# Linear Algebra - Vector Space
## Definition
for all vectors $\vec{u}$, $\vec{v}$, $\vec{w} \in V$, and $r$,$s\in\mathbb{R}$, if
| rule | explain | rule |explain |
| -------- | -------- | -------- |-------- |
| $\vec{v}+\vec{w}\in V$| close under addition | $r\vec{v}\in V$ |close under miltiplication|
|$r(\vec{v}+\vec{w}) = r\vec{v}+r\vec{w}$|distribute over vector addition|$(r+s)\vec{v} = r\vec{v}+s\vec{v}$|distribute over scalor addition|
|$(\vec{v}+\vec{w})+\vec{u}=\vec{v}+(\vec{w}+\vec{u})$|associative|$(rs)\vec{v}=r(s\vec{v})$|associative|
|$\vec{v}+\vec{w} = \vec{w}+\vec{v}$|commutative|$1\vec{v}=\vec{v}$|identity operation|
|exist $\vec{w}\in V$, $\vec{w}+\vec{v}=0$|additive inverse|$\vec{0}\in V$||
## Determine whether it is vector space
### Case 1
:::info
$V=\{\left[ {\begin{array}{} x\\y \end{array} } \right]\mid x^2+y^2\le1\}$. Is set V a vector space?
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Ans:
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V is not vector space because$\left[ {\begin{array}{} 1\\0 \end{array} } \right]+\left[ {\begin{array}{} 0\\1 \end{array} } \right]=\left[ {\begin{array}{} 1\\1 \end{array} } \right]\notin V$
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### Case 2
:::info
$V=\{\left[ {\begin{array}{} x\\y \end{array} } \right]
\mid x+y=0\}$. Is set V a vector space?
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Ans:
:::spoiler
V is a vector space because $\forall \vec{u}$, $\vec{v}$, $\vec{w} \in V$, and $r$,$s\in\mathbb{R}$ such that $\vec{v}+\vec{w}\in V$,$r\vec{v}\in V$, $\vec{v}+r\vec{w}$, $(r+s)\vec{v}=r\vec{v}+s\vec{v}$, $(\vec{v}+\vec{w})+\vec{u}=\vec{v}+(\vec{w}+\vec{u})$, $(rs)\vec{v}=r(s\vec{v})$, $\vec{v}+\vec{w} = \vec{w}+\vec{v}$, $1\vec{v}=\vec{v}$, exist $\vec{w}\in V$,$\vec{w}+\vec{v}=0$ and $\vec{0}\in V$
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### Case 3
:::info
$V=\{\left[ {\begin{array}{} x\\y \end{array} } \right]
\mid x+y=1\}$. Is set V a vector space?
:::
Ans:
:::spoiler
V is not vector space because$\left[ {\begin{array}{} 0\\0 \end{array} } \right]\notin V$
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### Case 4
:::info
$V=\{\left[ {\begin{array}{} x\\y \end{array} } \right]
\mid y>0\}$. Is set V a vector space?
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Ans:
:::spoiler
V is not vector space because$\left[ {\begin{array}{} 0\\0 \end{array} } \right]\notin V$
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