{%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.