练习:正交补和极小化问题

目录

1 6.C.1

设\(v_{1},\ldots ,v_{m}\in V\),证明\(\{v_{1},\ldots ,v_{m}\}^{\bot}= (\mathrm{span}(v_{1},\ldots ,v_{m}))^{\bot}\)

这个证明依赖于\( \langle v,u \rangle =0 \),则显然\( \langle \lambda v,u \rangle = 0 \)

2 6.C.2

设\(U\)是\(V\)的有限维子空间,证明\(U^{\bot} = \{0\}\),当且仅当\(U=V\)

显然,当\(U=V\)时,\(U^{\bot} = \{0\}\)。

另一方面,当\(U^{\bot} = \{0\}\),又因为\(V = U\oplus U^{\bot}\),所以\(U=V\)

3 6.C.3

设\(U\)是\(V\)的子空间,设\(u_{1},\ldots ,u_{m}\)是\(U\)的基,且\(u_{1},\ldots ,u_{m},w_{1},\ldots ,w_{n}\)是\(V\)的基。证明若对\(V\)的上述基应用格拉姆施密特过程得到组\(e_{1},\ldots ,e_{m},f_{1},\ldots ,f_{n}\),则\(e_{1},\ldots ,e_{m}\)是\(U\)的规范正交基,\(f_{1},\ldots ,f_{n}\)是\(U^{\bot}\)的规范正交基。

格拉姆施密特过程: 设\(v_{1},\ldots ,v_{m}\)是\(V\)中线性无关组,设\(e_{1} = v_{1}/ \| v_{1} \| \),对于\(j=2,\ldots ,m\),有:

\begin{equation} \label{eq:1} v_{j} = \frac{ v_{j} - \langle v_{j},e_{1} \rangle e_{1} -\ldots - \langle v_{j},e_{j-1} \rangle e_{j-1} }{ \| v_{j} - \langle v_{j},e_{1} \rangle e_{1} -\ldots - \langle v_{j},e_{j-1} \rangle e_{j-1} \| } \end{equation}

则\(e_{1},\ldots ,e_{m}\)是\(V\)的规范正交基,使得对于\(j=1,\ldots ,m\)有:

\begin{equation} \label{eq:2} \mathrm{span}(v_{1}\ldots ,v_{j}) = \mathrm{span}(e_{1},\ldots ,e_{j}) \end{equation}

根据格拉姆施密特过程,我们有:\( \mathrm{span}(e_{1},\ldots ,e_{m}) = \mathrm{span}(u_{1},\ldots ,u_{m}) = U \)。因为\(e_{1},\ldots ,e_{m}\)是蒸饺向量组,且其维度和\(U\)的基的维度相同,则\(e_{1},\ldots ,e_{m}\)是\(U\)的一组基。又因为\(V = U\oplus U^{\bot}\)。因为\(e_{1},\ldots ,e_{m},f_{1},\ldots ,f_{n}\)是一组正交向量组,所以:

\begin{equation} \label{eq:3} \mathrm{span}(f_{1},\ldots ,f_{n}) \subset U^{\bot} \end{equation}

因为\(\dim V = \dim U + \dim U^{\bot}\),且\(\dim \mathrm{span}(f_{1},\ldots ,f_{n}) = n \)。所以\( \mathrm{span}(f_{1},\ldots ,f_{n}) = U^{\bot} \) 因此\(f_{1},\ldots ,f_{n}\)是\(U^{\bot}\)的一组规范正交基。

4 6.C.4

给定\(\mathbf{R}^{4}\)的子空间:

\begin{equation} \label{eq:4} U = \mathrm{span}((1,2,3,-4),(-5,4,3,2)) \end{equation}

求\(U\)的一个规范正交基和\(U^{\bot}\)的一个规范正交基。

通过对\((1,2,3,-4),(-5,4,3,2)\)执行格拉姆施密特正交化过程。

  1. 令\(u_{1} = (1,2,3,-4)\),则\( \| u_{1} \| = \sqrt{30} = 5.4772 \),所以\[e_{1} = u_{1}/ \| u_{1} \|= (0.1826,0.3652,0.5477,-0.7303) \]
  2. 令\(u_{2} = (-5,4,3,2)\),则\(e_{2} = \frac{u_{2} - \langle u_{2},e_{1} \rangle e_{1} }{ \| u_{2} - \langle u_{2},e_{1} \rangle e_{1} \| } = (-0.7020, 0.5106,0.3556,0.3465) \)

然后我在\(e_{1},e_{2}\)基础上添加两个向量\(w_{1} = (0,0,1,0),w_{2} = (0,0,0,1)\),然后对此进行格拉姆施密特计算:得\(f_{1} = (0,1976,-0.5038,0.7573,0.3655)\); \(f_{2} = (0.6594,0.5935,0.0000,0.4615)\)

这个计算过程相当繁琐,即使使用matlab来完成也显得不怎么简洁,更遑论使用手工计算。要区分哪些是自己能做的,哪些是计算机可以代劳的。

5 6.C.5

设\(V\)是有限维的且\(U\)是\(V\)的子空间。证明\(P_{U^{\bot}} = I - P_{U}\),这里\(I\)是\(V\)上的恒等算子。

证明两个算子相等,一种做法是对于\(V\)内的任意元素\(v\),都有\(P_{U^{\bot}}(v) = (I - P_{U})(v)\)。我们令\(v = u + w\),其中\(u\in U,w\in U^{\bot}\),则有\(P_{U^{\bot}}(v) = w\),又因为\(P_{U}(v) = u=v-w\),得证。

6 6.C.6

设\(U\)和\(W\)均为\(V\)的有限维子空间。证明\(P_{U}P_{W} = 0\)当且仅当对所有\(u\in U\),\(w\in W\)均有\( \langle u,w \rangle = 0 \)

假设对所有\(u\in U\),\(w\in W\)均有\( \langle u,w \rangle = 0 \),则对于\(v=u+w+x,u\in U,w\in W,x\in V-U-W\)有\(P_{W}(v) = w\),根据已知\(w\in U^{\bot}\),所以\(P_{u}w = 0\),即\(P_{U}P_{W}v = 0\),由于\(v\)的任意性,则\(P_{U}P_{W} = 0\)

放过来,假设\(P_{U}P_{W} = 0\),则对于任意的\(v = u+w+x,u\in U,w\in W,x\in V-U-W\),都有\(P_{U}P_{W}v = 0\),显然\(P_{W}v = w\),即有\(P_{U}w = 0\),即\(w\)在\(U\)中的投影是\(0\)则有对\(u\in U\)都有\( \langle u,w \rangle = 0\)

7 6.C.7

设\(V\)是有限维的,\(P\in \mathcal{L}(V)\),使得\(P^{2} = P\),且\(\mathrm{null} P \)中的向量与\( \mathrm{range}P \)中的向量都正交,证明有\(V\)的子空间\(U\)使得\(P = P_{U}\)

令\(v\in V\)且可以写成\(v= Pv + (v-Pv)\)。令\(U = \mathrm{range}P\),则有:\(Pv \in U\),另一方面\(P(v-Pv) = Pv - P^{2}v = 0\),所以\(v-P{v}\in \mathrm{null}P\). 又因为\(nullP\)中的向量和\(range P\)中的向量互相垂直,则\(v-Pv \in U^{\bot}\)。所以\(Pv = P_{U}v\),即\(P=P_{U}\)

8 6.C.11

在\(\mathbf{R}^{4}\)中设\(U= \mathrm{span}((1,1,0,0),(1,1,1,2))\),求\(u\in U\)使得\( \| u - (1,2,3,4) \| \)最小。

这个问题显然是要求:\((1,2,3,4)\)到\(U\)的投影。这是一个极小化问题。

我们知道根据\(V = U\oplus U^{\bot}\)可以把\(v\)分解为\(v = u + w,u\in U,w\in U^{\bot}\). \(v\)在\(U\)中的投影可以表示为\(u = \langle v,e_{1} \rangle e_{1} + \ldots + \langle v,e_{m} \rangle e_{m} \),其中\(e_{1},\ldots ,e_{m}\)是\(U\)的规范正交基。

我们首先把\((1,1,0,0),(1,1,1,2)\)进行规范正交化。得到:\(e1 = (0.7071, 0.7071, 0,0),e2 = (0,0,0.4472,0.8944)\)。

然后我们得到\(v =(1,2,3,4)\)在\(e_{1},e_{2}\)上的投影。\( \langle v,e_{1} \rangle e_{1} = (1.5,1.5,0,0) \)和\( \langle v,e_{2} \rangle e_{2} = (0,0,2.2,2.4)\)。

然后我们得到\(v\)在\(U\)上的投影\( (1.5,1.5,2.2,2.4) \)

9 6.C.12

求\(p\in \mathcal{P}_{3}(\mathbf{R})\)使得\(p(0) =0,p^{'}(0) =0\),而且:

\begin{equation} \label{eq:5} \int_{0}^{1} |2+3x-p(x)|^{2}dx \end{equation}

最小。

这是个极小化问题。且\(U = span(x^{2},x^{3})\),\(v = 2+3x\),我们要求\(v\)向\(U\)的投影。按照顺序:

  1. 对\(x^{2},x^{3}\)对内积\(\int_{0}^{1}f(x)g(x)dx\)进行规范正交化\(e_{1},e_{2}\)。
  2. 求\(2+3x\)在这个规范正交化基上的投影。\( \langle v,e_{1} \rangle e_{1} \) 和\( \langle v,e_{2} \rangle e_{2} \)
  3. 求\(2+3x\)在\(U\)上的投影\( P_{U}v = \langle v,e_{1} \rangle e_{1} + \langle v,e_{2} \rangle e_{2} \)