Hermitian
Defintion 1. Let $T \in \mathcal{L}(V)$ where $V$ is an inner product space. We say that $T$ is hermitian (or self-adjoint) if $T = T^*$.
위와 같은 조건을 만족시켰을 때 선형 연산자가 hermitian이라고 부른다. 자명하게 선형 연산자 $T$가 hermitian일 조건은 $[T]_{\beta}$가 hermitian일 조건과 동치이다. ($\beta$는 orthonormal basis)
선형 연산자가 normal일 조건을 생각해본다면, hermitian이면 normal임을 쉽게 알 수 있다.
Lemma
Lemma. Let $T$ be a hermitian operator on a finite-dimensional inner product space $V$. Then
(a) Every eigenvalue of $T$ is real.
(b) If $V$ is a real inner product space, then the characteristic polynomial of $T$ splits.
Proof. (a) Let $x$ be an eigenvector of $T$. Then $T(x) = \lambda x$ and $T^*(x) = \overline{\lambda} x$. Since $T$ is hermitian, $\lambda = \overline{\lambda}$. Thus $\lambda \in \mathbb{R}$.
(b) Let $n = \dim(V)$, $\beta$ be an orthonormal basis for $V$, and $A = [T]_{\beta}$. Then $A$ is hermitian. Let $L_A$ be a left-multiplication transformation on $\mathbb{C}^n$. Note that $L_A$ is hermitian because $[L_A]_{\gamma} = A$, where $\gamma$ is the standard ordered basis for $\mathbb{C}^n$. Thus the eigenvalues of $L_A$ is real.
By the fundamental theorem of algebra, the characteristic polynomial of $L_A$ splits into factors of the form $t - \lambda$. Since each $\lambda$ is real, the polynomial splits over $\mathbb{R}$. But $\det(L_A - tI) = \det(A - tI_n) = \det(T - tI)$. Therefore the characteristic polynomial of $T$ splits. $\blacksquare$
Theorem 1
Theorem 1. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensioanl real inner product space. Then $T$ is hermitian $\iff$ there exists an orthonormal basis $\beta$ for $V$ consisting of eigenvectors of $T$.
Proof. $(\Longleftarrow)$
Since $V$ is real inner product space, $[T]^*_{\beta} = [T]^t_{\beta}$. Because $[T]_{\beta}$ is a diagonal matrix, $[T]_{\beta} = [T]^t_{\beta} = [T]^*_{\beta}$. Thus $T$ is hermitian.
$(\Longrightarrow)$
By Lemma, the characteristic polynomial of $T$ splits. Then by Schur's Theorem, we have an orthonormal basis $\beta$ for $V$ that $[T]_{\beta}$ is upper triangular.
Note that $A = [T]_{\beta} = [T^*]_{\beta} = [T]^*_{\beta} = A^*$. This means that $A$ is a diagnoal matrix. Thus $\beta$ consists of eigenvectors of $T$. $\blacksquare$
Normal인 선형 연산자가 항상 대각화 가능했듯이, hermitian 선형 연산자도 항상 그러하다. hermitian일 조건이 중요한 이유는 Lemma (a) 때문인데, 고유값이 항상 실수임이 보장되기 때문에 양자역학과 같은 물리학 분야에서 중요하게 다뤄진다.