Lemma 2

Let A be an involutive algebra and $x\in A$. The elements

are hermitian and satisfy $x=x_1+ix_2$. If $x=y_1+iy_2$ with $y_1$ and $y_2$ hermitian, then $x_1=y_1$ and $x_2=y_2$. Moreover, the element $x$ is normal if and only if $x_1$ and $x_2$ are permutable.

The first two assertions follow from Lemma 1 of I, p. 95. One computes that $xx^*-x^*x= 2i(x_2x_1-x_1x_2)$, hence $x$ is normal if and only if $x_1$ and $x_2$ are permutable.

Let A be a unital involutive algebra. In order that $x\in A$ be invertible, it is necessary and sufficient that $x^*$ be so, and one then has $(x^*)^{-1}= (x^{-1})^*$. Since $(x-\lambda e)^*=x^*-\lambda e$ for every $\lambda \in \mathbf{C}$, one deduces that Sp$_A(x^*) =$ Sp$_A(x)$.

Let A be an involutive algebra and $\widetilde{A}$ the algebra deduced from A by adjunction of a unit element. There exists in $\widetilde{A}$ a unique involution extending that of A, given by $(\lambda , x)^*= (\lambda , x^*)$ for $\lambda \in \mathbf{C}$ and $x\in A$. If $x\in A$, one has Sp$'_A(x^*) =$ Sp$'_A(x)$.

Let A and B be involutive algebras. By a morphism of A into B is meant an algebra morphism $\varphi$ of A into B such that $\varphi (x^*) =\varphi (x)^*$ whatever $x$ and $y$ in A may be. The identity mapping of A is a morphism of involutive algebras. If C is an involutive algebra and $\pi : B\rightarrow C$ a morphism of involutive algebras, then $\pi \circ \varphi$ is a morphism of involutive algebras. If $\varphi$ is an algebra isomorphism, then $\varphi^{-1}$ is a morphism of involutive algebras, and one says that $\varphi$ is an isomorphism of involutive algebras.

By prop. 1 of I, p. 95, if A and B are involutive algebras, an algebra morphism $\varphi$ of A into B is a morphism of involutive algebras if and only if $\varphi (A_h)\subset B_h$. By an involutive subalgebra of A is meant a self-adjoint subalgebra. The center of A is an involutive subalgebra. If $A_1$ is a self-adjoint two-sided ideal of A, the involution of A defines, by passing to the quotient, an involution in the algebra $A/A_1$, and the canonical mapping of A onto $A/A_1$ is a morphism of involutive algebras.

Let A be an involutive algebra. The radical of A is equal to the radical of the opposite algebra (A, VIII, p. 431, prop. 7), and is therefore self-adjoint.

Let A be an involutive algebra. If $M\subset A$ is self-adjoint, its commutant $M'$ is an involutive subalgebra of A. If $x\in A$, the bicommutant of $\{x, x^*\}$ is an involutive subalgebra containing $x$ and $x^*$, and this subalgebra is commutative if and only if $x$ is normal (n$^o5$ of I, p. 5).