Normal Operator
Defintion 1. Let $T \in \mathcal{L}(V)$ where $V$ is an inner product space. We say that $T$ is normal if $TT^* = T^*T$.
위와 같은 조건을 만족시켰을 때 선형 연산자가 normal, 즉 정규하다고 부른다. 자명하게 선형 연산자 $T$가 normal일 조건은 $[T]_{\beta}$가 normal일 조건과 동치이다. ($\beta$는 orthonormal basis)
Theorem 1
Theorem 1. Let $T$ be a normal operator on $V$ where $V$ is an inner product space. Then the following statements are true.
(a) $||T(x)|| = ||T^*(x)||, \forall x \in V$.
(b) $T - cI$ is normal, $\forall c \in F$.
(c) If $x$ is an eigenvector of $T$, then $x$ is also an eigenvector of $T^*$. In fact, if $T(x) = \lambda x$, then $T^*(x) = \overline{\lambda} x$.
(d) If $\lambda_1$ and $\lambda_2$ are distinct eigenvalues of $T$ with corresponding eigenvectors $x_1$ and $x_2$, then $x_1$ and $x_2$ are orthogonal.
Proof. (a) Note that $||T(x)||^2 = \langle T(x), T(x) \rangle = \langle x, T^*T(x) \rangle = \langle x, TT^*(x) \rangle = \langle T^*(x), T^*(x) \rangle$ = $||T^*(x)||^2, \forall x \in V$. Thus $||T(x)|| = ||T^*(x)||$.
(b) Note that $(T - cI)(T - cI)^* = (T - cI)(T^* - \overline{c}I) = TT^* - \overline{c}T - cT^* + |c|^2 \\ = T^*T - cT^* - \overline{c}T + |c|^2 = (T^* - \overline{c}I)(T- cI) = (T- cI)^*(T - cI).$ Thus $T - cI$ is normal.
(c) Note that $(T - \lambda I)(x) = \mathbf{0}$. Then $0 = ||(T - \lambda I)(x)|| = ||(T - \lambda I)^*(x)|| = ||T^*(x) - \overline{\lambda}x||$. Thus $T^*(x) = \overline{\lambda}x$.
(d) $0 = \langle x_1, \mathbf{0} \rangle = \langle x_1, T(x_2) - \lambda_2 x_2 \rangle = \langle x_1, T(x_2) \rangle - \overline{\lambda_2} \langle x_1, x_2 \rangle = \langle T^*(x_1), x_2 \rangle \\ - \overline{\lambda_2} \langle x_1, x_2 \rangle = \langle \overline{\lambda_1} x_1, x_2 \rangle - \overline{\lambda_2} \langle x_1, x_2 \rangle = (\overline{\lambda_1} - \overline{\lambda_2})\langle x_1, x_2 \rangle.$ Since $\lambda_1 \neq \lambda_2$, $\overline{\lambda_1} \neq \overline{\lambda_2}.$ Thus $\langle x_1, x_2 \rangle = 0$. $\blacksquare$
Theorem 2
Theorem 2. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional complex inner product space. Then $T$ is normal $\iff$ there exists an orthonormal basis for $V$ consisting of eigenvectors of $T$.
Proof. $(\Longleftarrow)$
Suppose that such basis $\beta$ exists. Then $[T]_{\beta}$ is diagonal matrix and so is $[T]_{beta}^*$. Thus $[T]_{\beta}[T]_{\beta}^* = [T]_{\beta}^*[T]_{\beta}$.
$(\Longrightarrow)$
By the fundamental theorem of algebra, the chracteristic polynomial of $T$ splits. Then by Schur's Theorem, there exists an orthonormal basis $\beta = \{v_1, ..., v_n\}$ for $V$ such that $[T]_{\beta} = A$ is upper triangular.
Clearly $v_1$ is an eigenvector of $T$. Assume that $v_1, ..., v_{k-1}$ are eigenvectors of $T$. We claim that $v_k$ is also an eigenvector of $T$.
Consider any $j < k$, and let $\lambda_j$ denote the eigenvalue of $T$ corresponding to $v_j$. Then $T^*(v_j) = \overline{\lambda}v_j$. Since $A$ is upper triangular, we have $T(v_k) = A_{1k}v_1 + \cdots + A_{kk}v_k$ and $A_{jk} = \langle T(v_k), v_j \rangle = \langle v_k, T^*(v_j) \rangle = \langle v_k, \overline{\lambda_j}v_j \rangle$ $= \lambda_j \langle v_k, v_j \rangle = 0$. Thus $T(v_k) = A_{kk}v_k$. By induction on $k$, all the vectors in $\beta$ are eigenvectors of $T$. $\blacksquare$
선형 연산자는 특성 다항식을 풀어서 고유값을 구한 뒤, 각 고유값의 고유공간의 기저를 찾고 합집합시켜주면 대각화시킬 수 있었다. 이때 전제는 대각화 가능해야 하다는 것인데, 그렇다면 구체적으로 대각화 가능할 조건은 무엇인지 알 필요가 있으며, 정규인 선형 연산자는 항상 대각화 가능하다는 것이 결론이다.