---
# System prepended metadata

title: 線性函數

---

{%hackmd 5xqeIJ7VRCGBfLtfMi0_IQ %}

# 線性函數

Linear function

##### Exercise 5

嵌射顧名思義有點像是把定義域嵌入到對應域之中﹐所以很多性質都會被保留下來。  

<!-- eng start -->
The name "injection" indicates that the domain is injected, or embedded, into the codomain, so many properties of the domain preserve.
<!-- eng end -->

##### Exercise 5(a)

令 $f: U\rightarrow V$ 為一線性函數。  
證明若 $f$ 是嵌射且  
$\alpha = \{\bu_1,\ldots,\bu_k\}$ 為 $U$ 中的一線性獨立集﹐  
則 $f(\alpha)$ 是 $V$ 中的一線性獨立集。  

<!-- eng start -->
Let $f: U\rightarrow V$ be a linear function.  Show that if $f$ is injective and $\alpha = \{\bu_1,\ldots,\bu_k\}$ is linearly independent in $U$, then $f(\alpha)$ is linearly independent in $V$.  
<!-- eng end -->

:::warning

- [x] $\alpha$ is linear --> $\alpha$ is linearly independent
:::

##### Exercise 5(a) -- answer here:
We set $c_1f({\bf u}_1) + \cdots +c_kf({\bf u}_k) = \bf0$.
Because the function $f$ is linear, $f(c_1{\bf u}_1 + \cdots + c_k\bu_k) = \bf0$.
Accroding to Exercise 4, we know that $\ker(f) = \{ \bzero \}$, because function $f$ is injective.
And so, we know $c_1\bu_1 + \cdots + c_k{\bf u}_k = \bf0$.
Also, $\alpha$ is linearly independent, so $c_1 = \cdots = c_k = 0$.
Therefore, $f(\alpha)$ is a linearly independent set in $V$.



# 將矩陣視為線性函數

Matrix as a linear function

##### Exercise 2

對以下各矩陣 $A$：  

1. 說明 $f_A$ 的作用。  
2. 寫出 $\range(f_A)$ 及 $\rank(f_A)$。  
3. 寫出 $\ker(f_A)$ 及 $\nul(f_A)$。  
4. 判斷 $f_A$ 是否可逆；若可逆﹐其反函數為何？

<!-- eng start -->
For each of the following matrices:  

1. Describe the effect of $f_A$.  
2. Find $\range(f_A)$ and $\rank(f_A)$.  
3. Find $\ker(f_A)$ and $\nul(f_A)$.  
4. Is $f_A$ invertible?  If yes, what is the inverse function?  
<!-- eng end -->
---
##### Exercise 2(e)

$$
    A = \begin{bmatrix}  
     0 & 1 & 0 \\
     0 & 0 & 1 \\
     1 & 0 & 0 \\
    \end{bmatrix}.
$$  

:::warning

- [x] replace all your `{\bf e}` with `\be` : the effects are the same but the latter one is easier
- [x] $\range(f) = \vspan ... = \mathbb{R}^3$
- [x] $\rank(f) = \dim(\range(f)) = 3$.
- [x] Since $\rank(f) = 3$ and $\rank(f) + \nul(f) = 3$, we have $\nul(f) = 0$.  Consequently, $\ker(f) = \{\bzero\}$.  
:::

##### Exercise 2(e) -- answer here:
The function $f_A$ will replace ${\be}_1$, ${\be}_2$ and ${\be}_3$ with ${\be}_3$, ${\be}_1$, and ${\be}_2$, respectively.

$$\operatorname{range}(f) = \vspan\left\{ \begin{bmatrix} 
 0 \\
 0 \\
 1 \\
\end{bmatrix}
, \begin{bmatrix}  
  1 \\
  0 \\
  0 \\
\end{bmatrix} , \begin{bmatrix}  
  0 \\
  1 \\
  0 \\
\end{bmatrix}
\right\}= \mathbb{R}^3,
$$

$\rank(f) = \dim(\range(f)) = 3$.  

Since $\rank(f) = 3$ and $\rank(f) + \nul(f) = 3$, we have $\nul(f) = 0$.  Consequently, $\ker(f) = \{\bzero\}$.

The function $f_A$ is invertible.  
And the inverse function of $f_A$ will replace ${\be}_3$, ${\be}_1$, and ${\be}_2$ with ${\be}_1$, ${\be}_2$ and ${\be}_3$, respectively.
