Unitary, Orthogonal Operator

Unitary, Orthogonal

Definition 1. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional inner product space over $F$. If $||T(x)|| = ||x||, \forall x \in V$, we call $T$ a unitary operator if $F = \mathbb{C}$ and call $T$ an orthogonal operator if $F = \mathbb{R}$.

유한차원의 경우 unitary, 혹은 othogonal, 즉 유니터리 혹은 직교 연산자라고 부르며, 무한차원의 경우 metric을 보존한다는 점을 강조하기 위해 isometry라고 부른다. 

    자명하게 선형 연산자 $T$가 unitary 혹은 orthogonal일 조건은 $[T]_{\beta}$가 unitary 혹은 orthogonal일 조건과 동치이다. ($\beta$는 orthonormal basis)

Lemma

Lemma. Let $U$ be a hermitian operator on a finite-dimensional inner product space $V$. If $\langle x, U(x) \rangle = 0, \forall x \in V$, then $U = T_0$.

Proof. Let $\beta$ be an orthonormal basis for $V$ consisting of eigenvectors. Then $\forall x \in \beta$, $U(x) = \lambda x$ for some $\lambda \in F$. Thus $\langle x, U(x) \rangle = \overline{\lambda} ||x|| = 0$. Thus $\lambda = 0$, so $U(x) = 0 = T_0(x), \forall x \in \beta$. This means that $U = T_0.$ $\blacksquare$

Theorem 1

Theorem 1. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional inner product space. Then the following statements are equivalent.
(a) $TT^* = T^*T = I$.
(b) $\langle T(x), T(y) \rangle = \langle x, y \rangle, \forall x, y \in V$.
(c) If $\beta$ is an orthonormal basis for $V$, then $T(\beta)$ is an orthonormal basis for $V$.
(d) There exists an orthonormal basis $\beta$ for $V$ such that $T(\beta)$ is an orthonormal basis for $V$.
(e) $||T(x)|| = ||x||, \forall x \in V$.

Proof. (a) $\Longrightarrow$ (b)
$\forall x, y \in V$, $\langle T(x), T(y) \rangle = \langle x, T^*T(y) \rangle = \langle x, y \rangle$.
(b) $\Longrightarrow$ (b)
Let $\beta = \{v_1, ..., v_n \}$. Then $\langle T(v_i), T(v_j) \rangle = \langle v_i, v_j \rangle = \delta_{ij}$. Thus $T(\beta)$ is orthonormal basis for $V$.
(c) $\Longrightarrow$ (d)
Clear.
(d) $\Longrightarrow$ (e)
Let $\beta = \{v_1, ..., v_n \}$. Then $\forall x \in V$, $x = \sum_{i=1}^n \langle x, v_i \rangle v_i$. Thus we have $$||T(x)||^2 = \langle T(x), T(x) \rangle = \langle \sum_{i=1}^n \langle x, v_i \rangle T(v_i), \sum_{j=1}^n \langle x, v_j \rangle T(v_j) \rangle \\ = \sum_{i=1}^n \langle x, v_i \rangle \overline{\langle x, v_i \rangle} = \langle x, x \rangle = ||x||^2$$ by Parseval's Identity. Thus $||T(x)|| = ||x||, \forall x \in V$. 
(e) $\Longrightarrow$ (a) 
Note that $\langle x, x \rangle = ||x||^2 = ||T(x)||^2 = \langle T(x), T(x) \rangle = \langle x, T^*T(x) \rangle, \forall x \in V$. Then $\langle x, x - T^*T(x) \rangle = \langle x, (I - T^*T)(x) \rangle = 0, \forall x \in V$. 
Since $(I - T^*T)^* = I - T^*T$, by Lemma, $I - T^*T = T_0$, so $T^*T = I$.
Let $A = [T^*]_{\beta}$ and $B = [T]_{\beta}$. Then $AB = I$, so $BA = I = [TT^*]_{\beta}$. Thus $T^*T = TT^* = I$. $\blacksquare$

    즉 unitary 혹은 orthogonal operator는 놈과 내적을 보존하므로 자명하게 orthonormal basis도 보존한다.

    Orthonormal basis의 벡터는 놈의 값이 1이고, unitary 혹은 orthogoanl operator는 놈을 보존하므로 자명하게 모든 고유값의 절댓값의 크기는 1임을 알 수 있다.

Corollary 1

Corollary 1. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional real inner product space. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value 1 $\iff$ $T$ is both hermitian and orthogonal.

Proof. ($\Longrightarrow$)
By Theorem 1, $T$ is hermitian. We have $TT^*(v) = T(\overline{\lambda}v) = \lambda \overline{\lambda} v = |\lambda|^2 v = v, \forall v \in \beta$. Then $TT^* = I$. In similar way, we can show that $T^*T = I$. Thus by Theorem 1, $T$ is orthogonal.
($\Longleftarrow$) Since $T$ is hermitian, by Theorem 1, there exists an orthonormal basis $\beta$ for $V$ consisting of eigenvectors. Then $||T(v)|| = ||\lambda v|| = |\lambda| ||v|| = |\lambda| = ||v|| = 1, \forall v \in \beta$. Thus every eigenvalue corresponding to $v \in \beta$ has absolute value 1. $\blacksquare$

Corollary 2

Corollary 2. Let $T \in \mathcal{L}(V)$ where $V$ is a finite-dimensional complex inner product space. Then $V$ has an orthonormal basis of eigenvectors of $T$ with corresponding eigenvalues of absolute value 1 $\iff$ $T$ is unitary.

Proof. We can prove in similar manner to Corollary 1. $\blacksquare$