线代第六章定义&定理整理(不更新了)

Chapter 6

6.1 Inner Products and Norms

Definition (inner product).

Let V be a vector space over F. An inner product on V is a function that assigns, to every ordered pair of vectors x and y in V, a scalar in F, denoted \(⟨x,y⟩\), such that for all x, y, and z in V and all c in F, the following hold:

(a) \(⟨x + z,y⟩ = ⟨x,y⟩ + ⟨z,y⟩.\)

(b) $⟨cx,y⟩=c⟨x,y⟩. $

(c) \(\overline{⟨x, y⟩} = ⟨y, x⟩,\) where the bar denotes complex conjugation.

(d) \(⟨x,x⟩>0\) if \(x \neq 0\).

Definition (conjugate transpose).

Let \(A ∈ M_{m×n}(F)\). We define the conjugate transpose or adjoint of A to be the \(n×m\) matrix \(A^∗\) such that \((A^∗)_{ij} = \overline{A_{ji}}\) for all \(i,j\).

Definition (inner product space).

A vector space \(V\) over \(F\) endowed with a specific inner product is called an inner product space. If \(F = C\), we call V a complex inner product space, whereas if \(F = R\), we call \(V\) a real inner product space.

Definition of some inner products.

Frobenius Inner product: \(\langle A, B\rangle=\operatorname{tr}\left(B^{*} A\right) \text { for } A, B \in M_{n\times n}(F).\)

实际上就是\(\langle A, B\rangle=\sum_{i}\sum_{j}A_{ij}\overline{B_{ij}}\)

Standard inner product on \(F^n\): \(x=\left(a_{1}, a_{2}, \ldots, a_{n}\right)\) and \(y=\left(b_{1}, b_{2}, \ldots, b_{n}\right)\) in \(\mathrm{F}^{n}\), \(\langle x, y\rangle=\sum_{i=1}^{n} a_{i} \bar{b}_{i}\).

实际上和Frobenius inner product是一个东西。

H of continuous complex-valued functions defined on the interval \([0, 2π]\): \(\langle f, g\rangle=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(t) \overline{g(t)} d t\).

Theorem 6.1.

Let V be an inner product space. Then for x, y, z ∈ V and c ∈ F , the following statements are true.

(a) \(⟨x,y + z⟩\) = \(⟨x,y⟩\) + \(⟨x,z⟩\).

(b) \(⟨x,cy⟩=\overline c⟨x,y⟩\).

(c) \(⟨x,0⟩ = ⟨0,x⟩ = 0\).

(d) \(⟨x,x⟩=0\) if and only if \(x=0\).

(e) If \(⟨x,y⟩=⟨x,z⟩\) for all \(x∈V\), then \(y=z\).

性质(a)和(b)统称conjugate linear,注意不要漏写共轭。

Definition (norm).

Let \(V\) be an inner product space. For \(x ∈ V\), we define the 􏰉norm or length of \(x\) by \(\|x\|= ⟨x, x⟩\).

Theorem 6.2.

Let \(V\) be an inner product space over \(F\). Then for all \(x, y ∈ V\) and \(c ∈ F\) , the following statements are true.

(a) \(\|cx\|= |c|·\|x\|.\)

(b) \(\|x\|=0\) if and only if \(x=0\). In any case, \(\|x\|≥0\).

(c) (Cauchy–Schwarz Inequality)\(|⟨x,y⟩|≤\|x\|·\|y\|\).

(d) (Triangle Inequality) \(\|x + y\| ≤ \|x\| + \|y\|\).

证明

(c)

\(y=0\)显然成立,假设\(y \neq 0\)。对于任意\(c \in F\),有

\[\begin{aligned} 0 \leq\|x-c y\|^{2} &=\langle x-c y, x-c y\rangle=\langle x, x-c y\rangle- c\langle y, x-c y\rangle \\ &=\langle x, x\rangle-\bar{c}\langle x, y\rangle- c\langle y, x\rangle+ c \bar{c}\langle y, y\rangle \end{aligned} \]

\(c=\frac{\langle x, y\rangle}{\langle y, y\rangle}\),则有\(0 \leq\langle x, x\rangle-\frac{|\langle x, y\rangle|^{2}}{\langle y, y\rangle}=\|x\|^{2}-\frac{|\langle x, y\rangle|^{2}}{\|y\|^{2}}\),所证不等式成立。

(d)

\[\begin{aligned}\|x+y\|^{2} &=\langle x+y, x+y\rangle=\langle x, x\rangle+\langle y, x\rangle+\langle x, y\rangle+\langle y, y\rangle \\ &=\|x\|^{2}+2 \Re\langle x, y\rangle+\|y\|^{2} \\ & \leq\|x\|^{2}+2|\langle x, y\rangle|+\|y\|^{2} \\ & \leq\|x\|^{2}+2\|x\| \cdot\|y\|+\|y\|^{2} \\ &=(\|x\|+\|y\|)^{2} \end{aligned} \]

Definition (orthogonal, unit vector, orthonormal).

Let \(V\) be an inner product space. Vectors \(x\) and \(y\) in \(V\) are orthogonal (perpendicular) if \(⟨x, y⟩ = 0\).

A subset \(S\) of \(V\) is orthogonal if any two distinct vectors in \(S\) are orthogonal.

A vector \(x\) in \(V\) is a unit vector if \(\|x\| = 1\).

Finally, a subset \(S\) of \(V\) is orthonormal if \(S\) is orthogonal and consists entirely of unit vectors.

6.2 The Gram–Schmidt Process and Orthogonal Complements

Definition (orthonormal basis).

Let V be an inner product space. A subset of V is an orthonormal basis for V if it is an ordered basis that is orthonormal.

Theorem 6.3.

Let V be an inner product space and \(S = {v_1, v_2, . . . , v_k}\) be an orthogonal subset of V consisting of nonzero vectors. If \(y ∈ span(S)\), then

\[y=\sum_{i=1}^{k} \frac{\left\langle y, v_{i}\right\rangle}{\left\|v_{i}\right\|^{2}} v_{i} \]

证明:设\(y = \sum_{i=1}^ka_iv_i\),写出\(\langle y, v_i\rangle\)表达式即可得出。

Corollary 1.

If, in addition to the hypotheses of Theorem 6.3, S is orthonormal and y ∈ span(S), then

\[y=\sum_{i=1}^{k} \left\langle y, v_{i}\right\rangle v_{i} \]

Corollary 2.

Let V be an inner product space, and let S be an orthogonal subset of V consisting of nonzero vectors. Then S is linearly independent.

Theorem 6.4. (the Gram–Schmidt process)

Let V be an inner product space and \(S = \{w_1, w_2, \ldots, w_n\}\) be a linearly independent subset of V. Define \(S′ = \{v_1, v_2, \ldots, v_n\}\), where \(v_1 = w_1\) and

\[v_{k}=w_{k}-\sum_{j=1}^{k-1} \frac{\left\langle w_{k}, v_{j}\right\rangle}{\left\|v_{j}\right\|^{2}} v_{j} \quad \text { for } 2 \leq k \leq n. \]

Then S′ is an orthogonal set of nonzero vectors such that span(S′) = span(S).

Theorem 6.5.

Let V be a nonzero finite-dimensional inner product space. Then V has an orthonormal basis \(\beta\). Furthermore, if \(\beta = \{v_1,v_2,...,v_n\}\) and x ∈ V, then

\[x=\sum_{i=1}^{n}\left\langle x, v_{i}\right\rangle v_{i} \]

证明:用 Gram–Schmidt 把 orthogonal basis 构造出来,再 normalize 即可。至于\(x=\sum_{i=1}^{n}\left\langle x, v_{i}\right\rangle v_{i}\) 实际上就是 Thereom 6.3 Corollary 1.

Corollary.

Let V be a finite-dimensional inner product space with an orthonormal basis $ = \beta = {v_1, v_2, \ldots, v_n}$. Let T be a linear operator on V, and let A = \([T]_\beta\). Then for any i and j, $$Aij = \langle T(vj),vi\rangle.$$

Definition (Fourier coefficients).

Let $\beta $ be an orthonormal subset (possibly infinite) of an inner product space V, and let \(x ∈ V\). We define the Fourier coefficients of \(x\) relative to \(\beta\) to be the scalars \(⟨x, y⟩\), where \(y ∈ β\).

Definition (orthogonal complement).

Let S be a nonempty subset of an inner product space V. We define \(S^\perp\) to be the set of all vectors in V that are orthogonal to every vector in S; that is, \(S^{\perp}=\{x \in V:\langle x, y\rangle= 0 \text { for all } y \in S\}\). The set \(s^\perp\) is called the orthogonal complement of S.

注意
  • S可以是任意集合,不一定是 subspace;

  • \(0 \in S\), \(S\cap S^\perp = \{0\}\); 否则\(S\cap S^\perp = \O\).

Theorem 6.6.

Let \(W\) be a finite-dimensional subspace of an inner product space \(V\), and let \(y∈V\). Then there exist unique vectors \(u∈W\) and \(z\in W^\perp\) such that \(y=u+z\). Furthermore, if\({v_1,v_2,\ldots,v_k}\)is an orthonormal basis for \(W\), then

\[u=\sum_{i=1}^{k}\left\langle y, v_{i}\right\rangle v_{i} \]

证明:直接令\(u=\sum_{i=1}^{k}\left\langle y, v_{i}\right\rangle v_{i}\),令\(z = y - u\),只需证\(z\in W^\perp\). 对任意 \(j\), 有

\[\begin{aligned}\left\langle z, v_{j}\right\rangle &=\left\langle\left(y-\sum_{i=1}^{k}\left\langle y, v_{i}\right\rangle v_{i}\right), v_{j}\right\rangle=\left\langle y, v_{j}\right\rangle-\sum_{i=1}^{k}\left\langle y, v_{i}\right\rangle\left\langle v_{i}, v_{j}\right\rangle \\ &=\left\langle y, v_{j}\right\rangle-\left\langle y, v_{j}\right\rangle= 0 \end{aligned} \]

下证 unique. 假设\(y = u + z = u' + z', u' \in W, z' \in W^\perp\), 则\(u - u'\in W, z - z' \in W^\perp\), \(u - u' = z - z' \in W \cap W^\perp = \{0\}.\)

Corollary (orthogonal projection).

The vector \(u = \sum_{i=1}^{k}\left\langle y, v_{i}\right\rangle v_{i}\) is the unique vector in \(W\) that is “closest” to \(y\); that is, for any \(x ∈ W\),$ |y − x| ≥ |y − u|$, and this inequality is an equality if and only if \(x = u\). \(u\) is called the orthogonal projection of \(y\) on \(W\).

证明:

注意到\(\langle u-x, z\rangle = 0\)

\(\|y-x\|^2 = \|u + z - x\|^2 = \|u-x\|^2 + \|z\|^2 \ge \|z\|^2\)

Theorem 6.7.

Suppose that \(S=\left\{v_{1}, v_{2}, \ldots, v_{k}\right\}\) is an orthonormal set in an n-dimensional inner product space \(V\). Then

(a) S can be extended to an orthonormal basis \(\{v_1, v_2, \ldots, v_k, v_{k+1}, \ldots, v_n\}\) for \(V\).

(b) If \(W = span(S)\), then \(S_1 = \{v_{k+1}, v_{k+2}, \ldots, v_n\}\) is an orthonormal basis for \(W^\perp\).

(c) If \(W\) is any subspace of \(V\), then \(dim(V) = dim(W) + dim(W^\perp)\).

证明:

(a) 先 extend,然后用 Gram–Schmidt process.

(b) 显然\(S_1 \subseteq W^\perp\), 只需证\(span(S_1) = W^\perp\). \(\forall x = \sum_{i = 1}^{n}a_iv_i \in W^\perp, \langle x, v_i\rangle = 0\) for \(1 \le i \le k\), 所以\(x = \sum_{i = k + 1}^{n}a_iv_i \in span(S_1).\)

(c) 由(b)显然。

6.3 The Adjoint of A Linear Operator

Theorem 6.8.

Let \(V\) be a finite-dimensional inner product space over \(F\), and let \(g: V → F\) be a linear transformation. Then there exists a unique vector \(y ∈ V\) such that \(g(x) = ⟨x, y⟩\) for all \(x ∈ V\).

Let \(\beta=\left\{v_{1}, v_{2}, \dots, v_{n}\right\}\) be an orthonormal basis for V, then

\[y = \sum_{i=1}^n\overline{g(v_i)}v_i. \]

证明:

先证存在,直接令\(y = \sum_{i=1}^n\overline{g(v_i)}v_i,\) 可以计算出\(\forall 1 \le j \le n, \langle v_j, y\rangle = \langle v_j, \sum_{i=1}^n\overline{g(v_i)}v_i\rangle = \sum_{i=1}^n g(v_i)\langle v_j, v_i\rangle = g(v_j).\) 根据\(g\)是 linear transformation, \(g(x) = \langle x, y \rangle.\)

再证唯一,假设\(\forall x\)\(\langle x, y'\rangle = g(x) = \langle x, y\rangle\),那么由于\(x\)的任意性,\(y' = y\)(Theorem 6.1 (e))。

Theorem 6.8 为 \(T^*\)的定义做了准备工作,只有证明了\(y\)的唯一性,才能定义出一个映射。

Theorem 6.9 (Definition of adjoint).

Let \(V\) be a finite-dimensional inner product space, and let \(T\) be a linear operator on \(V\). Then there exists a unique function \(T^*: V → V\) such that \(⟨T(x), y⟩ = ⟨x, T^*(y)⟩\) for all \(x, y ∈ V\). Furthermore, \(T^*\) is linear. \(T^*\) is called the adjoint of \(T\).

证明:

先证唯一存在,\(\forall y \in V\), 定义\(g(x) = \langle T(x), y \rangle\), 则根据 Theorem 6.9, 存在唯一\(z \in V\)使得\(\forall x \in V\)\(g(x) = \langle x, z \rangle\), 定义\(T^*(y) = z\), 即有\(\langle T(x), y\rangle = \langle x, T^*(y)\rangle.\)

然后证linear,这个根据 inner product 的 conjugate linear 性质可以容易地写出。

Theorem 6.10.

Let V be a finite-dimensional inner product space, and let \(β\) be an orthonormal basis for \(V\). If \(T\) is a linear operator on \(V\), then \(\left[\mathrm{T}^{*}\right]_{\beta}=[\mathrm{T}]_{\beta}^{*}.\)

证明: \(B_{i j}=\left\langle\mathrm{T}^{*}\left(v_{j}\right), v_{i}\right\rangle=\overline{\left\langle v_{i}, \mathrm{T}^{*}\left(v_{j}\right)\right\rangle}=\overline{\left\langle\mathrm{T}\left(v_{i}\right), v_{j}\right\rangle}=\overline{A_{j i}}=\left(A^{*}\right)_{i j}\)

Corollary.

Let \(A\) be an \(n × n\) matrix. Then \(L_{A^*} = (L_A)^*\).

Theorem 6.11.

Let \(V\) be an inner product space, and let \(T\) and \(U\) be linear operators on \(V\). Then

\[\begin{array}{l}{\text { (a) }(\mathrm{T}+\mathrm{U})^{*}=\mathrm{T}^{*}+\mathrm{U}^{*}} \\ {\text { (b) }(c \mathrm{T})^{*}=\bar{c} \mathrm{T}^{*} \text { for any } c \in F} \\ {\text { (c) }(\mathrm{TU})^{*}=\mathrm{U}^{*} \mathrm{T}^{*} ;} \\ {\text { (d) } \mathrm{T}^{* *}=\mathrm{T}} \\ {\text { (e) } \mathrm{I}^{*}=\mathrm{I}}\end{array} \]

Corollary.

以上对 linear operator 的性质,对矩阵也成立。

Let \(A\) and \(B\) be \(n × n\) matrices. Then

\[\begin{array}{l}{\text { (a) }(A+B)^{*}=A^{*}+B^{*}} \\ {\text { (b) }(c A)^{*}=\bar{c} A^{*} \text { for all } c \in F} \\ {\text { (c) }(A B)^{*}=B^{*} A^{*}} \\ {\text { (d) } A^{* *}=A} \\ {\text { (e) } I^{*}=I}\end{array} \]

这些性质的证明既可以转化为左乘、用linear operator的性质做,也可以直接用矩阵adjoint的定义。

Lemma 1.

Let \(A \in \mathbb{M}_{m \times n}(F), x \in F^{n},\) and \(y \in F^{m}\). Then

\[\langle A x, y\rangle_{m}=\left\langle x, A^{*} y\right\rangle_{n}. \]

证明:直接用标准$ F^n \(向量内积的定义。\)\(\langle A x, y\rangle_{m}=y^{*}(A x)=\left(y^{*} A\right) x=\left(A^{*} y\right)^{*} x=\left\langle x, A^{*} y\right\rangle_{n}.\)$

Lemma 2.

Let \(A \in \mathbb{M}_{m \times n}(F)\). Then \(rank(A^*A) = rank(A)\).

证明: 可证\(N(L_{A^*A}) = N(L_A)\),即\(A^*Ax = 0 \Leftrightarrow Ax = 0.\) 右推左显然成立,下证左推右。若\(A^*Ax = 0\), 则$0 = x*AAx = (Ax)^Ax = \langle Ax, Ax \rangle, $ 所以\(Ax = 0\).

Corollary.

If \(A\) is an \(m \times n\) matrix such that \(rank(A) = n\), then \(A^*A\) is invertible.

Theorem 6.12 (Least Squares Approximation,最小二乘法) .

Let \(A ∈ M_{m×n} (F)\) and \(y ∈ F^m\) . Then there exists \(x_0 ∈ F^n\) such that \((A^*A)x_0 = A^*y\) and \(∥Ax_0 −y∥ ≤ ∥Ax−y∥\) for all \(x ∈ F^n\). Furthermore, if \(rank(A) = n\), then \(x_0 = (A^*A)^{−1}A^*y\).

证明:

\(Ax \in R(A)\), 而在\(R(A)\)中存在唯一的离\(y\)最近的向量\(Ax_0\),这里的\(x_0\)即为所求。由 Theorem 6.6, \(Ax_0 - y \in R(A)^\perp.\) 现在求\(R(A)^\perp\)。若\(z \in R(A)^\perp,\) \(\forall x \in V,\)\(\langle A^*z, x \rangle = \langle z, Ax \rangle = 0\)。由于\(x\)任意性,\(A^*z = 0\),即\( z \in N(A^*).\) 反过来亦可推出若\(z \in N(A^*)\)则有\(z \in R(A)^\perp.\) 所以\(R(A)^\perp = N(A^*).\) 因为\(Ax_0 - y \in R(A)^\perp = N(A^*)\),所以有\(A^*(Ax_0 - y) = 0\), 若\(rank(A) = n\),则有\(x_0 = (A^*A)^{−1}A^*y.\)

Theorem 6.13 (Minimal Solution to Systems of Linear Equations,线性方程组的最小解)

A solution s to \(Ax = b\) is called a minimal solution if \(∥s∥ ≤ ∥u∥\) for all other solutions \(u\).

Let \(A \in \mathbb{M}_{m \times n}(F)\) and \(b ∈ F^m\). Suppose that \(Ax = b\) is consistent. Then the following statements are true.

(a) There exists exactly one minimal solution \(s\) of \(Ax = b\), and \(s ∈ R(L_{A^*})\).

(b) The vector s is the only solution to \(Ax = b\) that lies in \(R(L_{A^*})\); that is, if u satisfies \(\left(A A^{*}\right) u=b\), then \(s = A^*u\).

证明(a):对于任意解\(x\),可将\(x\)分解为\(x = s + y\),其中\(y \in N(A), s \in N(A)^\perp = R(A^*).\) \(Ax = As + Ay = As + 0 = As\), 所以\(s\)也是\(Ax = b\)的解;而由于\(\langle s, y \rangle = 0\),有\(\|x\| = \|s + y\| = \sqrt{\|s\|^2 + \|y\|^2} \ge \|s\|\),当且仅当\(y = 0\)\(x = s\)时取等,所以s是唯一最小解.

证明(b):假设\(R(L_{A^*})\)中存在另一解\(v\),则\(s - v \in N(A) \cap N(A)^\perp = {0}\), 所以\(s =v\).

6.4 Normal And Self-Adjoint Operators

Lemma.

Let \(T\) be a linear operator on a finite-dimensional inner product space \(V\). If \(T\) has an eigenvector, then so does \(T^*\).

证明:设\(v\)\(T\)的y一个eigenvector,则\(0 = \langle0, x\rangle = \langle(T - \lambda I)(v), x\rangle = \langle v, (T - \lambda I)^*(x)\rangle = \langle v, (T^*-\overline\lambda I)(x)\rangle\)

所以\(R(T^*-\overline\lambda I) = {v}^\perp \subsetneqq V\), 则\(N(T^*-\overline\lambda I) \neq\{0\}\)中的任意vector都是对应eigenvalue为\(\overline\lambda\)的eigenvector。

Theorem 6.14 (Schur).

Let \(T\) be a linear operator on a finite-dimensional inner product space \(V\). Suppose that the characteristic polynomial of \(T\) splits. Then there exists an orthonormal basis \(β\) for \(V\) such that the matrix \([T]_β\) is upper triangular.

证明:

用数学归纳法。设\(dim(V) = n\)\(n=1\)时显然成立。假设对\(n-1\)成立,则对n,由于特征多项式split,\(T\)有至少一个eigenvalue,则\(T^*\)也有至少一个eigenvalue,设为\(\lambda\),它对应至少一个unit vector \(z\). 设\(W = span(\{z\})\),下证\(W^\perp\)是T-invariant.

\(\forall y \in W^\perp, x = cz \in W\) where \(c \in F\)\(\langle T(y), x = cz \rangle = \langle y, T^*(cz) \rangle = \overline{c\lambda}\langle y, z \rangle = 0.\) 所以\(T(y) \in W^\perp.\) 所以\(W^\perp\)是T-invariant,可以定义\(T_{W^\perp}\),而\(dim(W^\perp) = n - 1\),应用假设可得存在一个\(W^\perp\)的orthonormal basis \(\gamma\) 使得\([T_{W^\perp}]_\gamma\)是上三角矩阵,显然\(z\)垂直于\(\gamma\)中的每个向量,令\(\beta = \gamma \cup \{z\}\),则\(\beta\)是orthonormal basis, 且\([T]_β\)也是上三角矩阵。

Definitions (normal).

Let \(V\) be an inner product space, and let \(T\) be a linear operator on \(V\). We say that \(T\) is normal if \(TT^*= T^*T\). An \(n×n\) real or complex matrix \(A\) is normal if \(AA^* = A^*A.\)

Theorem 6.15.

Let \(V\) be an inner product space, and let \(T\) be a normal operator on \(V\). Then the following statements are true.

(a) \(\|\mathrm{T}(x)\|=\left\|\mathrm{T}^{*}(x)\right\|\) for all \(x \in V\).

(b) \(T−cI\) isnormal for every \(c∈F\).

(c) If \(x\) is an eigenvector of \(T\), then \(x\) is also an eigenvector of \(T^*\). In fact, if \(T(x) = λx\), then \(T^*(x) = \overline λx\).

(d) If \(λ_1\) and \(λ_2\) are distinct eigenvalues of \(T\) with corresponding eigenvectors \(x_1\) and \(x_2\), then \(x_1\) and \(x_2\) are orthogonal.

证明:

(c) 令 \(U = T - \lambda I\),则\(U^* = T^* - \overline\lambda I\),且根据(b),\(U\)也normal。根据(a)有\(0 \|U(x)\| = \|U^*(x)\| = \|T^*(x) - \overline\lambda x\|\)

(d) \(\lambda_{1}\left\langle x_{1}, x_{2}\right\rangle=\left\langle\lambda_{1} x_{1}, x_{2}\right\rangle=\left\langle T\left(x_{1}\right), x_{2}\right\rangle=\left\langle x_{1}, T^{*}\left(x_{2}\right)\right\rangle =\left\langle x_{1}, \overline{\lambda_{2}} x_{2}\right\rangle=\lambda_{2}\left\langle x_{1}, x_{2}\right\rangle\)

由于\(\lambda_1 \neq \lambda_2\), 有\(\left\langle x_{1}, x_{2}\right\rangle=0\).

Theorem 6.16.

Let \(T\) be a linear operator on a finite-dimensional complex inner product space \(V\). Then \(T\) is normal if and only if there exists an orthonormal basis for \(V\) consisting of eigenvectors of T.

证明:

假设\(T\) normal,则根据代数基本定理,\(T\)的特征多项式在复数域上split;根据Schur定理,存在一个\(V\)的 orthonormal basis \(\beta = \{v_1, v_2, \ldots, v_n\}\)使得\([T]_\beta\)是上三角矩阵。则\(v_1\)一定是eigenvalue。假设\(v_1, v_2, \ldots, v_{k-1}\) 都是eigenvalue,那么\(1 \le j < k\), \(A_{jk} = \langle T(v_k) ,v_j\rangle = \langle v_k, T^*(v_j)\rangle = \langle v_k, \overline{\lambda_j} v_j\rangle = 0\),所以\(v_k\)也是eigenvalue。

假设存在全是eigenvector的orthonormal basis \(\beta\),则\([T]_{\beta}\)\([T^*]_\beta\)都是对角矩阵,满足交换律,所以\(T\) is normal​.

注意:

对于复数域上的向量空间,normal => diagonalizable;

然而对于实数域上的向量空间则不一定,例如旋转矩阵。

Definition (self-adjoint).

Let T be a linear operator on an inner product space V. We say that T is self-adjoint (Hermitian) if T = \(T^*\). An \(n × n\) real or complex matrix \(A\) is self-adjoint (Hermitian) if \(A = A^*\).

Lemma.

Let T be a self-adjoint operator on a finite-dimensional inner product space V. Then

(a) Every eigenvalue of T is real.
(b) Suppose that V is a real inner product space. Then the characteristic polynomial of T splits.

证明:

(a) \(\lambda x = T(x) = T^*(x) = \overline\lambda x\)

Theorem 6.17.

Let \(T\) be a linear operator on a finite-dimensional real inner product space $$V$$. Then \(T\) is self-adjoint if and only if there exists an orthonormal basis \(β\) for \(V\) consisting of eigenvectors of \(T\).

证明:假设T是self-adjoint, 根据Shur定理,找一组orthonormal basis \(\beta\)使\([T]_\beta = A\)是上三角矩阵,而\(A^* = [T]^*_\beta = [T^*]_\beta = [T]_\beta = A\). 所以\(A^*\)也是上三角矩阵,所以A是对角矩阵。

6.5 Unitary & Orthogonal Operators and their Matrices

Definitions (unitary/orthogonal operator, isometry).

If \(\|T(x)\| = \|x\|\) for all \(x \in V\), we call T a unitary operator if \(F = C\) and an orthogonal operator if \(F = R\).

In the infinite-dimensional case, an operator satisfying \(\|T(x)\| = \|x\|\) for all \(x \in V\) is called an isometry.

Lemma.

Let U be an self-adjoint operator on a finite-dimensional inner product space V. If \(\langle x, U(x)\rangle = 0\) for all \(x \in V\), then \(U = T_0\).

证明:若\(U(x) = \lambda x\), 则\(0 = \langle x, U(x)\rangle = \lambda\langle x, x\rangle\).

Theorem 6.18:

4 equivalent statements about unitary/othogonal operators:

Let T be a linear operator on a finite-dimensional inner product space V. Then the following statements are equivalent.

(a) \(TT^* = T^*T = I\).

(b) \(\langle T(x), T(y)\rangle = \langle x, y\rangle\) for all \(x, y \in V\).

(c) If \(\beta\) is an orthonormal basis for V, then \(T(\beta)\) is an orthonormal basis for \(V\).

(d) \(\|T(x)\| = \|x\|\) for all \(x \in V\).

证明:

(a) -> (b): \(\langle T(x), T(y)\rangle = \langle x, T^*T(y)\rangle = \langle x, y\rangle\)

(b) -> (c) 显然

(c) -> (d): \(\|x\|^2 = \langle \sum_{i=1}^na_iv_i, \sum_{j=1}^na_jv_j\rangle = \sum_{i=1}^n\sum_{j=1}^na_i\overline{a_j}\langle v_i, v_j\rangle = \sum_{i=1}^n\sum_{j=1}^na_i\overline{a_j}\delta_{ij} = \sum_{i=1}^n\|a_i\|^2\)

(d) -> (a): 将x在\(\beta\)下表示,T(x)在\(T(\beta)\)下表示,可得二者范数相同。

Corollary 1 & 2:

\(|\lambda| = 1 \Leftrightarrow\) orthonormal/unitary.

Let T be a linear operator on a finite-dimensional complex [real] inner product V. Then V has an orthonormal basis of eigenvectors of T with corresponding eigenvalues of absolute value 1 if and only if T is both unitary[self-adjoint and orthogonal].

Definition (unitarily equivalent).

\(A = P^*BP\) where \(P\) is unitary. (The definition of orthogonally equivalent is similar.)

Theorem 6.19 & 6.20.

Complex[real] matrix \(A\) is normal[symmetrix] iff \(A\) is unitarily equivalent to a complex[real] diagonal matrix.

证明(Complex):

If \(A = P^*DP\), then $$AA* = (P*DP)(PD^P) = P*DDP = P*DDP = P*DPP^DP = A^*A$$.

充分性前文已证。

Theorem 6.21 (Schur) 舒尔定理的矩阵表示.

\(A\in M_{n\times n}(F)\), and the characteristic polynomial of \(A\) splits. If F = C[R], then A is unitarily[orthogonally] equivalent to a complex[real] upper triangular matrix.

6.6 Orthogonal Projections & Spectral Theorem

Definition (orthogonal projection)

We say \(T\) is an orthogonal projection if \(R(T)^\perp = N(T)\) and \(N(T)^\perp = R(T)\).

Theorem 6.24.

Let V be an inner product space, and let T be a linear operator on V. Then T is an orthogonal projection iff T has an adjoint \(T^*\) and \(T^2=T=T^*\).

Theorem 6.25 (The Spectral Theorem).

T is a linear operator on a finite-dimensional inner product space V over F with the distinct eigenvalues \(\lambda_1,\lambda_2,\ldots,\lambda_k\). Assume that T is normal is F = C and that T is self-adjoint if F = R. For each \(i(1 \le i \le k)\), Let \(W_i\) be the eigenspace of T corresponding to the eigenvalue \(\lambda_i\), and let \(T_i\) be the orthogonal projection of \(V\) on \(W_i\). Then:

(a) \(V = W_1\oplus W_2 \oplus \ldots \oplus W_k\).

(b) if \(W_i'\) denotes the direct sum of the subspaces \(W_j\) for \(j \neq i\), then \(W_i^\perp = W_i'\).

(c) \(T_iT_j = \delta_{ij}T_i\) for \(1 \le i, j \le k\).

(d) \(I = T_1 + T_2 + \ldots + T_k\).

(e) \(T = \lambda_1T_1 + \lambda_2T_2 + \ldots + \lambda_kT_k\).

The set \(\{\lambda_1, \lambda_2, \dots, \lambda_k\}\) is called the spectrum of \(T\), the sum \(I = T_1 + T_2 + \ldots + T_k\) is called the resolution of the identity operator induced by \(T\), and the sum \(T = \lambda_1T_1 + \lambda_2T_2 + \ldots + \lambda_kT_k\) is called the spectral decomposition of T.

\[[T]_\beta = \begin{pmatrix}\ \lambda_1I_{m_1}&O&\cdots&O\\ O&\lambda_2I_{m_2}&\cdots&O \\ \vdots&\vdots&&\vdots\\ O&O&\cdots&\lambda_kI_{m_k}\end{pmatrix} \]

Corollary 1.

If F = C, then T is normal iff \(T^* = g(T)\) for some polynomial \(g\).

Corollary 2.

If F = C, then T is unitary iff T is normal and all \(|\lambda| = 1\).

Corollary 3.

If F = C and T is normal, then T is self-adjoint iff every eigenvalue of T is real.

Corollary 4.

Let T be as is the spectral theorem with spectral decomposition \(T = \lambda_1T_1 + \lambda_2T_2 + \ldots + \lambda_kT_k\), Then each \(T_j\) is a polynomiao in T.

posted @ 2019-11-25 15:13  胡小兔  阅读(1389)  评论(5编辑  收藏