User:SammyGr/Self-adjoint element

In mathematics, an element of a *-algebra is called self-adjoint if it is the same as its adjoint element.

Definition
Let $$\mathcal{A}$$ be a *-algebra. An element $$a \in \mathcal{A}$$ is called self-adjoint if $a = a^*$.

The set of self-adjoint elements is referred to as $\mathcal{A}_{sa}$.

A subset $$\mathcal{B} \subseteq \mathcal{A}$$ that is closed under the involution *, i.e. $$\mathcal{B} = \mathcal{B}^*$$, is called self-adjoint.

A special case from particular importance is the case where $$\mathcal{A}$$ is a complete normed *-algebra, that satisfies the C*-identity ($$\left\| a^*a \right\| = \left\| a \right\|^2 \ \forall a \in \mathcal{A}$$), which is called a C*-algebra.

Especially in the older literature on *-algebras and C*-algebras, such elements are often called hermitian. Because of that the notations $$\mathcal{A}_h$$, $$\mathcal{A}_H$$ or $$H(\mathcal{A})$$ for the set of self-adjoint elements are also sometimes used, even in the more recent literature.

Examples

 * Each positive element of a C*-algebra is self-adjoint.
 * For each element $$a$$ of a *-algebra, the elements $$aa^*$$ and $$a^*a$$ are self-adjoint, since * is an involutive antiautomorphism.
 * For each element $$a$$ of a *-algebra, the real and imaginary parts $\operatorname{Re}(a) = \frac{1}{2} (a+a^*)$ and $\operatorname{Im}(a) = \frac{1}{2 \mathrm{i} } (a-a^*)$  are self-adjoint, where $$\mathrm{i}$$ denotes the imaginary unit.
 * If $$a \in \mathcal{A}_N$$ is a normal element of a C*-algebra $$\mathcal{A}$$, then for every real-valued function $$f$$, which is continuous on the spectrum of $$a$$, the continuous functional calculus defines a self-adjoint element $f(a)$.

Criteria
Let $$\mathcal{A}$$ be a *-algebra. Then:


 * Let $$a \in \mathcal{A}$$, then $$a^*a$$ is self-adjoint, since $$(a^*a)^* = a^*(a^*)^* = a^*a$$. A similarly calculation yields that $$aa^*$$ is also self-adjoint.
 * Let $$a = a_1 a_2$$ be the product of two self-adjoint elements $a_1,a_2 \in \mathcal{A}_{sa}$. Then $$a$$ is self-adjoint if $$a_1$$ and $$a_2$$ commutate, since $$(a_1 a_2)^* = a_2^* a_1^* = a_2 a_1$$ always holds.
 * If $$\mathcal{A}$$ is a C*-algebra, then a normal element $$a \in \mathcal{A}_N$$ is self-adjoint if and only if its spectrum is real, i.e. $\sigma(a) \subseteq \R$.

In *-algebras
Let $$\mathcal{A}$$ be a *-algebra. Then:


 * Each element $$a \in \mathcal{A}$$ can be uniquely decomposed into real and imaginary parts, i.e. there are uniquely determined elements $$a_1,a_2 \in \mathcal{A}_{sa}$$, so that $$a = a_1 + \mathrm{i} a_2$$ holds. Where $a_1 = \frac{1}{2} (a + a^*)$ and $a_2 = \frac{1}{2 \mathrm{i}} (a - a^*)$ .|undefined
 * The set of self-adjoint elements $$\mathcal{A}_{sa}$$ is a real linear subspace of $\mathcal{A}$. From the previous property, it follows that $$\mathcal{A}$$ is the direct sum of two real linear subspaces, i.e. $\mathcal{A} = \mathcal{A}_{sa} \oplus \mathrm{i} \mathcal{A}_{sa}$.
 * If $$a \in \mathcal{A}_{sa}$$ is self-adjoint, then $$a$$ is normal.
 * The *-algebra $$\mathcal{A}$$ is called a hermitian *-algebra if every self-adjoint element $$a \in \mathcal{A}_{sa}$$ has a real spectrum $\sigma(a) \subseteq \R$.

In C*-algebras
Let $$\mathcal{A}$$ be a C*-algebra and $$a \in \mathcal{A}_{sa}$$. Then:


 * For the spectrum $$\left\| a \right\| \in \sigma(a)$$ or $$-\left\| a \right\| \in \sigma(a)$$ holds, since $$\sigma(a)$$ is real and $$r(a) = \left\| a \right\|$$ holds for the spectral radius, because $$a$$ is normal.
 * According to the continuous functional calculus, there exist uniquely determined positive elements $$a_+,a_- \in \mathcal{A}_+$$, such that $$a = a_+ - a_-$$ with $a_+ a_- = a_- a_+ = 0$. For the norm, $$\left\| a \right\| = \max(\left\|a_+\right\|,\left\|a_-\right\|)$$ holds. The elements $$a_+$$ and $$a_-$$ are also referred to as the positive and negative parts. In addition, $$|a| = a_+ + a_-$$ holds for the absolute value defined for every element $
 * For every $$a \in \mathcal{A}_+$$ and odd $$n \in \mathbb{N}$$, there exists an uniquely determined $$b \in \mathcal{A}_+$$ that satisfies $$b^n = a$$, i.e. an unique $$n$$-th root, as can be shown with the continuous functional calculus.