Subspace本身并不是一个很难的概念，就是一个自己也是space的子集合。其检验可以简化为Theorem 1， 即 $\alpha,\beta \in W,c\in F\Leftrightarrow c\alpha+\beta\in W$， 在很多其他的书中，这个定理用于subspace的定义。Example 6 中给出了一些例子，包括zero subspace，symmetric，Hermitan（self-adjoint）空间等。Example 7 说明homogeneous system的解的集合是一个subspace，并且还推广了一个矩阵的带数乘的分配律，即 $A(dB+C)=d(AB)+AC$，当然里面矩阵的乘法要有意义。
从Theorem 2开始，用一个很新的方式看待某一类特殊的subspace，即subspace spanned by S，Theorem 2首先说明，V的任意subspace的交集还是一个subspace，由此可知，包含S的subspace有一个最小的subspace（不断地取S的subspace的交集），并因此将subspace spanned by S定义出来：所有包含S的subspace的交。这一定义的好处是允许S有无限元素（实际上是无限维元素），当S只包含有限个vector时，subspace spanned by S是传统上大家比较熟悉的样子。Theorem 3即阐述了：subspace spanned by S实际上是S中vector所有的linear combination的集合，注意这里同样没有限制S是有限的，这就是这一种subspace定义方式的好处。
接下来有一个新的定义即集合的和或者说subspace的和，实质是每个集合中任取一个vector并相加而组成的集合。如果 $W_1,W_2,\dots,W_k$是 $V$的subspace，那么 $W=W_1+W_2+\dots+W_k$也是一个subspace（按定义可证），并且 $W=\text{subspace spanned by }W_1\cup\dots \cup W_k$，因为采用Theorem 3的类似证法，任意包含 $W_1\cup\dots \cup W_k$的subspace都会包含 $W$。
最后是几个例子，其中Example 10和11比较精彩。Example 10提出了row-space的概念，其是矩阵的row vectors spanned的subspace，为未来引入矩阵的rank打下基础。Example 11是一个无限维的subspace的例子，并且为第四章介绍多项式打下基础。

Exercises

1. Which of the following sets of vectors $\alpha =(a_1,\dots,a_n)$ in $R^n$ are subspaces of $R^n(n\geq3)$?

( a ) all $\alpha$ such that $a_1\geq0$;
( b ) all $\alpha$ such that $a_1+3a_2=a_3$;
( c ) all $\alpha$ such that $a_2=a_1^2$;
( d ) all $\alpha$ such that $a_1a_2=0$;
( e ) all $\alpha$ such that $a_2$ is rational.
Solution:
(a) No, since $(1,0,\dots,0)$ satisfies, but $-(1,0,\dots,0)=(-1,0,\dots,0)$ doesn’t.
(b) Yes, for $\alpha=(a_1,\dots,a_n ),\beta=(b_1,\dots,b_n)$, if $a_1+3a_2=a_3,b_1+3b_2=b_3$, then any $c\in R$, the vector $c\alpha+\beta=(ca_1+b_1,\dots,ca_n+b_n)$ has the property
$(ca_1+b_1 )+3(ca_2+b_2 )=c(a_1+3a_2 )+b_1+3b_2=ca_3+b_3$
( c ) No, since $(1,1,\dots,0)$ satisfies, but $2(1,1,\dots,0)=(2,2,\dots,0)$ doesn’t.
(d) No, since $(1,0,\dots,0)$ and $(0,1,\dots,0)$ satisfies, but $(1,1,\dots,0)=(1,0,\dots,0)+(0,1,\dots,0)$ doesn’t.
(e) No, since $(0,1,\dots,0)$ satisfies, but $\sqrt{2} (0,1,\dots,0)=(0,\sqrt{2},\dots,0)$ doesn’t.

2. Let $V$ be the (real) vector space of all functions $f$ from $R$ into $R$. Which of the following sets of functions are subspaces of $V$?

( a ) all $f$ such that $f(x^2)=f(x)^2$;
( b ) all $f$ such that $f(0)=f(1)$;
( c ) all $f$ such that $f(3)=1+f(-5)$;
( d ) all $f$ such that $f(-1)=0$;
( e ) all $f$ which are continuous.
Solution:
(a) No, consider $f(x)=1$ if $x\geq 0$ and $f(x)=-1$ if $x<0$, then $f(x^2 )=1=f(x)^2$, but $2f(x)$ doesn’t satisfy $(2f)(x^2 )=(2f(x))^2$ since $(2f)(x^2 )=2$, while $(2f(x))^2=4$.
(b) Yes, if $f,g$ are functions s.t. $f(0)=f(1),g(0)=g(1)$, then
$(cf+g)(0)=(cf)(0)+g(0)=cf(0)+g(0)=cf(1)+g(1)=(cf+g)(1),\quad \forall c\in R$
( c ) No, if f satisfies $f(3)=1+f(-5)$, then $2f$ doesn’t, since
$(2f)(3)=2f(3)=2+2f(-5)=2+(2f)(-5)\neq 1+(2f)(-5)$
(d) Yes, if if $f,g$ are functions s.t. $f(-1)=g(-1)=0$, then
$(cf+g)(-1)=cf(-1)+g(-1)=0+0=0,\quad \forall c\in R$
(e) Yes, since $cf+g$ is continuous for any continuous functions $f,g$.

3.Is the vector $(3,-1,0,-1)$ in the subspace of $R^5$ spanned by the vectors $(2,-1,3,2),(-1,1,1,-3)$ and $(1,1,9,-5)$?

Solution: It’s equivalent to determine whether the system of equations
$\begin{cases}2x_1-x_2+x_3=3\\-x_1+x_2+x_3=-1\\3x_1+x_2+9x_3=0\\2x_1-3x_2-5x_3=-1\end{cases}$
has solutions or not. Since
$A'=\begin{bmatrix}2&-1&1&3\\-1&1&1&-1\\3&1&9&0\\2&-3&-5&-1\end{bmatrix}\rightarrow\begin{bmatrix}-1&1&1&-1\\0&1&3&1\\0&4&13&-3\\0&-1&-3&-3\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&-1&1\\0&1&3&1\\0&4&13&-3\\0&0&0&-4\end{bmatrix}$
this system of equations has no solution, thus the answer is no.

4. Let $W$ be the set of all $(x_1,x_2,x_3,x_4,x_5)$ in $R^5$ which satisfy

$\begin{array}{c}2x_1-x_2+\frac{4}{3}x_3-x_4&=0\\x_1+\frac{2}{3}x_3-x_5&=0\\9x_1-3x_2+6x_3-3x_4-3x_5&=0\end{array}$

Find a finite set of vectors which spans $W$.

Solution: We have
$\begin{bmatrix}2&-1&4/3&-1&0\\1&0&2/3&0&-1\\9&-3&6&-3&-3\end{bmatrix}\rightarrow\begin{bmatrix}1&0&2/3&0&-1\\0&-1&0&-1&2\\3&-1&2&-1&-1\end{bmatrix}\\ \rightarrow\begin{bmatrix}1&0&2/3&0&-1\\0&1&0&1&-2\\0&-1&0&-1&2\end{bmatrix}\rightarrow\begin{bmatrix}1&0&2/3&0&-1\\0&1&0&1&-2\\0&0&0&0&0\end{bmatrix}$
The solutions to the system can be written as $x_1=x_5-\frac{2}{3} x_3,x_2=2x_5-x_4$. The finite set of vectors which spans $W$ can be found as $(-2,0,3,0,0),(0,-1,0,1,0),(1,2,0,0,1)$

5. Let $F$ be a field and let $N$ be a positive integer $(n\geq2)$. Let $V$ be the vector space of all $n\times n$ matrices over $F$. Which of the following sets of matrices $A$ in $V$ are subspaces of $V$?

( a ) all invertible $A$;
( b ) all non-invertible $A$;
( c ) all $A$ such that $AB=BA$, where $B$ is some fixed matrix in $V$;
( d ) all $A$ such that $A^2=A$.
Solution:
(a) No, since $A=\begin{bmatrix}1&\\&1\end{bmatrix},B=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ are invertible matrices, but $B-A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is not.
(b) No, since $A=\begin{bmatrix}1&0\\0&0\end{bmatrix},B=\begin{bmatrix}0&0\\0&1\end{bmatrix}$ are not invertible, but $A+B=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ is.
( c ) Yes, since for $A,C$ belongs to the set, we have $(dA+C)B=d(AB)+CB=d(BA)+BC=B(dA+C),\forall d\in F$
(d) No, since if $A=\begin{bmatrix}1&0\\0&1\end{bmatrix}$, then $A^2=A$, but $3A=\begin{bmatrix}3&0\\0&3\end{bmatrix}$ and $(3A)^2=9I=3(3A)$.

6. ( a ) Prove that the only subspaces of $R^1$ are $R^1$ and the zero subspace.

( b ) Prove that a subspace of $R^2$ is $R^2$, or the zero subspace, or consists of all scalar multiples of some fixed vector in $R^2$. (The last type of subspace is, intuitively, a straight line throuth the origin.)

( c ) Can you describe the subspace of $R^3$?

Solution:
(a) It’s apparent that $R^1$ and the zero space are subspaces of $R^1$, assume there’s $W\subset R^1$ which is a subspace different from $R^1$ and the zero space, then there’s $a\in W,a\neq 0$ and $b\notin W$, let $k=b/a$, then $b=ka$, which means $b\in W$ since $W$ is a subspace, this is a contradiction.
(b) It’s apparent that $R^2$ and the zero space are subspaces of $R^2$, if $W=\{kv:v\in R^2 \}$, then it’s easy to prove $W$ is a subspace of $R^2$. Assume there’s $U\subset R^2$ which is a subspace, but not of the three types above, then $0\in U$ and $\exists u\in U,\exists v\notin U$, since $U$ is not of the last type, we can find $u'\in U$ s.t. $u'$ is not a scalar multiple of $u$, thus we must be able to find $x_1,x_2\in R$ s.t. $x_1 u+x_2 u'=v$, this means $v\in U$ and thus a contradiction.
( c ) The subspaces of $R^3$ is $R^3$, the zero subspace, all forms of $\{kv+lu:u,v\in R^3,k,l\in R\}$.

7. Let $W_1$ and $W_2$ be subspaces of a vector space $V$ such that the set-theoretic union of $W_1$ and $W_2$ is also a subspace. Prove that one of the spaces $W_i$ is contained in the other.

Solution: Let $W=W_1\cup W_2$, then $W$ is a subspace of $V$, assume $\exists a\in W_1\backslash W_2,b\in W_2\backslash W_1$, then $a,b\in W$, thus $a+b\in W$, which means $(a+b\in W_1 )\vee (a+b\in W_2 )$, but this means
$(b=(a+b)-a\in W_1 )\vee (a=(a+b)-b\in W_2 )$
and either leads to a contradiction, thus either $W_1\backslash W_2=\empty$ or $W_2\backslash W_1=\empty$, the conclusion follows.

8. Let $V$ be the vector space of all functions from $R$ into $R$; let $V_e$ be the subset of even functions, $f(--x)=f(x)$; let $V_o$ be the subset of odd functions, $f(-x)=-f(x)$.

( a ) Prove that $V_e$ and $V_o$ are subspaces of $V$.
( b ) Prove that $V_e+V_o=V$.
( c ) Prove that $V_e\cap V_o=\{0\}$.
Solution:
(a) first let $f,g\in V_e$, then $f$ and $g$ are even functions, for $c\in R$ we have
$(cf+g)(-x)=cf(-x)+g(-x)=cf(x)+g(x)=(cf+g)(x)$
thus $cf+g\in V_e$ and $V_e$ is a subspace of $V$. Similarly, if $f,g\in V_o$, then
$(cf+g)(-x)=cf(-x)+g(-x)=-cf(x)-g(x)=-(cf+g)(x)$
thus $cf+g\in V_o$ and $V_o$ is a subspace of V.
(b) let $\forall h\in V$, define
$f(x)=\dfrac{h(x)+h(-x)}{2},\quad g(x)=\dfrac{h(x)-h(-x)}{2}$
then $f(-x)=f(x),g(-x)=-g(x)$, so $f\in V_e,g\in V_o$, and $f+g=h$.
( c ) If $h\in V_e\cap V_o$, then $h(-x)=h(x)=-h(x)$, thus $h(x)=0,\forall x\in R$.

9. Let $W_1$ and $W_2$ be subspaces of a vector space $V$ such that $W_1+W_2=V$ and $W_1\cap W_2=\{0\}$. Prove thet for each vector $\alpha$ in $V$ there are unique vectors $\alpha_1$ in $W_1$ and $\alpha_2$ in $W_2$ such that $\alpha = \alpha_1+\alpha_2$.

Solution: Let $α\in V$, since $V=W_1+W_2$, we can find $α_1\in W_1,α_2\in W_2 s.t. α=α_1+α_2$.
To prove uniqueness, suppose there’s $β_1\in W_1,,β_2\in W_2$ s.t. $α=β_1+β_2$, then
$α_1+α_2=β_1+β_2 \Rightarrow α_1-β_1=β_2-α_2$
since $α_1-β_1\in W_1$ and $β_2-α_2\in W_2$, we know that $α_1-β_1\in W_1\cap W_2,β_2-α_2\in W_1\cap W_2$, thus $α_1-β_1=β_2-α_2=0$, which means $α_1=β_1,β_2=α_2$.