---
# System prepended metadata

title: Lecture 3
tags: [' NCTU OCW', ' 莊重', Linear Algebra]

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tags: Linear Algebra, NCTU OCW, 莊重
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# Lecture 3

## Objectives

1. Definition
2. Verification
3. Properties

### Definition

$\text{Let }V\text{ be a vector space over a field }F\text{ and }W\subset V\text{.}\\
\text{Then }W\text{ is a }\textit{subspace}\text{ of }V\\
\text{if }W\text{ is a vector space over }F\text{ with the same two operations.}$

Observation:
checking if the 10 axioms apply is troublesome; is there a faster way to verify $W$?
Intuitively we have that $\{1,2,5,6,7,8\}$ must also apply if $W\subset V$

#### Theorem 1.3
$W\text{ is a subspace of }V\\
\iff\begin{equation}
\begin{cases}
\vec{0}\in W & \text{(1)}\\
x+y\in W\text{ whenever }x,y\in W & \text{(2)}\\
\alpha x\in W\text{ whenever }\alpha\in F,~ x\in W & \text{(3)}
\end{cases}
\end{equation}$

__remark__:
In fact, if $W\neq\phi$, then we don't need to check if $\text{(1)}$ of theorem 1.3 holds.
*// this remark is left as exercise*

__proof__: (theorem 1.3)
($\rightarrow$)
__2__ and __3__ is trivial, since $W$ is vector space.
__1__? the problem is "$\vec{0}$" may not be the same in $V$ and in $W$.
the theorem wants $\vec{0}_V$ of $V$ is the same element of $\vec{0}_W$ of $W$ is the same one.
notice we use the same operations, and $W$ by definition of vector space is non-empty, we could apply *elimination* in $V$:
$$\begin{equation}\begin{cases}
\vec{0}_V&+x=x \\
\vec{0}_W&+x=x
\end{cases}\end{equation}
\forall x\in W$$
we conclude that $\vec{0}_V=\vec{0}_W$. __1__ holds.
($\leftarrow$)
notice we only need to check validity of axiom $4$. That is, existence of anti element of operator $+$ in $W$.
$\forall x\in W\subset V,~ \exists \left(-x\right)\in V \text{ s.t. }x+\left(-x\right)=\vec{0}.$
claim: $\left(-x\right)\in W$
notice $\left(-x\right)=\left(-1\right)x$ (check lecture 2)
by __3__, we know the claim holds.
The proof is hence complete. $\text{Q.E.D.}$

##### Examples of subspaces (theorem 1.3)

- $V=P(F)=\{a_nx^n+\cdots+a_1x+a_0\vert ~n\in\mathbb{N},~ a_i\in F,~ i\in\left(\mathbb{N}\cup \{0\} \right)\}$.
  All polynomials.
  $W=P_n(F)=\{a_nx^n+\cdots+a_1x+a_0\vert ~ a_i\in F,~ i\in\left(\mathbb{N}\cup\{0\}\right),~ i\leq n\}$.
  All polynomials with degree less or equal to $n$.
  $V$ is a vector space *// easy to check*
  Is $W$ subspace of $V$?
  1 holds, $\vec{0}=0\in W$. 2 and 3 both hold, which is trivial.
  We conclude, by theorem 1.3, $W$ __*IS*__ subspace of $V$.
- $M_{n\times m}\left(F\right)\overset{\Delta}{=}\{\left(a_{ij}\right)_{n\times m}\vert a_{ij}\in F\}$
  then if $V=M_{n\times m}$ and $W=\{\left(a_{ij}\right)_{n\times m}\vert a_{ii}=0,~ \forall i\in\{1,2,\cdots,n\}\}$
  then $W$ is subspace of $V$.
- (__diagonal matrix__)
  $W=\{\left(a_{ij}\right)_{n\times n}\vert a_{ij}=0,~ \forall i,j\in\{1,2,\cdots,n\},~ i\neq j\}$
  then $W$ is also subspace of $V$.
- $W=\{\left(a_{ij}\right)_{n\times n}\vert a_{ij}\geq0,~ \forall i,j\in\{1,2,\cdots,n\}\}$
  then $W$ is __NOT__ subspace of $V$.
  notice scalar multiplication is not closed.
- $V=\mathbb{R}^2\\
   W=\{\left(x,mx\right)\vert~ x\in\mathbb{R},~ m\text{ is some constant in }\mathbb{R}\}$
   (observe: $W$ is a straight line with slope equals to $m$ in the 2-D Euclidean space)
  then $W$ is subspace of $V$.
- $V=\mathbb{R}^2\\
   W=\{\left(x,mx+b\right)\vert~ x\in\mathbb{R},~ m\text{ is some constant in }\mathbb{R},~ \\
   b\text{ is some constant in }\mathbb{R}\}$
   (observe: $W$ is a straight line with slope equals to $m$ in the 2-D Euclidean space)
  then $W$ is __NOT__ a vector space, and hence not subspace of $V$.
  __observation__:
    In Euclidean space $\mathbb{R}^n$, subspace must contain $\left(0,\cdots,0\right)$
    
#### Theorem 1.4

Suppose $W_\alpha$ is subspace of $V$, $\alpha\in\Lambda$, $V$ vector space, then we have:
$$\left(\bigcap\limits_{\alpha\in\Lambda}W_\alpha\right) \text{ is a subspace of }V$$
notice we don't limit $\lvert\Lambda\rvert$.

__proof__:
suppose $x,y\in\left(\bigcap\limits_{\alpha\in\Lambda}W_\alpha\right)\\
\implies x,y\in W_\alpha,~ \forall \alpha\in\Lambda\\
\implies x+y\in W_\alpha,~ \forall \alpha\in\Lambda\\
\implies x+y\in\left(\bigcap\limits_{\alpha\in\Lambda}W_\alpha\right)$

We can do similar treatment to $\alpha x ~\forall x\in W$; hence we conclude theorem holds. $\text{Q.E.D.}$

__remark__:
$\cup$ is not as good as $\cap$ in this case:
consider $y=x$ and $y=-x$ in 2-D Euclidean space.

__remark__:
$\textit{Trivial Subspace}$:
suppose $V$ is vector space, then $V$ and $\{\vec{0}_V\}$ are both *Trivial Subspace* of $V$.

#### Definition

$S_1+S_2\overset{\Delta}{=}\{\left(x+y\right)\vert~ x\in S_1,~ y\in S_2\}$

##### Lemma

suppose $W_1,W_2$ subspaces of $V$, we have:
1. $W_1+W_2$ is subspace of $V$
2. $W_1\cap W_2$ is subspace of $V$
3. $W_1\cup W_2$ is NOT NECESSARILY subspace of $V$
4. $W_1+ W_2$ is the __*smallest*__ subspace containing $W_1$ and $W_2$,
   i.e. if $W$ subspace of $V$ and $\left(W_1\cup W_2\right)\subset W$, then $\left(W_1+W_2\right)\subset W$
   notice if we let $x$ be $\vec{0}$, we could easily see that $W_1\cup W_2 \subset W_1 + W_2$
5. $W_1\cup W_2\text{ is subspace of }V
   \iff \left(W_1\subset W_2\right)\lor \left(W_2\subset W_1\right)$
   (*// left as exercise*)
   
## Exercises

1. $W_1\cup W_2\text{ is subspace of }V
   \iff \left(W_1\subset W_2\right)\lor \left(W_2\subset W_1\right)$

### Proof

1. $\rightarrow$
   proof by contradiction;
   suppose $\left({\neg{\left(W_1\subset W_2\right)}}\right) \land \left({\neg{\left(W_1\subset W_2\right)}}\right)$
   since subspaces of vector spaces is never empty, we know: (via *theorem 1.4*)
   $\left(\left(W_1\setminus W_2\right)\neq \phi\right)\land
   \left(\left(W_2\setminus W_1\right)\neq \phi\right)\land
   \left(\left(W_1\cap W_2\right)\neq \phi\right)$
   let $x\in\left(W_1\setminus W_2\right)$, $y\in\left(W_2\setminus W_1\right)$,
   it's obvious that $\left((x+y)\notin W_1\right)\land \left((x+y)\notin W_2\right)$
   we conclude that operator $+$ is not closed in the set $W_1\cup W_2$.
   $\leftarrow$
   Trivial.
   Hence the proposition holds.
   $\text{Q.E.D.}$