irreducible implies the commutant consists of multiples of identity?

I was trying to solve exercises (4) on Page 59 of the book "A short course on spectral theory", William Avreson.

Let $A$ be a Banach star-algebra. A representation $\pi\in$rep$(A,H)$ is said to be irreducible if the only closed $\pi(A)$-invariant subspaces of $H$ are the trivial ones $\{0\}$ and $H$. Show that $\pi$ is irreducible if and only if the commutant of $\pi(A)$ consists of scalar multiples of the identity operator.

I know the irreducible representation may have another definition. Namely, if any projection commutes with $\pi(A)$ then $\pi$ is irreducible. But the question is how to prove its commutant consists of multiples of identity if given it is irreducible? (This one is much stronger.)

Solution 1:

Let $P$ be a projection in the commutant $\pi(A)'$. Then $PH$ is a subspace, invariant for $\pi(A)$. Since $\pi$ is irreducible, $PH$ is either $0$ or $H$, which implies that $P=0$ or $P=I$.

The above shows that the only projections in $\pi(A)'$ are $0$ and $I$. As a von Neumann algebra is the norm-closed span of its projections, $\pi(A)=\mathbb C I$.