= Solution
A <character of an algebra> is a nonzero algebra homomorphism $\varphi:A\to\mathbb C$. The <character space> is
$$
\Phi_A=\{\varphi:A\to\mathbb C:\varphi\text{ is a character}\}.
$$
For a unital algebra, $\varphi(1)=1$. Moreover $\varphi(a)\in\sigma_A(a)$, since applying $\varphi$ shows that $a-\varphi(a)1$ cannot be invertible. In a Banach algebra,
$$
|\varphi(a)|\leq r(a)\leq\|a\|.
$$
Thus $\|\varphi\|\leq1$, while $\varphi(1)=1$ gives equality.
If $A$ is commutative and $\lambda\in\sigma_A(a)$, the proper ideal generated by $a-\lambda1$ lies in a maximal ideal. The quotient by that maximal ideal is $\mathbb C$, and its quotient map is a character taking $a$ to $\lambda$. Therefore
$$
\sigma_A(a)=\{\varphi(a):\varphi\in\Phi_A\}.
$$
Since every Banach-algebra element has nonempty spectrum, $\Phi_A$ is nonempty.
For $X=Y\oplus Z$, let $P,Q$ be the complementary coordinate projections. A character must send each idempotent to zero or one, and $P+Q=I$ forces their values to be different. Choose mutually inverse isomorphisms between $Y$ and $Z$ and regard them as off-diagonal operators $S,T$ on $X$. Then
$$
ST=P,\qquad TS=Q.
$$
Multiplicativity would give $\varphi(P)=\varphi(S)\varphi(T)=\varphi(T)\varphi(S)=\varphi(Q)$, a contradiction. Hence $\Phi_{\mathcal B(X)}$ is empty. This is the <absence of characters on an operator algebra with isomorphic complementary summands>.
If $a=a^*$ in a unital C*-algebra, then $e^{ita}$ is unitary for real $t$. If $\lambda\in\sigma(a)$, the spectral mapping theorem gives $e^{it\lambda}\in\sigma(e^{ita})$, whose modulus is one. Varying positive and negative $t$ forces $\operatorname{Im}\lambda=0$, so $\sigma(a)\subset\mathbb R$.
If $B\subseteq A$ is a unital C*-subalgebra and $x\in B$ is normal, spectral permanence holds. Indeed, when $x-\lambda$ is invertible in $A$, the positive normal element
$$
(x-\lambda)^*(x-\lambda)
$$
has spectrum bounded away from zero. Continuous functional calculus uniformly approximates its reciprocal by polynomials, placing the reciprocal in $B$. It follows that $(x-\lambda)^{-1}\in B$. Thus $\sigma_B(x)=\sigma_A(x)$.
The <Commutative Gelfand--Naimark theorem> says that the Gelfand transform is an isometric unital star-isomorphism
$$
A\cong C(\Phi_A)
$$
for every commutative unital C*-algebra. If $a$ is positive, continuous functional calculus for the function $t\mapsto\sqrt t$ on $\sigma(a)\subset[0,\infty)$ gives a positive $b$ with $b^2=a$. Pointwise uniqueness in $C^*(1,a)\cong C(\sigma(a))$ gives the unique positive square root. This is the <positive square root in a C-star algebra>[positive square root in a C*-algebra].
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