内积空间的算符及谱分解定理
adjoint map
定义
对于线性映射 \(T:V \rightarrow W\) 定义： \(T^*:W \rightarrow V\)，st :
\[ \left<Tv,w\right> = \left<v, T^* w\right> \]
左边是 \(W\) 空间的内积，右边是 \(V\) 空间的内积。
首先证明 \(T^*\) 是个映射：
当给定 \(T\) 和 \(w\) 之后，定义 \(V\) 上的泛函，使得 \(\varphi(v) = \left<Tv,w\right>\) (显然可证是个泛函)，然后由 Riesz Representation Theorem可知，一定存在唯一的\(v'\)，st \(\varphi(v) = \left<v,v'\right>\)，于是自然定义\(T^* w = v'\)，由唯一性可知这是一个合法映射。
性质：
简单易证 \(T^*\) 是个线性映射，且具有以下性质
- \((S+T)^* = S^*+T^*\)
- \((\lambda S)^* = \bar \lambda S^*\)
- \((T^{*})^* = T\)
- \((ST)^* = T^* S^*\)
映射空间性质：
\[ \begin{aligned} range(T^*) = null(T)^{\bot} \\ range(T) = null(T^*)^{\bot} \\ null(T^*) = range(T)^{\bot} \\ null(T) = range(T^*)^{\bot} \end{aligned} \]
自伴随算符(self-adjoint operators)
定义
Definition self-adjoint An operator \(T \in \mathcal{L}(V)\) is called self-adjoint if \(T=T^*\). In other words \(T \in \mathcal{L}(V)\) is self-adjoint if and only if \(\langle Tv,w\rangle\equiv\langle v,Tw\rangle\) for all \(w\in V\).
性质
Eigenvalues of self-adjoint operators are real.
Suppose \(V\) is a complex inner product space and \(T\in\mathcal{L}(V\) ). Then \(T\) is self-adjoint if and only if \(\langle Tv,v\rangle\in\mathbb{R}\) for every \(v \in V\)
(hint: \(\langle T u,w\rangle=\frac{\langle T(u+w),u+w\rangle-\langle T(u-w),u-w\rangle}{4}+ \frac{ \langle T(u+i\,w),u+i\,w \rangle- \langle T(u-i\,w),u-i\,w \rangle}{4}i\) )
(同时说明任意含\(T\)映射内积\(\langle Tv,w\rangle\)可以被 \(\langle Tu,u\rangle\) 这种形式改写)
Normal 算符
定义
Definition normal An operator on an inner product space is called normal if it commutes with its adjoint In other words, \(T\in\mathcal{L}(V)\) is normal if \(TT^{*}=T^{*}T\).
性质
7.20 \({T}\) is normal if and only if \(||Tv|| = ||T^ * v||\) for all \(v \in V\) （这也就说明 \(T\) 和 \(T^*\) 的null空间相同)。
7.21 For \(T\) normal and \(T\) has a eigenvector with eigenvalue \(\lambda\). Then \(v\) is also an eigenvector of \(T^*\) with eigenvalue \(\bar\lambda\).
(hint: \(T-\lambda I\) 也是 normal 的，借助此证明）
7.22 Suppose \(T\in\mathcal{L}(V)\) is normal. Then eigenvectors of \(T\) corresponding to distinct eigenvalues are orthogonal.
(hint: \((\lambda_1-\lambda_2)\left<v_1|v_2\right>\))
Complex Spectral Theorem
Suppose \(F=C\) and \(T∈ \mathcal{L}(V)\). Then the following are equivalent:
- \(T\) is normal.
- \(V\) has an orthonormal basis consisting of eigenvectors of \(T\).
- \(T\) has a diagonal matrix with respect to some orthonormal basis of \(V\).
(hint: Schur’s Theorem: Suppose \(V\) is a finite-dimensional complex vector space and \(T\) is linear operator Then has an upper-triangular matrix with respect to some orthonormal basis of V)
Real Spectral Theorem
(证明略复杂，见书，不想掌握)
Suppose \(\mathcal F\) is \(R\) and \(T \in \mathcal L(V)\). Then the following are equivalent:
- \(T\) is self-adjoint.
- \(V\) has an orthonormal basis consisting of eigenvectors of \(T\).
- \(T\) has a diagonal matrix with respect to some orthonormal basis of \(V\).
半正定算符
Definition
positive operator : An operator \(T\in \lambda(V)\) is called positive if \({T}\) is self-adjoint and $ Tv,v $ for all \(v\in V\).
An operator \(R\) is called a square root of an operator T if \(R^2 = T\).
性质
7.35 Let \(T \in \mathcal L (V)\). Then the following are equivalent: (a) T is positive; (b) T is self-adjoint and all the eigenvalues of T are nonnegative; (c) T has a positive square root; (d) T has a self-adjoint square root; (e) there exists an operator \(R \in \mathcal L (V)\) such that \(T =R^*R\).
7.36 Every positive operator on \(V\) has a unique positive square root.
isometry operator
定义
An operator \(S\in\mathcal{L}(V)\) is called an isometry if \(\|S\nu\|=\|\nu\|\) for all \(v ∈ V\). In other words, an operator is an isometry if it preserves norms.
性质
7.42 Suppose \(S ∈ \mathcal{L}(V)\). Then the following are equivalent:
- \(S\) is an isometry.
- \(\left< Su,Sv\right> =\left< u,v\right>\) for all \(u,v \in V\).
- \(Se_1,...,Se_n\) is orthonormal for every orthonormal list of vectors \(e_1,...,e_n\) in \(V\).
- there exists an orthonormal basis \(e_1,...,e_n\) of \(V\) such that \(S e_1,\ldots,S e_n\) is orthonormal.
- \(S^*S = 1\).
- \(SS^*=1\).
- \(S^*\) is an isometry.
- \(S\) is invertible and \(S^{-1}=S\).
7.43 Suppose \(V\) is a complex inner product space and \(S∈ \mathcal{L}(V)\). Then the following are equivalent:
- \(S\) is an isometry.
- There is an orthonormal basis of \(V\) consisting of eigenvectors of \(S\) whose corresponding eigenvalues all have absolute value 1.
（hint: 谱分解定理）