---
# System prepended metadata

title: Untitled

---

{%hackmd 5xqeIJ7VRCGBfLtfMi0_IQ %}

##### Exercise 3

依照各小題的步驟來證明子空間的兩個條件等價。  

1. $V = \vspan(S)$ for some vectors $S$.  
2. $V$ is a nonempty subset that is closed under scalar multiplication and vector addition.

並用這些條件來判斷一個集合是否為子空間。

<!-- eng start -->
Use the given instructions to show the following statements are equivalent.

1. $V = \vspan(S)$ for some vectors $S$.  
2. $V$ is a nonempty subset that is closed under scalar multiplication and vector addition.

Then you may use the equivalent conditions to check if a set is a subspace.
<!-- eng end -->

##### Exercise 3(a)

證明若條件 1 成立則條件 2 成立。

<!-- eng start -->
Show that Condition 1 implies Condition 2.
<!-- eng end -->

Sample:  
Suppose $V = \vspan(S)$ for some $S\subseteq\mathbb{R}^n$.  
We verify each of the requirements in condition 2.  

**nonempty**  
Since $\vspan(S)$ always contains zero vector, $V$ is nonempty.  

**closed under scalar multiplication**  
Suppose $\bv\in V$ and $k$ a scalar.  
Then $\bv$ can be written as $c_1\bu_1 + \cdots + c_k\bu_k$ for some vectors $\bu_i\in S$ and scalars $c_i$.  
Then $k\bv = kc_1\bu_1 + \cdots + kc_k\bu_k$.  
Since $k$ and $c_i$ are both scalars.
So $k\bv\in\vspan(S) = V$.  

**closed under vector addition**  
Suppose $\bv_1,\bv_2\in V$.  
Then $\bv_1$ can be written as $a_1\bu_1 + \cdots + a_k\bu_k$, for some vectors $\bu_i\in S$ and scalars $a_i$,  
and $\bv_2$ can be written as  $b_1\bu_1 + \cdots + b_k\bu_k$, for some vectors $\bu_i\in S$ and scalars $b_i$.  
Then $\bv_1 + \bv_2 = (a_1 + b_1)\bu_1 + \cdots + (a_k + b_k)\bu_k = c_1\bu_1 + \cdots + c_k\bu_k$ for scalars $a_i+b_i=c_i$.  
So $\bv_1 + \bv_2 \in\vspan(S) = V$.  

##### Exercise 3(b)

證明若條件 2 成立則條件 1 成立。

<!-- eng start -->
Show that Condition 2 implies Condition 1.
<!-- eng end -->

Sample:  
Suppose $V$ is a nonempty subset of $\mathbb{R}^n$ and is closed under scalar multiplication and vector addition.  
It is enough to show that $V = \vspan(V)$.  

**$V\subseteq\vspan(V)$**  
Each element in $V$ is linear combination of $V$ and is in $\vspan(V)$, so $V\subseteq\vspan(V)$.

:::warning

- [x] As the result, $V$ is closed under scalar multiplication and vector addition. Therefore, $\bu\in V$. --> Since $V$ is closed under scalar multiplication and vector addition, we know $\bu\in V$.
:::

**$\vspan(V)\subseteq V$**  
Let $\bu$ be an element of $\vspan(V)$.  
Then $\bu$ can be written as $c_1\bu_1 + \cdots + c_k\bu_k$ for some vectors $\bu_i\in V$ and scalars $c_i$.  
Since $V$ is closed under scalar multiplication and vector addition, we know $\bu\in V$.

##### Exercise 3(c)

判斷 $\emptyset$ 是否為一子空間。

<!-- eng start -->
Check if $\emptyset$ is a subspace.
<!-- eng end -->

Since $\emptyset$ does not match "$V$ is a nonempty subset" in Condition 2, $\emptyset$ is not a subspace.  

##### Exercise 3(d)

判斷 $\{\bzero\}$ 是否為一子空間。

<!-- eng start -->
Check if $\{\bzero\}$ is a subspace.
<!-- eng end -->

Because we vacuously define $\vspan(\emptyset) = \{\bzero\}$, $\{\bzero\}$ is a subspace.

