# 常見的向量空間

This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/).

{%hackmd 5xqeIJ7VRCGBfLtfMi0_IQ %}

from lingeo import random_int_list
from linspace import vtop, vtom

## Main idea

Let $\mathcal{P}_d$ be the set of all polynomials of degree at most $d$ (with real coefficients).
Let $+$ and $\cdot$ be the regular polynomial addition and scalr multiplication, respectively.
Then $(\mathcal{P}_d, +, \cdot)$ is a vector space, usually denoted as just $\mathcal{P}_d$.

It is known that $\beta = \{1, x, \ldots, x^d\}$ is a basis of $\mathcal{P}_d$, so $\dim(\mathcal{P}_d) = d + 1$.
The zero vector in $\mathcal{P}_d$ is $0 = 0 + 0x + \cdots + 0x^d$.
We usually call $\beta$ the **standard basis** of $\mathcal{P}_d$.

It is easy to see that essentially every polynomial stores $d+1$ coefficients, and its behavior is very similar to a vector in $\mathbb{R}^{d+1}$.
When $p = a_0 + a_1x + \cdots + a_dx^d$, we define $\operatorname{ptov}(p) = (a_0,\ldots,a_d)$.

Let $\mathcal{M}_{m,n}$ be the set of all $m\times n$ matrices (over $\mathbb{R}$).
Let $+$ and $\cdot$ be the regular matrix addition and scalr multiplication, respectively.
Then $(\mathcal{M}_{m,n}, +, \cdot)$ is a vector space, usually denoted as just $\mathcal{M}_{m,n}$.

Let $E_{ij}$ be the matrix whose $ij$-entry is $1$ while other entries are $0$.
It is known that $\beta = \{E_{11}, \ldots, E_{1n}, \ldots, E_{m1}, \ldots1, E_{mn}\}$ is a basis of $\mathcal{M}_{m,n}$, so $\dim(\mathcal{M}_{m,n}) = mn$.
The zero vector in $\mathcal{M}_{m,n}$ is $O_{m,n}$, the $m\times n$ zero matrix.
We usually call $\beta$ the **standard basis** of $\mathcal{M}_{m,n}$.

Again, every $m\times n$ matrix stores $mn$ coefficients, and behaves almost the same as a vector in $\mathbb{R}^{mn}$.
When $M = \begin{bmatrix} a_{ij} \end{bmatrix}$, we define $\operatorname{mtov}(p) = (a_{11},\ldots,a_{1n},\ldots,a_{m1},\ldots,a_{mn})$.

## Side stories
- gcd of polynomials
- matrix equation

## Experiments

##### Exercise 1

執行以下程式碼。
己知 $A{\bf v} = {\bf b}$。

### code
set_random_seed(0)
print_ans = False
m,n = 4,3
A = matrix(m, random_int_list(m*n))
v = vector(random_int_list(n))
b = A * v

print("A =")
show(A)
print("v = (c1, c2, c3) =", v)
print("b =", b)

print("p1 =", vtop(A.transpose()[0]))
print("p2 =", vtop(A.transpose()[1]))
print("p3 =", vtop(A.transpose()[2]))

print("M1, M2, M3 =")
pretty_print(vtom(A.transpose()[0],2,2), ", ", 
             vtom(A.transpose()[1],2,2), ", ", 
             vtom(A.transpose()[2],2,2)
            )

if print_ans:
    print("c1p1 + c2p2 + c3p3 =", vtop(b))
    print("c1M1 + c2M2 + c3M3 =")
    show(vtom(b,2,2))

:::warning

- [x] 以seed(0)為例： --> 以 `seed(0)` 為例：
:::

以 `seed(0)` 為例：
$$
A = \begin{bmatrix}
-4&3&5\\ -5&-5&0\\ 3&-3&3\\ 4&-4&-3\end{bmatrix},$$
$$v=(c_1,c_2,c_3)=(2,4,-4),$$

$$b=(-16,-30,-18,4),$$

$$p_1 = 4x^3+3x^2-5x-4, p_2 = -4x^3-3x^2-5x+3, p_3 = -3x^3+3x^2+5,
$$
$$
M_1,M_2,M_3=\begin{bmatrix} -4&-5\\ 3&4\end{bmatrix},\begin{bmatrix} 3&-5\\ -3&-4\end{bmatrix},\begin{bmatrix} 5&0\\ 3&-3\end{bmatrix}.
$$

##### Exercise 1(a)

求 $c_1p_1 + c_2p_2 + c_3p_3$。

$= 2(4x^3+3x^2-5x-4)+4(-4x^3-3x^2-5x+3)-4(-3x^3+3x^2+5)$

$= 4x^3-18x^2-30x-16.$

##### Exercise 1(b)

求 $c_1M_1 + c_2M_2 + c_3M_3$。

$=2\begin{bmatrix} -4&-5\\ 3&4\end{bmatrix}+4\begin{bmatrix} 3&-5\\ -3&-4\end{bmatrix}-4\begin{bmatrix} 5&0\\ 3&-3\end{bmatrix}$

$=\begin{bmatrix} -16&-30\\ -18&4\end{bmatrix}.$

## Exercises

##### Exercise 2

令 $U$ 為 $\mathcal{P}_3$ 中所有 $p_1 = (x+1)(x+2)$ 的倍式、
$V$ 為 $\mathcal{P}_3$ 中所有 $p_2 = (x+1)(x+3)$ 的倍式。

##### Exercise 2(a)

說明 $U$ 和 $V$ 都是 $\mathcal{P}_3$ 的子空間、
並判斷它們的維度。

:::warning

- [x] 我不懂你的 component 是什麼意思。

First, $U$ contains the zero polynomial $0 = p_1\cdot 0$.

Let $f$ be an element in $U$.
Then $f$ can be written as $p_1q$ for some polynomial $q$ of degree at most $1$.
Thus, for any scalar $c$, $cp = ???$, so $cp$ is also an element in $U$.

On the other hand, let $f$ and $g$ be two elements in $U$.
Then $f = p_1q_f$ and $g = p_1q_g$ for some polynomials $q_f$ and $q_g$ of degree at most $1$.
Thus, $f + g = ???$, so $f + g$ is also an element in $U$.

Therefore, $U$ is a subspace.
:::

Ans:
First, $U$ contains the zero polynomial $0 = p_1\cdot 0$.

Let $f$ be an element in $U$.
Then $f$ can be written as $qp_1$ for some polynomials $q$ of degree at most $1$.
Thus, for any scalar $c$, $cf = (cq)p_1$, which is also an element in $U$.

On the other hand, let $f$ and $g$ be elements in $U$.
Then $f = q_fp_1$ and $g=q_gp_1$ for some polynomials $q_f$ and $q_g$ of degree at most $1$, and $f+g=q_fp_1+q_gp_1=(q_f+q_g)p_1$, which is also an element in $U$.

Therefore, $U$ is a subspace in $\mathcal{P}_3$.

Elements in $U$ can be all expressed as $(c_1+c_2x)p_1=c_1p_1+(c_2x)p_1$.
We find that $p_1$ and $xp_1$ are two independent elements spanning the vector space $U$.

Also, each elements $(c_1+c_2x)p_1$ in $U$ can be expressed as $\operatorname{ptov}((c_1+c_2x)p_1)$ in $\mathbb{R}^4$.
We find that $\operatorname{ptov}((c_1+c_2x)p_1)$ can be viewed as the linear combination of two independent vectors $c_1\cdot\operatorname{ptov}(p_1)+c_2\cdot\operatorname{ptov}(xp_1)$.
These two independent vectors $\operatorname{ptov}(p_1)$ and $\operatorname{ptov}(xp_1)$ span the subspace which contains all possible $\operatorname{ptov}((c_1+c_2x)p_1)$.

Therefore, the dimension of $U$ is $2$.

The arguments above hold for $V$ as well.

<!-- $p_1$ is one of the component in $U$, now let $c$ be any real constant, then $cp_1$ including $0$($0p_1$) is also in both the set $U$ and $\mathcal{P}_3$, so $U$ is a subspace in $\mathcal{P}_3$.
The argument also holds for $V$ with one of its component $p_2$. -->

##### Exercise 2(b)

令 $A$ 為一個 $4\times 2$ 矩陣其各行向量為 $\operatorname{ptov}(p_1)$ 和 $\operatorname{ptov}(xp_1)$、
$B$ 為一個 $4\times 2$ 矩陣其各行向量為 $\operatorname{ptov}(p_2)$ 和 $\operatorname{ptov}(xp_2)$。
令 ${\bf b} = (1,1,0,0)$。
求 $\begin{bmatrix} A & B \end{bmatrix}{\bf x} = {\bf b}$ 中 ${\bf x}$ 的一個解﹐
並找到兩個一次以下的多項式 $a$ 和 $b$
使得 $ap_1 + bp_2 = 1 + x$。

:::warning

- [x] ${\bf x_p}+{\bf r}{\bf x_n}$ --> $\bx_p + r\bx_n$ （純量、下標 不要粗體）
- [x] choosing $r = -1$ to give a solution $x = (2,1,-1,-1)$, then --> By choosing $r = -1$, we get a solution $\bx = (2,1,-1,-1)$. Then
:::

Ans:
Let
$$
A = \begin{bmatrix}
2&0\\
3&2\\
1&3\\
0&1\\
\end{bmatrix}
,B = \begin{bmatrix}
3&0\\
4&3\\
1&4\\
0&1\\
\end{bmatrix},
$$

and

$$
\begin{bmatrix}
A&B
\end{bmatrix}{\bf x} =
\begin{bmatrix}
2&0&3&0\\
3&2&4&3\\
1&3&1&4\\
0&1&0&1\\
\end{bmatrix}
\begin{bmatrix}
c_1\\
c_2\\
c_3\\
c_4\\
\end{bmatrix}.
$$

We find that the equation above is corresponding to the polynomial operation $(2+3x+x^2)(c_1+c_2x)+(3+4x+x^2)(c_3+c_4x)$, so the matrix multiplication $\begin{bmatrix} A&B \end{bmatrix}{\bf x} = {\bf b}$ can be viewed as polynomial equation $ap_1+bp_2$ producing $\operatorname{vtop}({\bf b})=(1+x)$.
Solutions to $\begin{bmatrix} A&B \end{bmatrix}{\bf x} = {\bf b}$ are

$$
{\bf x} = {\bf x}_p+r{\bf x}_n =
\begin{bmatrix} -1\\0\\1\\0 \end{bmatrix}+
r\begin{bmatrix} -3\\-1\\2\\1 \end{bmatrix}.
$$

By choosing $r = -1$, we get a solution $\bx = (2,1,-1,-1)$. Then polynomials $a$ and $b$ are $c_1+c_2x = 2+x$ and $c_3+c_4x = -1+-x$.

##### Exercise 2(c)

找出一個 $\mathcal{P}_3$ 中的多項式 $q$
使得它無法利用 $a,b\in\mathcal{P}_1$ 寫成 $q = ap_1 + bp_2$ 的形式。

:::warning

- [ ] 你誤會了喔，$\mathcal{P}_1$ 是所有次數最高為 1 的多項式集合，所以 $a,b$ 可以有 $x$ 項。這題要用上一題的矩陣。
:::

Ans:
The statement implies that $\operatorname{ptov}(q)$ is not in the column space of $\begin{bmatrix} A&B \end{bmatrix}$, so $\operatorname{ptov}(q)$
could be a nonzero vector in the left nullspace, plus possibly any vector in the column space of $\begin{bmatrix} A&B \end{bmatrix}$.

By calculating the basis of $\ker(\begin{bmatrix} A&B \end{bmatrix}\trans)$ through row operations, we have that $\{{\bf l}\}$ is a basis of $\ker(\begin{bmatrix} A&B \end{bmatrix}\trans)$.
Let $\bx$ be any vector in $\mathbb{R}^4$.
Then $q$ can be chosen as any vector satisfying

$$
\operatorname{ptov}(q)=c{\bf l} + \begin{bmatrix} A&B \end{bmatrix}\bx, c\neq0.
$$

By choosing $c=1$ and $\bx=(1,1,1,0)$, we get $\operatorname{ptov}(q)=(4,10,4,2)$, and the polynomial $p$ is $4+10x+4x^2+2x^3$.

<!--$a,b\in\mathcal{P}_1$ implies that $a$ and $b$ are constants, and all combinations $ap_1 + bp_2$ span a subspace $Q$ in $\mathcal{P}_3$.
Let $Q$ be the column space of

$$
\begin{bmatrix}
\operatorname{ptov}(p_1) & \operatorname{ptov}(p_2)
\end{bmatrix} =
\begin{bmatrix}
2&3\\
3&4\\
1&1\\
\end{bmatrix}.
$$

Then $q$ can be any vectors in $\mathbb{R}^3$ containing vectors in $Q^\perp$(spanned by $(1,-1,1)$), such as

$$
\begin{bmatrix} 2\\3\\1 \end{bmatrix} +
\begin{bmatrix} 1\\-1\\1 \end{bmatrix} =
\begin{bmatrix} 3\\2\\2 \end{bmatrix}
.$$ -->

##### Exercise 3

令 $A$ 為 $3\times 3$ 的全 $1$ 矩陣。
令
$$V = \{ X\in\mathcal{M}_{3,3} : AX = O \}.
$$

##### Exercise 3(a)

說明 $V$ 是 $\mathcal{M}_{3,3}$ 的子空間、
並判斷它的維度。

:::warning

- [x] 存在 ${M}_A,{M}_B \in V$. ... 所以, --> 若 ${M}_A,{M}_B \in V$. ... 則
:::

Ans:

（1）若 ${M}_A,{M}_B \in V$.

$${M}_A= \begin{bmatrix}
{a}_{11} & {a}_{12} & {a}_{13}\\
{a}_{21} & {a}_{22} & {a}_{23}\\
{a}_{31} & {a}_{32} & {a}_{33}\\
\end{bmatrix},
{M}_B= \begin{bmatrix}
{b}_{11} & {b}_{12} & {b}_{13} \\
{b}_{21} & {b}_{22} & {b}_{23} \\
{b}_{31} & {b}_{32} & {b}_{33} \\
\end{bmatrix}.$$

則,

$$A{M}_A=
\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix} \begin{bmatrix}
{a}_{11} & {a}_{12} & {a}_{13}\\
{a}_{21} & {a}_{22} & {a}_{23}\\
{a}_{31} & {a}_{32} & {a}_{33}\\
\end{bmatrix}=
\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix},$$

$$A{M}_B=
\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}
\begin{bmatrix}
{b}_{11} & {b}_{12} & {b}_{13} \\
{b}_{21} & {b}_{22} & {b}_{23} \\
{b}_{31} & {b}_{32} & {b}_{33} \\
\end{bmatrix}=
\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}.$$

得
$$
{a}_{11}+{a}_{21}+{a}_{31}=0,\\
{a}_{12}+{a}_{22}+{a}_{32}=0,\\
...$$
以此類推。

$A({M}_A+{M}_B)=
\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}
\begin{bmatrix}
{a}_{11}+{b}_{11}&{a}_{12}+{b}_{12}&{a}_{13}+{b}_{13}\\
{a}_{21}+{b}_{21}&{a}_{22}+{b}_{22}&{a}_{23}+{b}_{23}\\
{a}_{31}+{b}_{31}&{a}_{32}+{b}_{32}&{a}_{33}+{b}_{33}\\
\end{bmatrix}.$

$${a}_{11}+{b}_{11}+{a}_{21}+{b}_{21}+{a}_{31}+{b}_{31}=0,\\
({a}_{11}+{a}_{21}+{a}_{31})+({b}_{11}+{b}_{21}+{b}_{31})=0,\\
0+0=0.$$
以此類推至各項得出，
$$A({M}_A+{M}_B)=O.
$$
所以， ${M}_A+{M}_B \in V$。

:::warning

- [x] 存在 $k \in R$, --> 令 $k\in\mathbb{R}$，則
- [x] $k*0$ --> $k\cdot 0$
:::

（2）令 $k\in\mathbb{R}$，
$$
A(k{M}_A)=
\begin{bmatrix} 1&1&1\\1&1&1\\1&1&1\end{bmatrix}
\begin{bmatrix}
k{a}_{11}&k{a}_{12}&k{a}_{13}\\
k{a}_{21}&k{a}_{22}&k{a}_{23}\\
k{a}_{31}&k{a}_{32}&k{a}_{33}
\end{bmatrix}.
$$
$$k{a}_{11}+k{a}_{21}+k{a}_{31}
\\=k({a}_{11}+{a}_{21}+{a}_{31})
\\=k\cdot 0=0.
$$
以此類推至各項得出，
$$A(k{M}_A)=O.
$$
則，$k{M}_A \in V$。

:::warning

- [x] 加一個（3）因為 $AO = O$，所以 $O\in V$。
:::

（3）因為 $AO = O$，所以 $O\in V$。

##### Exercise 3(b)

把 $X$ 寫成
$$X = \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{bmatrix}.
$$
找出一個 $9\times 9$ 的矩陣 $\Psi$
使得 $AX = O \iff \Psi\operatorname{mtov}(X) = {\bf 0}$。

:::success
Great!
:::

Ans:

$$\Psi\operatorname{mtov}(X)=
\begin{bmatrix}
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1\\
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1\\
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1\\
\end{bmatrix}
\begin{bmatrix}
a\\b\\c\\d\\e\\f\\g\\h\\i
\end{bmatrix}={\bf 0}
$$
所以，
$$\Psi=\begin{bmatrix}
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1\\
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1\\
1&0&0&1&0&0&1&0&0\\
0&1&0&0&1&0&0&1&0\\
0&0&1&0&0&1&0&0&1\\
\end{bmatrix}.
$$

##### Exercise 3(c)

求出 $V$ 的一組基底。

:::warning

- [ ] $V$ 裡面都是矩陣...，應該要把上一題的 $\Psi$ 的 $\ker(\Psi)$ 求出基底，再轉回矩陣。
:::

Ans:

計算$\Psi$的$\beta_k$得到,

$$\ker(\Psi) =
\vspan\left\{
\begin{bmatrix}
-1\\0\\0\\1\\0\\0\\0\\0\\0
\end{bmatrix},
\begin{bmatrix}
0\\-1\\0\\0\\1\\0\\0\\0\\0
\end{bmatrix}
\begin{bmatrix}
0\\0\\-1\\0\\0\\1\\0\\0\\0
\end{bmatrix}
\begin{bmatrix}
-1\\0\\0\\0\\0\\0\\1\\0\\0
\end{bmatrix}
\begin{bmatrix}
0\\-1\\0\\0\\0\\0\\0\\1\\0
\end{bmatrix}
\begin{bmatrix}
0\\0\\-1\\0\\0\\0\\0\\0\\1
\end{bmatrix}
\right\}.
$$

因此，

$$
V = \vspan\left\{
\begin{bmatrix}
-1&0&0\\1&0&0\\0&0&0
\end{bmatrix},
\begin{bmatrix}
0&-1&0\\0&1&0\\0&0&0
\end{bmatrix},
\begin{bmatrix}
0&0&-1\\0&0&1\\0&0&0
\end{bmatrix},
\begin{bmatrix}
-1&0&0\\0&0&0\\1&0&0
\end{bmatrix},
\begin{bmatrix}
0&-1&0\\0&0&0\\0&1&0
\end{bmatrix},
\begin{bmatrix}
0&0&-1\\0&0&0\\0&0&1
\end{bmatrix}
\right\}
$$

同時可驗證此集合獨立，所以此集合為 $V$ 的基底。

##### Exercise 4(a)

令 $V = \{ A \in \mathcal{M}_{n,n} : A^\top = A \}$
為所有對稱矩陣所形成的集合。
驗證 $V$ 為 $\mathcal{M}_{n,n}$ 的一個子空間、
並求出它的維度。

:::warning

- [x] $\mathcal{O}_{n,n}$ --> $O$
- [x] We may write $O^\top = {O}$.
since $A^\top = A$, we have ${A}\in V$.
In the other words, ${O}\in V$.
-->
Since $O^\top = {O}$ we have ${O}\in V$.
- [x] If ${A}$, ${B}\in V$, ${\bf r}\in R$, then ${\bf r}A$, ${\bf r}B\in V$. --> If $A\in V$ and $r\in \mathbb{R}$, then $rA\in V$.
:::

**Ans:**
**1. Claim:** ${O}\in V$.
Let ${O}$ be the $n\times n$ zero matrix in $\mathcal{M}_{n,n}$.
Since $O^\top = {O}$, we have ${O}\in V$.

**2. Claim:** If $A\in V$ and $r\in \mathbb{R}$, then $rA\in V$.
Let ${A}\in V$, ${r}\in \mathbb{R}$.
Since ($rA)^\top = rA^\top = rA$,
we have $rA\in V$.

**3. Claim:** If ${A}$, ${B}\in V$, then ${A + B}\in V$.
Let ${A},{B}\in V$.
Since ($A + B)^\top = A^\top + B^\top = {A} + {B}$,
we have ${A + B}\in V$.

As a result, $V$ is a subspace in $\mathcal{M}_{n,n}$.

Since $A^\top = A$, an $n\times n$ matrix $A$ is sysmetric where its entry $a_{ij}$ = $a_{ji}$, for all $i,j$ such that $1\le i$, $j \le n$.
Let $E_{ij}$ be the matrix whose $ij$-entry and $ji$-entry are $1$ while other entries are $0$.
Hence, $\beta = \{E_{11}, E_{22}, \ldots, E_{nn}, E_{12}+E_{21}, \ldots, E_{n-1,n}+E_{n,n-1}\}$ is a basis of ${V}$,
so $\dim(V) = \frac {n(n + 1)}{2}$.

##### Exercise 4(b)

令 $V = \{ A \in \mathcal{M}_{n,n} : A^\top = -A \}$
為所有反對稱矩陣所形成的集合。
驗證 $V$ 為 $\mathcal{M}_{n,n}$ 的一個子空間、
並求出它的維度。

**Ans:**
**1. Claim:** ${O}\in V$.
Let ${O}$ be the $n\times n$ zero matrix in $\mathcal{M}_{n,n}$.
Since $O^\top = {-O}$, we have ${O}\in V$.

**2. Claim:** If $A\in V$ and $r\in \mathbb{R}$, then $rA\in V$.
Let ${A}\in V$, ${r}\in \mathbb{R}$.
Since ($rA)^\top = rA^\top = r(-A) = - (rA)$,
we have $rA\in V$.

**3. Claim:** If ${A}$, ${B}\in V$, then ${A} + {B}\in V$.
Let ${A},{B}\in V$.
Since ($A + B)^\top = A^\top + B^\top = - ({A + B})$.
we have ${A} + {B}\in V$.

As a result, $V$ is a subspace in $\mathcal{M}_{n,n}$.

Since $A^\top = -A$, an $n\times n$ matrix $A$ is skew-sysmetric where its entry $a_{ij}$ = $-a_{ji}$, for all $i,j$ such that $1\le i$, $j \le n$ and all of the main diagonal entries must be zeroes, that is $a_{ii} = 0$
Let $E_{ij}$ be the matrix whose $ij$-entry is $1$, $ji$-entry is $-1$, while other entries are $0$.
Hence, $\beta = \{E_{12}-E_{21}, \ldots, E_{n-1,n}-E_{n,n-1}\}$ is a basis of ${V}$,
so $\dim(V) = \frac {n(n - 1)}{2}$.

:::warning
- [x] 你選的基底的元素都不是對稱矩陣
:::

##### Exercise 5

考慮向量空間 $\mathcal{P}_3$。
令
$$D = \begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}.
$$

:::warning
第 5 題目前空白
:::

##### Exercise 5(a)

令 $p$ 為 $\mathcal{P}_3$ 中的一個多項式
而 $p'$ 為其微分。
驗證 $\operatorname{ptov}(p') = D\operatorname{ptov}(p)$。

答:

令 $p=a+bx+cx^2+dx^3$，其中$a,b,c,d \in \mathbb{R}$。
則 $p'=b+2cx+3dx^2$。
$$D\operatorname{ptov}(p)=
\begin{bmatrix}0 & 1 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
a\\b\\c\\d \end{bmatrix}=\begin{bmatrix}b\\2c\\3d\\0\end{bmatrix}.
$$
則 $\operatorname{ptov}(p')=\begin{bmatrix}b\\2c\\3d\\0\end{bmatrix}=D\operatorname{ptov}(p)$。

##### Exercise 5(b)

求 $\mathcal{P}_3$ 中所有 $p' = 0$ 的 $p$。

答:

令 $p=a+bx+cx^2+dx^3$，其中 $a,b,c,d \in \mathbb{R}$。
若 $p'=b+2cx+3dx^2=0$。
則 $b,c,d=0$，且 $a$ 為任意實數。
此時 $\mathcal{P}_3$ 中的 $p=a$，其中$a \in \mathbb{R}$。

##### Exercise 5(c)

求 $\mathcal{P}_3$ 中所有 $p' = x$ 的 $p$。

答:

令 $p=a+bx+cx^2+dx^3$，其中 $a,b,c,d \in \mathbb{R}$。
若 $p'=b+2cx+3dx^2=x$。
則 $2c=1$，$b,d=0$，且 $a$ 為任意實數。
此時 $\mathcal{P}_3$ 中的 $p=a+\frac{1}{2}x^2$，其中 $a \in \mathbb{R}$。

:::info
目前分數 6.5
:::