form $cI$ which commutes with $\rho(g)$ for all $g$.