Commutative $ \mathbb{C} $-algebra with involution. The following is a problem in the Professor’s lecture notes; I don’t know how to prove it:

A complex $ ^{*} $-algebra $ A $ is commutative if and only if the set of self-adjoint elements of $ A $ forms a real sub-algebra of $ A $ if and only if every element of $ A $ is normal.

 A: Let $ A $ be a $ ^{*} $-algebra over $ \mathbb{C} $, and denote its set of self-adjoint elements by $ A_{\text{sa}} $.


Theorem 1 $ A $ is commutative if and only if $ A_{\text{sa}} $ is a real sub-algebra of $ A $.

Proof: Suppose that $ A $ is commutative. Then for all $ a,b \in A_{\text{sa}} $ and $ \lambda \in \mathbb{R} $, we have


*

*$ (a + b)^{*} = a^{*} + b^{*} = a + b $,

*$ (ab)^{*} = b^{*} a^{*} = ba = ab $ and

*$ (\lambda \cdot a)^{*} = \overline{\lambda} \cdot a^{*} = \lambda \cdot a $.
We thus see that $ A_{\text{sa}} $ is closed under addition, multiplication and $ \mathbb{R} $-scalar multiplication. Therefore, $ A_{\text{sa}} $ is a real sub-algebra of $ A $.
Conversely, suppose that $ A_{\text{sa}} $ is a real sub-algebra of $ A $. Then for all $ a,b \in A_{\text{sa}} $, we have $ ab \in A_{\text{sa}} $, so $ (ab)^{*} = ab $. However, we also have $ (ab)^{*} = b^{*} a^{*} = ba $. Hence, all pairs of self-adjoint elements of $ A $ commute. As $ A $ can be decomposed as $ A_{\text{sa}} + i \cdot A_{\text{sa}} $, it follows that all pairs of elements of $ A $ commute, thereby making $ A $ a commutative $ ^{*} $-algebra. $ \quad \spadesuit $


Theorem 2 $ A $ is commutative if and only if every element of $ A $ is normal.

Proof: If $ A $ is commutative, then every element of $ A $ is obviously normal.
Conversely, suppose that every element of $ A $ is normal. Then for all $ a,b \in A_{\text{sa}} $, we have
\begin{align}
       (a + ib)^{*} (a + ib) &= (a + ib) (a + ib)^{*}, \\
            (a - ib)(a + ib) &= (a + ib)(a - ib), \\
(a^{2} + b^{2}) + i(ab - ba) &= (a^{2} + b^{2}) + i(ba - ab), \\
                     ab - ba &= ba - ab, \\
                         2ab &= 2ba, \\
                          ab &= ba.
\end{align}
Hence, all pairs of self-adjoint elements of $ A $ commute. As $ A $ can be decomposed as $ A_{\text{sa}} + i \cdot A_{\text{sa}} $, it follows that all pairs of elements of $ A $ commute, thereby making $ A $ a commutative $ ^{*} $-algebra. $ \quad \spadesuit $
