Ch.4 Vector Spaces

Let $V$ be the line with slope 2 passing through the origin
$V=\left\{t\begin{pmatrix}1\\2\end{pmatrix}\middle|t\in\mathbb{R}\right\}$
It is very easy to check that given $\vec{v}_1,\vec{v}_2,\vec{v}_3\in V$, all 10 conditions can be satisfied.

$P=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\in\mathbb{R}^3\middle|2x+y+3z=0\right\}$
Varify the two closure conditions, (1) and (6), because the others are similarly easily checked
Condition (1): sum of two vectors in the plane makes another vector in the plane
Let $\vec{p}_1=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}$ and $\vec{p}_2=\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$.
The sum is $\vec{p}_1+\vec{p}_2=\begin{pmatrix}x_1+x_2\\y_1+y_2\\z_1+z_2\end{pmatrix}$, which can be plugged in
For condition (6), show multiplying by a scalar yields a vector in the plane. Easy to check.

The set of all $n$-tall vectors (column vectors with $n$ components) is a vector space under natural operations.
All 10 conditions are easy to verify.

Consider the set of quadratic polynomials $P_2\left\{a_0+a_1x+a_2x^2\middle|a_0,a_1,a_2\in\mathbb{R}\right\}$
Adding two polynomials gives you another polynomial.
Multiplying by a scalar gives another polynomial, therefore these conditions are satisfied.
Since vector spaces can have linear combinations, you can take a linear combination of quadratic polynomials to get another quadratic polynomial

The set $M_{m\times n}$ of all $m\times n$ matrices is a vector space with addition defined as follows:
For matrix $A,B\in M_{m\times n}$ and $C=A+B$,
$C_{i,j}=A_{i,j}+B_{i,j}$
and multiplication by a scalar with $C=rA$ by
$C_{i,j}=rA_{i,j}$
for $i=1,...,m$ and $j=1,...,n$
Meaning linear combinations of $m\times n$ matrices yields another $m\times n$ matrix

$\vec{v}=(1+0)\cdot\vec{v}=\vec{v}+(0\cdot\vec{v})$
Add additive inverse $\vec{w}$ such that $\vec{v}+\vec{w}=0$
$\vec{w}+\vec{v}=\vec{w}+\vec{v}+0\cdot\vec{v}$
$\vec{0}=\vec{0}+0\cdot\vec{v}$
$\vec{0}=0\cdot\vec{v}$

$(-1\cdot\vec{v})+(1\cdot\vec{v})=(-1+1)\cdot\vec{v}=0\cdot\vec{v}=\vec{0}$ by (1)

$r\cdot\vec{0}=r\cdot(0\cdot\vec{0})$ by (1), then
$r\cdot(0\cdot\vec{0})=(r\cdot0)\cdot\vec{0}=0\cdot\vec{0}=\vec{0}$

Strategy: prove $(1)\implies(3)\implies(2)\implies(1)$
$(1)\implies(3)$ is obvious because $S$ is a vector space
$(3)\implies(2)$ is obvious because "any number" can mean 2
all that remains is to prove $(2)\implies(1)$
Assume $S\subset V$ for a vector space $V$ ($S$ is nonempty) and that $S$ is closed under linear combinations of pairs of vectors.
If $S$ is closed under a linear combination of a pair of vectors, then for $s_1,s_2\in S$, $s_1+s_2\in S$.
For commutativity, for $s_1,s_2\in S$, $s_1+s_2\in S$ is equal to $s_1+s_2\in V$, which is equal to $s_2+s_1\in V$ since $V$ is a vector space, and this is in turn equal to $s_2+s_1\in S$.
Associativity is similar.
For the zero vector, the linear combination of $0\cdot s_1+0\cdot s_2=\vec{0}\in S$, which can be shown to be the zero vector in a similar way.
For additive inverse, since $\vec{0}\in S$, the linear combination $0\cdot\vec{0}+(-1)\cdot\vec{s}$ is in $S$, and is clearly the additive inverse.
The multiplication conditions are similar.

The vector space $\mathcal{P}_2=\{a_0+a_1x+a_2x^2|a_0,a_1,a_2\in\mathbb{R}\}$ has a subspace of linear polynomials $L=\{b_0+b_1x|b_0,b_1\in\mathbb{R}\}$
To prove, we only need to show it is closed under a linear combination of two members.
$r\cdot(b_0+b_1x)+s\cdot(c_0+c_1x)=(rb_0+sc_0)+(rb_1+sc_1)x\in L$
Note that this only works because we assumed $\mathcal{P}_2$ is a vector space.
If that isn't given, you must prove $\mathcal{P}_2$ is a vector space.

Subset of $\mathbb{R}^3$ is a plane
$P=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\middle|2x-y+z=0\right\}$
Since $\mathbb{R}^3$ is clearly a vector space, to show this is a subspace we only need to prove it is closed under linear combinations
Take two vectors $\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix},\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}\in P$
They satisfy
$2x_1-y_1+z_2=0\\2x_2-y_2+z_2=0$
The vector sum is $\begin{pmatrix}x_1+x_2\\y_1+y_2\\z_1+z_2\end{pmatrix}$, which must also satisfy the condition
$2(x_1+x_2)-(y_1+y_2)+(z_1+z_2)=2x_1-y_1+z_2+2x_2-y_2+z_2=0+0=0$
so $P$ is closed under addition.
Next take $\begin{pmatrix}x\\y\\z\end{pmatrix}\in P$ and $r\in \mathbb{R}$
Consider the condition on the scalar product $\begin{pmatrix}rx\\ry\\rz\end{pmatrix}$:
$2rx-ry+rz=r(2x-y+z)=r\cdot0=0$
so $P$ is closed under scalar multiplication
Therefore, $P$ is a subspace of $\mathbb{R}^3$

Consider this subset of $\mathbb{R}^3$:
$P=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\middle|2x-y+z=1\right\}$
Note that this is the identical to the subset in Example 4.7 except it equals $1$ instead of $0$.
Under vector addition of $\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix},\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}\in P$
the vector sum $\begin{pmatrix}x_1+x_2\\y_1+y_2\\z_2+z_2\end{pmatrix}$ does not satisfy the condition:
$2(x_1+x_2)-(y_1+y_2)+(z_1+z_2)=2x_1-y_1+z_1+2x_2-y_2+z_2=1+1=2$
A similar thing occurs in scalar multiplication, where it no longer equals $1$, so these are not closed under addition or scalar multiplication, so $P$ is not a subspace (it is still a subset)