线性空间

Vector Space

1.1 vector space

A vector space is a set (denoted by \(V\)) along with an addition and a scalar multiplication satisfying the following properties.

\[\begin{aligned} \forall &u,v,w \in V, a,b \in F \\\\&(1)u+v = v+u \\\\&(2)(u+v)+w = u+(v+w) \\\\ &(3)\exists \enspace 0, st\enspace 0+u = 0 \\\\ &(4)\exists \enspace h, st\enspace h+u = 0 \\\\ &(5)\exists \enspace 1, st\enspace 1u=u \\\\ &(6)a(u+v)=au+av,(a+b)u=au+bu \end{aligned}\]

1.2 \(F^S\)

\(F^S\) denotes a function from S to F, such as \(F^2\),$F^ $, $ R^{[1,2]}$, and \(F^n\) also can be expressed by \(\\{x_1,x_2,......, x_n:x_i \in F \enspace for \enspace i =1,2,3,...,n \\}\)

the scalar multiplication and addition are defined by:

\[\begin{aligned} &f,g \in F^S, x \in S \\\\&(f+g)(x)=f(x)+g(x) \\\\&(\lambda g)(x)= \lambda g(x) \end{aligned}\]

1.3 unique additive identity and unique additive inverse

prove:

(zero here means vector 0)

\[\begin{aligned} & suppose \enspace \exists 0,0',\forall v \in V, \\\\&0= v+(-v)=(0'+ v )+(-v)=0'+(v+-v)=0'+0=0' \\\\\&therefore \enspace 0=0' \\\\&suppose \enspace \exists w,w',st \enspace w+v=0,w'+v=0 \\\\&w = 0 + w=(w'+v)+w=w' \\\\&\therefore w = w' \end{aligned}\]

1.4 vector 0 and number 0

\[\begin{aligned} 0v&=(0+0)v=0v+0v \\\\ \therefore 0v &= 0 \\\\ a0_v&=a(v-v)=av -av=0_v \end{aligned}\]

1.5 -1 times v

\[\begin{aligned} 0_v&=(1+(-1))v=v+(-1)v \\\\0_v&=v+(-v) \\\\\therefore -1v& = -v \end{aligned}\]

1.6 subspace

A subset of V is a subspace if it inherites addition and multiplication defined on V and it is a vector space.

1.7 conditions for a subspace

U is a subspace of V if and only if:

\[\begin{aligned} &(1)0_v \in U \\\\ &(2)\forall u,w \in U\Rightarrow u+w \in U \\\\ &(3)\forall a\in F,v\in U, av \in U \end{aligned}\]

I'll deduce \(U\) is a linear space 1.1 from the assumptions above.

1.1(1),(2) is easy to get because U is the subset of V and condition(2) ensures \(v+w \in U\).

1.1(3) is equal to condition(2).

1.1(4): according to condition(3)，we could know \(-v = -1v \in U\) (see 1.5) , which means every element has its own additive inverse in \(U\)

1.1(5) \(1 \in F\)

1.1(6) can be seen as \(U\) is a subset of \(V\) and condition(2)(3) ensures \(av+bw \in U\).

conversely, easy to prove.

1.8 sum of subspaces

\(U_1,U_2,......,U_n\) are subsets of V, define sum of these subspaces by:

\[\begin{aligned} U_1+U_2......+U_n = \\{u_1+u_2+u_3...+u_n:u_i \in U_i,i=1,2,3...,n\\} \begin{aligned} If $U_1,U_2,......,U_n$ are subspaces of V and each element in $U_1+U_2......+U_n$ can only be expressed uniquely in $u_1+u_2+u_3+......+u_n$, we call $U_1+U_2......+U_n$ is a direct sum,we use $U_1 \oplus U_2 \oplus ..... U_n$ to denote it. ## 1.9 condition of direct sum $U_1,U_2,......,U_n$ is subspace of V, $U_1+U_2+U_3...+U_n$ is direct sum if and only if $0$ can only be expressed uniquely in $0_1+0_2+0_3......+0_n$ prove: \begin{aligned} &suppose \enspace 0 = 0_1+0_2+0_3......+0_n \\\\ &if\enspace \exists u = u_1+u_2...+u_n=u_1'+u_2'...u_n' \\\\ &0 = u-u=(u_1'-u_1)+(u_2'-u_2)...+(u_n'-u_n) \end{aligned}\]

Because 0 can only be expressed in one way, \(u_i = u_i'\), which means \(U_1+U_2...+U_n\) is a direct sum

1.10 direct sum of two subspace

\(U_1+U_2\) is a direct sum if and only if \(U_1 \cap U_2 =\\{0\\}\)

prove:

\[\begin{aligned} &if\enspace U_1 \cap U_2 = \{0\} \\\\&0 = u_1 + u_2 \\\\&u_1 \in U_1 u_2 \in U_2 \\\\&\because u_1 = -u_2 \in U_2 \\\\&\therefore u_1 \in U_1 \cap U_2 \\\\&\therefore u_1 = 0 \end{aligned}\]

\(U_1+U_2\) is a direct sum, reversely,

\[\begin{aligned} \\\\&U_1+U_2\enspace is\enspace direct\enspace sum \\\\&\forall u \in U_1 \cap U_2 \\\\ &if\enspace u\ne0 \enspace u = 0_1 +u = u + 0_2 \end{aligned}\]

Contradiction with assumptions, therefore \(u = 0\)

exercise

1.B

1.B.1

\(-(-v)=-(-1v)=-1(-1v)=(-1 \times -1)v=v\)

1.B.2

\(if\enspace a \ne 0,\frac{1}{a}0_v=0_v =(\frac{1}{a}\times av)=v\)

1.B.3

\[\begin{aligned} &suppose\enspace \exists\vec{x}'\enspace st\enspace 3\vec{x}'+\vec{v}=\vec{w} \\\\&3\vec{x}'+\vec{v} = \vec{w}=3\vec{x}+\vec{v} \\\\&\Rightarrow \vec{x}'-\vec{x}= 0 \end{aligned}\]

1.B.5

\[\begin{aligned} \exists0,st\enspace 0_v+u=u \Rightarrow 0v=0_v\enspace for\enspace \forall v \in V \\\\0_v=u-u=(1-1)u=0u \end{aligned}\]

1.B.6

\(R \cup \{\infty\}\cup \{-\infty\}\) is a vector space can be verified by checking the scalar multiplication and addition between infinity and constant.

1.C

1.C.3

\[\begin{aligned} &\forall f,g \in U =\{f:f'(-1)=3f(2),f \in \textbf{R}^{(-4,4)}\} \\\\&(f+g)'(-1)= f'(-1)+g'(-1) = 3f(2)+3g(2)= 3(f+g)(2) \\\\&\therefore f+g \in U \\\\& \\\\&(\lambda f)'(-1) = \lambda f'(-1)=3 (\lambda f)(2) \\\\&\therefore \lambda f \in U \\\\& \\\\& 0'(-1)=3\enspace 0(2)=0 \\\\&\therefore 0\enspace is\enspace in\enspace U \end{aligned}\]

1.C.6

\[\begin{aligned} (a)& a^3=b^3 \Rightarrow a = b\enspace when\enspace a,b \in \mathcal{R} \\\\(b)& (1,e^{\pi i2/3},0)+(e^{\pi i4/3},1,0) \notin U \end{aligned}\]

1.C.7

\(\\{(1,0),(0,0),(0,1)\\}\) is closed under the addition but it is not closed under the scalar multiplication

1.C.8

\((\lambda,2\lambda)\cup(\lambda,3\lambda) \lambda \in R\) closed under the scalar multiplication but it is not closed under the addition

1.C.10

\begin{aligned} u,w U_1 U_2 \u+w ,u U_1 \ u+w,u U_2 \u+w, uU_1 U_2 obviously, 0isinthesubsetU_1 U_2 \begin{aligned}

therefore \(U_1 \cap U_2\) is subspace satisfies the three conditions in 1.7, \(U_1 \cap U_2\) is a subspace.

1.C.11

using the same method in 1.C.10

1.C.12

from 1.C.10,we can infer that \(U_1 \in U_2 \Rightarrow U_1 \cup U_2\) is a subspace,

reversely,

\begin{aligned} U_1 U_2isasubspace \supposev U_1 / U_2 , u U_2 / U_1 (/heremeansquotientofsets) \U_1 U_2isasubspace \v+u U_1 U_2 \supposev+u U_1 \U_1isasubspace \u U_1 \contradicttotheassumption \v+u U_2 \v U_2 \contradicttotheassumption \Thenullhypothesisiswrong \U_1/U_2orU_2/U1isanemptyset \U_2 U_1 or U_1 U_2 \begin{aligned}

1.C.13

if there exists one subspace contains another, from 1.C.12 we can prove it easily.

and in another case, each of the three subspaces contains vector which is not contained by other subspaces:

\begin{aligned} _1 U_1 U_2 U_3 \_2 U_2 U_1 U_3 \_3 U_3 U_1 U_2 \begin{aligned}

\(u_1 + u_2 \in U_1 \cup U_2 \cup U_3\), using the samemethodin 1.C.12 ,\(u_1 +u_2 \in U_3 \notin U_2\enspace and\enspace \notin U_1\) can be deduced. and also \(u_1 +2u_2 \in U_3 \notin U_2\enspace and\enspace \notin U_1\), because \(U_3\) is a subspace of V, \(u_2 = (u_1+2u_2)-(u_1+u_2) \in U_3\), it is contradict to the assumption. so this case should be discarded.

1.C.18

It is an example set which contains 0 but does not have addictive inverses.

1.C.19 - 22

19: \(W=\{(x,y,0):x,y \in \R\},\enspace U_1=\{(0,y,z):y,z \in \R\},\enspace U_2=\{(0,0,z):z \in \R\}\) 20:\(\{(0,x,y,0):x,y \in F\}\) 21:\(\{(0,0,x,0,0):x \in F\},\{(0,0,0,x,0):x \in F\},\{(0,0,0,0,x):x \in F\}\)

1.C.23

\begin{aligned} prove: \V = U_1 W \vU_1, v Vandv W \alsoV =U_2W \v U_2 \U_1 U_2 \reverselyU_2 U_1 \U_2 = U_1 \begin{aligned}

1.C.24

the function is odd and even:

\(f(-x) =f(x)=-f(x) \Rightarrow f(x) = 0\\\therefore U_e \cup U_o = {0(x)} \Rightarrow U_e \oplus U_o\)

for all function \(g(x)\) in \(\R^\R\), can be expressed by \(w(x)+v(x):w(x)=\frac{g(x)+g(-x)}{2},v(x)=\frac{g(x)-g(-x)}{2}\), where \(w(x)\) is an even function, \(v(x)\) is an odd function.