p.19, example 1.35
(a) If $b\in \mathbb{F}$, then
$$ U\equiv\{(x_1,x_2,x_3,x_4)\in \mathbb{F}^4:x_3=5x_4+b\} $$
is a subspace of $\mathbb{F}^4$ if and only if $b=0$.
Suppose $b=0$. Then $U$ is a subspace of $\mathbb{F}^4$:
- $0\in U$ since $x_4=0\implies x_3=0$
- closed under addition: $u,v\in U\implies u+v=(u_1+v_1,u_2+v_2,5(u_4+v_4),u_4+v_5)\in U$
- closed under scalar multiplication: $c\in \mathbb{F},u\in U\implies cu=(cu_1,cu_2,5cu_4,cu_4)\in U$
If $b\neq 0$, then $0\notin U$ because either $x_4=0\implies x_3=b\neq0$ or $x_3=0\implies5x_4+b=0\implies x_4=\frac{b}{5}\neq0$.
(b) The set $U$ of continuous real-values functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$.
- $f\equiv0$ is continuous, hence $f\in U$
- $f,g\in U\implies (f+g)=f+g\in U$ since the sum of continuous functions is continuous
- $c\in \mathbb{F},f\in U\implies cf\in U$ since $cf$ is continuous
(c) The set $U$ of differentiable real-valued functions on $\mathbb{R}$ is a subspace of $\mathbb{R}^\mathbb{R}$.
- $f\equiv0$ is differentiable, hence $f\in U$
- $f,g\in U\implies (f+g)=f+g\in U$ since the sum of differentiable functions is differentiable
- $c\in \mathbb{F},f\in U\implies cf\in U$ since $cf$ is differentiable
(d) The set $U$ of differentiable real-valued functions $f$ on the interval $(0,3)$ such that $f’(2)=b$ is a subspace of $\mathbb{R}^{(0,3)}$ if and only if $b=0$.
Suppose $b=0$. Then
- $f\equiv0$ is differentiable and $f’(2)=0=b$, hence $f\in U$
- $f,g\in U\implies (f+g)=f+g\in U$ since the sum of differentiable functions is differentiable and $f’(2)+g’(2)=0+0=0$
- $c\in \mathbb{F},f\in U\implies cf\in U$ since $cf$ is differentiable and $cf’(2)=c0=0$
If $b\neq0$ and $f\equiv0$, then $f’(2)=0\neq b\implies f\notin U$.
(e) The set $U$ of all sequences of complex numbers with limit $0$ is a subspace of $\mathbb{C}^{\infty}$
- Define $0\equiv(0,0,…)$ to be the additive identity of $\mathbb{C}^{\infty}$. Then $0\in U$ since $\lim_{n\rightarrow\infty}0_n=0$.
- $f,g\in U\implies f+g\in U$ since $\lim_{n\rightarrow\infty}(f_n+g_n)=\lim_{n\rightarrow\infty}f_n+\lim_{n\rightarrow\infty}g_n=0+0=0$
- $c\in\mathbb{C},f_n\in U\implies\lim_{n\rightarrow\infty}cf_n=c\lim_{n\rightarrow\infty}f_n=0\implies cf_n\in U$
p.20, example 1.37
$$\begin{align*} U+W &= \{u+w:u\in U, w\in W\}\\ &= \{(x,0,0)+(0,y,0):(x,0,0)\in U, (0,y,0)\in W\}\\ &= \{(x,0,0)+(0,y,0):x,y\in \mathbb{F}\}\\ &= \{(x,y,0):x,y\in \mathbb{F}\} \end{align*}$$
General Example 1.1
Define
$$ U\equiv\{u=(u_1,u_2)\in\mathbb{R}^2: u_1=u_2\} \quad\quad\text{and}\quad\quad W\equiv\{w=(w_1,w_2)\in\mathbb{R}^2: w_1=-w_2\} $$
Then $U\cup W$ is not a subspace and $U\oplus W=\mathbb{R}^2$ is a subspace.
Proof $U$ is a subspace of $\mathbb{R}^2$:
- $0=(0,0)\in U$.
- if $x,y\in U$, then $x+y=(x_1,x_2)+(y_1,y_2)=(x_1+y_1,x_2+y_2)\in U$ since $x_1+y_1=x_2+y_2$.
- if $t\in\mathbb{R}$ and $u\in U$, then $tu=(tu_1,tu_2)\in U$ since $tu_1=tu_2$.
And similarly $W$ is a subspace of $\mathbb{R}^2$. In exercise 1.7, we showed that $U\cup W$ is not a subspace of $\mathbb{R}^2$ because its not closed under addition. More generally, in exercise 1.9, we showed that the union of two subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces is contained in the other. Then $U\cup W$ is not a subspace of $\mathbb{R}^2$ since $U\not\subseteq W$ and $W\not\subseteq U$.
Next we wish to show that $U+W=U\oplus W$. The text showed this to be true if and only if $U\cap W=\{0\}$, which is clear.
And lastly we wish to show that $U\oplus W=\mathbb{R}^2$. Clearly $U\oplus W\subset\mathbb{R}^2$ so let’s show that $\mathbb{R}^2\subset U\oplus W$.
First, concretely, pick an arbitrary point in $\mathbb{R}^2$, say $(-1,4)$. We will show that $(-1,4)\in U+W$. That is, we wish to find $u=(u_1,u_2)\in U$ and $w=(w_1,w_2)\in W$ such that $u+w=(-1,4)$ or
$$ u_1+w_1=-1\\ u_2+w_2=4 $$
But $u_2=u_1$ and $w_2=-w_1$. Hence we wish to find $u_1$ and $w_1$ such that
$$ u_1+w_1=-1\\ u_1-w_1=4 $$
$$ \begin{bmatrix}1&1&-1\\1&-1&4\end{bmatrix}\rightarrow\begin{bmatrix}1&1&-1\\0&-2&5\end{bmatrix}\rightarrow\begin{bmatrix}1&1&-1\\0&1&-\frac{5}{2}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{5}{2}-1\\0&1&-\frac{5}{2}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{3}{2}\\0&1&-\frac{5}{2}\end{bmatrix} $$
More generally, for any $(a,b)\in\mathbb{R}^2$, we wish to find $u_1$ and $w_1$ such that
$$ u_1+w_1=a\\ u_1-w_1=b $$
$$ \begin{bmatrix}1&1&a\\1&-1&b\end{bmatrix}\rightarrow\begin{bmatrix}1&1&a\\0&-2&b-a\end{bmatrix}\rightarrow\begin{bmatrix}1&1&a\\0&1&\frac{a-b}{2}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&a-\frac{a-b}{2}\\0&1&\frac{a-b}{2}\end{bmatrix}\rightarrow\begin{bmatrix}1&0&\frac{a+b}{2}\\0&1&\frac{a-b}{2}\end{bmatrix} $$
Hence $\mathbb{R}^2\subset U\oplus W$ and $U\oplus W=\mathbb{R}^2$.
$\blacksquare$