# 找一組好基底

Find a good basis

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

$\newcommand{\trans}{^\top}
\newcommand{\adj}{^{\rm adj}}
\newcommand{\cof}{^{\rm cof}}
\newcommand{\inp}[2]{\left\langle#1,#2\right\rangle}
\newcommand{\dunion}{\mathbin{\dot\cup}}
\newcommand{\bzero}{\mathbf{0}}
\newcommand{\bone}{\mathbf{1}}
\newcommand{\ba}{\mathbf{a}}
\newcommand{\bb}{\mathbf{b}}
\newcommand{\bc}{\mathbf{c}}
\newcommand{\bd}{\mathbf{d}}
\newcommand{\be}{\mathbf{e}}
\newcommand{\bh}{\mathbf{h}}
\newcommand{\bp}{\mathbf{p}}
\newcommand{\bq}{\mathbf{q}}
\newcommand{\br}{\mathbf{r}}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\by}{\mathbf{y}}
\newcommand{\bz}{\mathbf{z}}
\newcommand{\bu}{\mathbf{u}}
\newcommand{\bv}{\mathbf{v}}
\newcommand{\bw}{\mathbf{w}}
\newcommand{\tr}{\operatorname{tr}}
\newcommand{\nul}{\operatorname{null}}
\newcommand{\rank}{\operatorname{rank}}
%\newcommand{\ker}{\operatorname{ker}}
\newcommand{\range}{\operatorname{range}}
\newcommand{\Col}{\operatorname{Col}}
\newcommand{\Row}{\operatorname{Row}}
\newcommand{\spec}{\operatorname{spec}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\Vol}{\operatorname{Vol}}
\newcommand{\sgn}{\operatorname{sgn}}
\newcommand{\idmap}{\operatorname{id}}
\newcommand{\am}{\operatorname{am}}
\newcommand{\gm}{\operatorname{gm}}
\newcommand{\mult}{\operatorname{mult}}
\newcommand{\iner}{\operatorname{iner}}$

from lingeo import random_int_list, random_good_matrix

## Main idea

Let $A$ be an $n\times n$ matrix.
Recall that $f_A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is the function defined by $f_A(\bx) = A\bx$.
Using the standard basis $\mathcal{E}_n$, the matrix representation of $f_A$ is

$$
[f_A] = [f_A]_{\mathcal{E}_n}^{\mathcal{E}_n} = A.
$$

In constrast, the matrix representation of $f_A$ with respect to another basis $\beta$ is

$$
[f_A]_\beta^\beta = [\idmap]_{\mathcal{E}_n}^\beta [f_A] [\idmap]_\beta^{\mathcal{E}_n}.
$$

If we let $Q = [\idmap]_\beta^{\mathcal{E}_n}$, then the columns of $Q$ are the vectors in $\beta$.
Also, we can write, $[f_A]_\beta^\beta = Q^{-1}AQ$.

On a matrix level, we say $A$ and $B$ are **similar** if there is an invertible matrix $Q$ such that $B = Q^{-1}AQ$.
In other words, $A$ and $B$ are just the same linear function represented under different bases.

Suppose $\beta = \{\bv_1, \ldots, \bv_n\}$ has the nice property that

$$
A\bv_i = \lambda_i \bv_i
$$

for some values $\lambda_i$ and for $i = 1,\ldots, n$.

Then we know

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
| & ~ & | \\
[f_A(\bv_1)]_\beta & \cdots & [f_A(\bv_n)]_\beta \\
| & ~ & | \\
\end{bmatrix} =
\begin{bmatrix}
| & ~ & | \\
[\lambda_1\bv_1]_\beta & \cdots & [\lambda_n\bv_n]_\beta \\
| & ~ & | \\
\end{bmatrix} =
\begin{bmatrix}
\lambda_1 & ~ & ~ \\
~ & \ddots & ~ \\
~ & ~ & \lambda_n
\end{bmatrix},
$$

which is a diagonal matrix.
In this case, we say $A$ is **diagonalizable** .
Whenever $A\bv = \lambda\bv$ for some value $\lambda$ and some nonzero vector $\bv$,
we say $\lambda$ is an **eigenvalue** of $A$ and
$\bv$ is an **eigenvector** of $A$ with respect to $\lambda$.

Similarly, if $f: V\rightarrow V$ is a linear function and
$f(\bv) = \lambda \bv$ for some value $\lambda$ and some nonzero vector $\bv$,
then we say $\lambda$ is an eigenvalue of $f$ and
$\bv$ is an eigenvector of $f$ with respect to $\lambda$.

##### Proposition

Let $A$ be an $n\times n$ matrix. Then the following are equivalent.

- $A$ is diagonalizable.
- There is a basis $\beta$ of $\mathbb{R}^n$ that is composed of eigenvectors.
- There are an invertible matrix $Q$ and a diagonal matrix $D$ such that $D = Q^{-1}AQ$.

When $A$ is a real symmetric matrix (meaning $A\trans = A$), such a nice basis exists;
moreover, it can be chosen to be orthonormal, so the corresponding $Q$ has $Q\trans = Q^{-1}$.

##### Spectral theorem (vector version)

Let $A$ be an $n\times n$ symmetric matrix.
Then there is an orthonormal basis $\beta$ of $\mathbb{R}^n$ such that $[f_A]_\beta^\beta = D$ is a diagonal matrix.
That is, there is an orthogonal matrix $Q$ such that $Q^\top AQ = D$ is a diagonal matrix.

## Side stories

- Jordan canonical form
- algebra of $Q^{-1}AQ$
- matrix power
- simultaneously diagonalizable

## Experiments

##### Exercise 1

執行以下程式碼。
令 $\beta = \{\bv_1, \cdots, \bv_n\}$ 為 $Q$ 的各行向量。

<!-- eng start -->
Run the code below. Let $\beta = \{\bv_1, \cdots, \bv_n\}$ be the columns of $Q$.
<!-- eng end -->

### code
set_random_seed(0)
print_ans = False

n = 3
D = diagonal_matrix(random_int_list(n, 3))
Q = random_good_matrix(n,n,n, 2)
A = Q * D * Q.inverse()

pretty_print(LatexExpr("A ="), A)
pretty_print(LatexExpr("Q ="), Q)


if print_ans:
    v1, v2, v3 = Q.columns()
    u1, u2, u3 = D.columns()
    print("f_A(v1) =", A * v1)
    print("vector representation =", u1)
    print("f_A(v1) =", A * v2)
    print("vector representation =", u2)
    print("f_A(v1) =", A * v3)
    print("vector representation =", u3)
    print("matrix represesntation =")
    pretty_print(D)

When `seed = 5` 時，we get matrix$$
A = \begin{bmatrix}
-3 & -1 & -1\\
4 & -4 & -5\\
-4 & 2 & 3\\
\end{bmatrix}.
$$
and$$
Q = \begin{bmatrix}
1 & 0 & -1\\
-1 & 1 & 3\\
1 & -1 & -2\\
\end{bmatrix}.
$$

##### Exercise 1(a)

求 $f_A(\bv_1)$ 及 $[f_A(\bv_1)]_\beta$。

<!-- eng start -->
Find $f_A(\bv_1)$ and $[f_A(\bv_1)]_\beta$.
<!-- eng end -->
$Ans:$
$f_A(\bv_1)$ $=A\cdot\begin{bmatrix}
1\\
-1\\
1\\
\end{bmatrix}$ $=\begin{bmatrix}
-3\\
3\\
-3\\
\end{bmatrix}$。
Because $f_A(\bv_1)$ is represented by $\beta$ as $-3\cdot\bv_1+0\cdot\bv_2+0\cdot\bv_3$，
so $[f_A(\bv_1)]_\beta = \begin{bmatrix}
-3\\
0\\
0\\
\end{bmatrix}$ 。

##### Exercise 1(b)

求 $f_A(\bv_2)$ 及 $[f_A(\bv_2)]_\beta$。

<!-- eng start -->
Find $f_A(\bv_2)$ and $[f_A(\bv_2)]_\beta$.
<!-- eng end -->
$Ans:$
We can deduce from 1(a) that $f_A(\bv_2)=\begin{bmatrix}
0\\
1\\
-1\\
\end{bmatrix}$，$[f_A(\bv_2)]_\beta=\begin{bmatrix}
0\\
1\\
0\\
\end{bmatrix}$。

##### Exercise 1(c)

求 $f_A(\bv_3)$ 及 $[f_A(\bv_3)]_\beta$。

<!-- eng start -->
Find $f_A(\bv_3)$ and $[f_A(\bv_3)]_\beta$.
<!-- eng end -->
$Ans:$
We can deduce from 1(a) that $f_A(\bv_3)=\begin{bmatrix}
2\\
-6\\
4\\
\end{bmatrix}$，$[f_A(\bv_3)]_\beta=\begin{bmatrix}
0\\
0\\
-2\\
\end{bmatrix}$。
##### Exercise 1(d)

求 $[f_A]_\beta^\beta$。

<!-- eng start -->
Find $[f_A]_\beta^\beta$.
<!-- eng end -->
$Ans:$
We selected $[f_A(\bv_1)]_\beta$、$[f_A(\bv_2)]_\beta$、$[f_A(\bv_3)]_\beta$ to be the row vector of $[f_A]_\beta^\beta$，can get $[f_A]_\beta^\beta =\begin{bmatrix}
-3 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & -2\\
\end{bmatrix}$
:::info
What do the experiments try to tell you? (open answer)

...
:::

:::warning

Not answered.
:::

## Exercises

##### Exercise 2

對以下的矩陣 $A$ 及基底 $\beta$，
求出 $[f_A]_\beta^\beta$。

<!-- eng start -->
For each of the following matrices $A$ and bases $\beta$, find $[f_A]_\beta^\beta$.
<!-- eng end -->

##### Exercise 2(a)

$$
A = \begin{bmatrix}
5 & -1 \\
-1 & 5
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\right\}.
$$
Let $B$ be a matrix whose columns is made up of the vectors in $\beta$
then,

$$
AB = \begin{bmatrix}
5 & -1 \\
-1 & 5
\end{bmatrix}\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}=\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}[f_A]_\beta^\beta
$$

so

$$
[f_A]_\beta^\beta=B^{-1}AB=\begin{bmatrix}
4 & 0 \\
0 & 6
\end{bmatrix}.
$$

##### Exercise 2(b)

$$
A = \begin{bmatrix}
-1 & 5 \\
5 & -1
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\right\}.
$$

$Ans:$
Let the row vectors in $\beta$ be $\bv_1,\bv_2$,
and $f_A(\bv_1) = A\bv_1 = (4,4) = 4\bv_1 + 0\bv_2$，so $[f_A(\bv_1)]_\beta=(4,0)$。
Because $f_A(\bv_2) = A\bv_2 = (-6,6) = 0\bv_1 + -6\bv_2$，so $[f_A(\bv_2)]_\beta=(0,-6)$。
We know $$
[f_A]_\beta^\beta =
\begin{bmatrix}
| & ~ & | \\
[f_A(\bv_1)]_\beta & \cdots & [f_A(\bv_n)]_\beta \\
| & ~ & | \\
\end{bmatrix}。$$Finally, put the above vectors into row vectors to get $[f_A]_\beta^\beta =\begin{bmatrix}
4 & 0 \\
0 & -6\\
\end{bmatrix}$。

##### Exercise 2(c)

$$
A = \begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\right\}.
$$

$Ans:$
Let the row vectors in $\beta$ be $\bv_1,\bv_2$,
and $f_A(\bv_1) = A\bv_1 = (2,2) = 2\bv_1 + 0\bv_2$，so $[f_A(\bv_1)]_\beta=(2,0)$。
Because $f_A(\bv_2) = A\bv_2 = (0,0) = 0\bv_1 + 0\bv_2$，so $[f_A(\bv_2)]_\beta=(0,0)$。
We know

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
| & ~ & | \\
[f_A(\bv_1)]_\beta & \cdots & [f_A(\bv_n)]_\beta \\
| & ~ & | \\
\end{bmatrix}。
$$

Finally, put the above vectors into row vectors to get $[f_A]_\beta^\beta =\begin{bmatrix}
2 & 0 \\
0 & 0\\
\end{bmatrix}$。

##### Exercise 2(d)

$$
A = \begin{bmatrix}
0 & 1 \\
-6 & 5
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 3 \end{bmatrix},
\begin{bmatrix} 1 \\ 2 \end{bmatrix}
\right\}.
$$

$Ans:$
$f_A(\bv_1)=A\cdot\begin{bmatrix}
1\\
3\\
\end{bmatrix}$ $=\begin{bmatrix}
3\\
9\\
\end{bmatrix}$ ，
Because $f_A(\bv_1)$ is represented by $\beta$ as $3\cdot\bv_1+0\cdot\bv_2$，
so $[f_A(\bv_1)]_\beta = \begin{bmatrix}
3\\
0\\
\end{bmatrix}$ 。
$f_A(\bv_2)=A\cdot\begin{bmatrix}
1\\
2\\
\end{bmatrix}$ $=\begin{bmatrix}
2\\
4\\
\end{bmatrix}$ ，
Because $f_A(\bv_2)$ is represented by $\beta$ as $0\cdot\bv_1+2\cdot\bv_2$，
so $[f_A(\bv_2)]_\beta = \begin{bmatrix}
0\\
2\\
\end{bmatrix}$ 。

So put the above vector into a row vector to get $[f_A]_\beta^\beta =\begin{bmatrix}
3 & 0\\
0 & 2\\
\end{bmatrix}$ 。
##### Exercise 2(e)

$$
A = \begin{bmatrix}
0 & 1 \\
-4 & 0
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 2 \end{bmatrix},
\begin{bmatrix} 1 \\ -2 \end{bmatrix}
\right\}.
$$

$Ans:$
$f_A(\bv_1)=A\cdot\begin{bmatrix}
1\\
2\\
\end{bmatrix}$ $=\begin{bmatrix}
2\\
-4\\
\end{bmatrix}$ ，
Because $f_A(\bv_1)$ is represented by $\beta$ as $0\cdot\bv_1+2\cdot\bv_2$，
so $[f_A(\bv_1)]_\beta = \begin{bmatrix}
0\\
2\\
\end{bmatrix}$ 。
$f_A(\bv_2)=A\cdot\begin{bmatrix}
1\\
-2\\
\end{bmatrix}$ $=\begin{bmatrix}
-2\\
-4\\
\end{bmatrix}$ ，
Because $f_A(\bv_2)$ is represented by $\beta$ as $-2\cdot\bv_1+0\cdot\bv_2$,
so $[f_A(\bv_2)]_\beta = \begin{bmatrix}
-2\\
0\\
\end{bmatrix}$ 。
So put the above vector into a row vector to get $[f_A]_\beta^\beta =\begin{bmatrix}
0 & -2\\
2 & 0\\
\end{bmatrix}$。

##### Exercise 3

以下練習說明有些時候儘管矩陣無法對角化，
還是可以把 $[f_A]_\beta^\beta$ 化成一定簡單的形式。
未來會學到這些例子叫喬丹標準型。

對以下的矩陣 $A$ 及基底 $\beta$，
求出 $[f_A]_\beta^\beta$。

<!-- eng start -->
The following exercises demonstrate some examples where we can find a basis to simplify $[f_A]_\beta^\beta$ even if the matrix is not diagonalizable.

For each of the following matrices $A$ and bases $\beta$, find $[f_A]_\beta^\beta$.
<!-- eng end -->

##### Exercise 3(a)

$$
A = \begin{bmatrix}
\frac{1}{2} & -\frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2}
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\right\}.
$$

$Ans:$
$f_A(\bv_1)=A\cdot\begin{bmatrix}
1\\
1\\
\end{bmatrix}$ $=\begin{bmatrix}
0\\
0\\
\end{bmatrix}$ ，
Because $f_A(\bv_1)$ is represented by $\beta$ as $0\cdot\bv_1+0\cdot\bv_2$，
So $[f_A(\bv_1)]_\beta = \begin{bmatrix}
0\\
0\\
\end{bmatrix}$ 。
$f_A(\bv_2)=A\cdot\begin{bmatrix}
1\\
-1\\
\end{bmatrix}$ $=\begin{bmatrix}
1\\
1\\
\end{bmatrix}$ ，
Because $f_A(\bv_2)$ is represented by $\beta$ as $1\cdot\bv_1+0\cdot\bv_2$，
So $[f_A(\bv_2)]_\beta = \begin{bmatrix}
1\\
0\\
\end{bmatrix}$ 。

And $[f_A]_\beta^\beta =\begin{bmatrix}
0 & 1\\
0 & 0\\
\end{bmatrix}$ 。
##### Exercise 3(b)

$$
A = \begin{bmatrix}
0 & 1 \\
-4 & 4
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 2 \\ 4 \end{bmatrix},
\begin{bmatrix} -1 \\ 0 \end{bmatrix}
\right\}.
$$

$Ans:$

$f_A(\bv_1)=A\cdot\begin{bmatrix}
2\\
4\\
\end{bmatrix}$ $=\begin{bmatrix}
4\\
8\\
\end{bmatrix}$ ，
Because $f_A(\bv_1)$ is represented by $\beta$ as $2\cdot\bv_1+0\cdot\bv_2$，
So $[f_A(\bv_1)]_\beta = \begin{bmatrix}
2\\
0\\
\end{bmatrix}$ 。
$f_A(\bv_2)=A\cdot\begin{bmatrix}
-1\\
0\\
\end{bmatrix}$ $=\begin{bmatrix}
0\\
4\\
\end{bmatrix}$ ，
Because $f_A(\bv_2)$ is represented by $\beta$ as $1\cdot\bv_1+2\cdot\bv_2$，
So $[f_A(\bv_2)]_\beta = \begin{bmatrix}
1\\
2\\
\end{bmatrix}$ 。

And $[f_A]_\beta^\beta =\begin{bmatrix}
2 & 1\\
0 & 2\\
\end{bmatrix}$ 。
##### Exercise 3(c)

$$
A = \begin{bmatrix}
0 & 1 \\
-9 & 6
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 3 \\ 9 \end{bmatrix},
\begin{bmatrix} -1 \\ 0 \end{bmatrix}
\right\}.
$$

$f_A(\bv_1)=A\cdot\begin{bmatrix}
3\\
9\\
\end{bmatrix}$ $=\begin{bmatrix}
9\\
27\\
\end{bmatrix}$ ，
Because $f_A(\bv_1)$ is represented by $\beta$ as $3\cdot\bv_1+0\cdot\bv_2$，
So $[f_A(\bv_1)]_\beta = \begin{bmatrix}
3\\
0\\
\end{bmatrix}$ 。
$f_A(\bv_2)=A\cdot\begin{bmatrix}
-1\\
0\\
\end{bmatrix}$ $=\begin{bmatrix}
0\\
9\\
\end{bmatrix}$ ，
Because $f_A(\bv_2)$ is represented by $\beta$ as $1\cdot\bv_1+3\cdot\bv_2$，
So $[f_A(\bv_2)]_\beta = \begin{bmatrix}
1\\
3\\
\end{bmatrix}$ 。

And $[f_A]_\beta^\beta =\begin{bmatrix}
3 & 1\\
0 & 3\\
\end{bmatrix}$ 。
##### Exercise 4

對以下的矩陣 $A$ 及基底 $\beta$，
求出 $[f_A]_\beta^\beta$。

<!-- eng start -->
For each of the following matrices $A$ and bases $\beta$, find $[f_A]_\beta^\beta$.
<!-- eng end -->

##### Exercise 4(a)

$$
A = \begin{bmatrix}
4 & 0 & -1 \\
0 & 4 & -1 \\
-1 & -1 & 5
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix},
\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}
\right\}.
$$

$Ans:$
We set the row vector in $\beta$ be $\bv_1,\bv_2,\bv_3$,
then $f_A(\bv_1) = A\bv_1 = (3,3,3) = 3\bv_1 + 0\bv_2 + 0\bv_3$, so $[f_A(\bv_1)]_\beta=(3,0,0)$.
$f_A(\bv_2)=A\bv_2=(4,-4,0)= 0\bv_1 + 4\bv_2 + 0\bv_3$, so$[f_A(\bv_2)]_\beta=(0,4,0)$.
$f_A(\bv_3)=A\bv_3=(6,6,12)= 0\bv_1 + 0\bv_2 + 6\bv_3$, so$[f_A(\bv_3)]_\beta=(0,0,6)$.
And

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
| & ~ & | \\
[f_A(\bv_1)]_\beta & \cdots & [f_A(\bv_n)]_\beta \\
| & ~ & | \\
\end{bmatrix}.
$$

Finally,we put the above vectors into a set of row vectors to get $[f_A]_\beta^\beta =\begin{bmatrix}
3 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & 6
\end{bmatrix}$.

##### Exercise 4(b)

$$
A = \begin{bmatrix}
2 & -2 & 3 \\
-2 & 2 & 3 \\
3 & 3 & -3
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},
\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix},
\begin{bmatrix} 1 \\ 1 \\ -2 \end{bmatrix}
\right\}.
$$

$Ans:$
We set the row vector in $\beta$ be $\bv_1,\bv_2,\bv_3$ .
Then $f_A(\bv_1) = A\bv_1 = (3,3,3) = 3\bv_1 + 0\bv_2 + 0\bv_3$, so $[f_A(\bv_1)]_\beta=(3,0,0)$.
$f_A(\bv_2)=A\bv_2=(4,-4,0)= 0\bv_1 + 4\bv_2 + 0\bv_3$, so $[f_A(\bv_2)]_\beta=(0,4,0)$.
$f_A(\bv_3)=A\bv_3=(-6,-6,12)= 0\bv_1 + 0\bv_2 + (-6)\bv_3$, so $[f_A(\bv_3)]_\beta=(0,0,-6)$.

And

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
| & ~ & | \\
[f_A(\bv_1)]_\beta & \cdots & [f_A(\bv_n)]_\beta \\
| & ~ & | \\
\end{bmatrix}.
$$

Finally,we put the above vectors into a set of row vectors to get
$[f_A]_\beta^\beta =\begin{bmatrix}
3 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & -6
\end{bmatrix}$。

##### Exercise 4(c)

$$
A = \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
6 & -11 & 6
\end{bmatrix}
\quad\text{and}\quad
\beta = \left\{
\begin{bmatrix} 1 \\ 3 \\ 9 \end{bmatrix},
\begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix},
\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}
\right\}.
$$

$Ans:$
We set the row vector in $\beta$ be $\bv_1,\bv_2,\bv_3$.
Then $f_A(\bv_1) = A\bv_1 = (3,9,27) = 3\bv_1 + 0\bv_2 + 0\bv_3$, so $[f_A(\bv_1)]_\beta=(3,0,0)$.
$f_A(\bv_2)=A\bv_2=(2,4,8)= 0\bv_1 + 2\bv_2 + 0\bv_3$, so $[f_A(\bv_2)]_\beta=(0,2,0)$.
$f_A(\bv_3)=A\bv_3=(1,1,1)= 0\bv_1 + 0\bv_2 + 1\bv_3$, so $[f_A(\bv_3)]_\beta=(0,0,1)$.

And

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
| & ~ & | \\
[f_A(\bv_1)]_\beta & \cdots & [f_A(\bv_n)]_\beta \\
| & ~ & | \\
\end{bmatrix}.
$$

Finally,we put the above vectors into a set of row vectors to get $[f_A]_\beta^\beta =\begin{bmatrix}
3 & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 1
\end{bmatrix}$.

##### Exercise 5

令 $A$ 及 $B$ 為兩 $n\times n$ 矩陣，而 $Q$ 為一 $n\times n$ 可逆矩陣。

<!-- eng start -->
Let $A$ and $B$ be $n\times n$ matrices and $Q$ an $n\times n$ invertible matrix.
<!-- eng end -->

##### Exercise 5(a)

以矩陣運算說明

$$
Q^{-1}AQ + Q^{-1}BQ = Q^{-1}(A + B)Q.
$$

<!-- eng start -->
By matrix arithemic, show that

$$
Q^{-1}AQ + Q^{-1}BQ = Q^{-1}(A + B)Q.
$$
<!-- eng end -->

Ans:
We are already known that the matrix has a distribution law:
$$
A(B+C)=AB+AC, \quad (A+B)C=AC+BC
$$
Therefore, we can first propose $Q^{-1}$ from $Q^{-1}AQ + Q^{-1}BQ$, and we will get
$Q^{-1}(AQ+BQ).$
Then put $Q$ forward to get $Q^{-1}((A+B)Q)$, it is $Q^{-1}(A + B)Q.$
So we proved $Q^{-1}AQ + Q^{-1}BQ = Q^{-1}(A + B)Q.$

##### Exercise 5(b)

利用 $f_A + f_B = f_{A + B}$ 的性質說明

$$
Q^{-1}AQ + Q^{-1}BQ = Q^{-1}(A + B)Q.
$$

<!-- eng start -->
Use the fact $f_A + f_B = f_{A + B}$ to show that

$$
Q^{-1}AQ + Q^{-1}BQ = Q^{-1}(A + B)Q.
$$
<!-- eng end -->

**[由孫心提供]**
##### Exercise 5(b)-- answer here:
Since $Q^{-1}$ exist, the column vector of $Q^{-1}$ is a basis in the domain $\beta$.
To observe $f_A$ ， $f_A$ will send a vector $\bv$ onto $A\bv$, if we use $\beta$ to obserse, it can be regarded as sending $[\bv]_\beta$ onto $Q^{-1}AQ[\bv]_\beta$.
And using $\beta$ to observe $f_B\, \ f_{A+B}$ will get a similar conclusion.

According to the property of $f_A + f_B = f_{A + B}$, the equation can be obtained
$$
(Q^{-1}AQ + Q^{-1}BQ)[\bv]_\beta = Q^{-1}(A + B)Q[\bv]_\beta,\\
[\ (Q^{-1}AQ + Q^{-1}BQ) - Q^{-1}(A + B)Q\ ][\bv]_\beta = \bzero.
$$

Since this equation is true for any $\bv$, so $Q^{-1}AQ + Q^{-1}BQ = Q^{-1}(A + B)Q$.

##### Exercise 5(c)

以矩陣運算說明

$$
(Q^{-1}AQ)^n = Q^{-1}A^nQ.
$$

<!-- eng start -->
By matrix arithemic, show that

$$
(Q^{-1}AQ)^n = Q^{-1}A^nQ.
$$
<!-- eng end -->

**[由孫心提供]**
##### Exercise 5(c)-- answer here:
We already know that $QQ^{-1}=Q^{-1}Q=I$ ， $I$ identity matrix, so
$$
\begin{aligned}
(Q^{-1}AQ)^n &= (Q^{-1}AQ)(Q^{-1}AQ)...(Q^{-1}AQ) \\
&= Q^{-1}A(QQ^{-1})A(Q...Q^{-1})AQ \\
&= Q^{-1}A^{n}Q.
\end{aligned}
$$

##### Exercise 5(d)

利用 $f_A \circ \cdots \circ f_A = f_{A^n}$ 的性質說明

$$
(Q^{-1}AQ)^n = Q^{-1}A^nQ,
$$

其中 $f_A \circ \cdots \circ f_A$ 是指將 $f_A$ 函數和自己合成 $n$ 次。

<!-- eng start -->
Use the fact $f_A \circ \cdots \circ f_A = f_{A^n}$ to show that

$$
(Q^{-1}AQ)^n = Q^{-1}A^nQ,
$$

where $f_A \circ \cdots \circ f_A$ is the composition of $f_A$ with itself $n$ times.
<!-- eng end -->

**[由孫心提供]**
##### Exercise 5(d)-- answer here:
Since $Q$ is invertible, let $\beta$ be a basis of the column vector of $Q's$ domain.
To observe $f_A$, $f_A$ will send a vector $\bv$ to $A\bv$,and if we use $\beta$ to observe this action, it can be regarded as the expression $[\bv]_\beta$ sent to the notation $[f_A]_\beta^\beta[\bv]_\beta$ 。
That is, multiply $[\bv]_\beta$ before $Q^{-1}AQ$ 。
And $f_A \circ \cdots \circ f_A$ refers to combining the $f_A$ function with itself $n$ times, which is equivalent to sending $\bv$ $n$ times.
Then if we use $\beta$ to observe, it is to multiply $[\bv]_\beta$ before $(Q^{-1}AQ)^n$.

Observe $f_{A^n}$, $f_{A^n}$ is to send a vector $\bv$ to $A^n\bv$, if you use $\beta$ to observe, it is in $[ \bv]_\beta$ is multiplied by $Q^{-1}A^nQ$.

Then the equation can be obtained
$$
(Q^{-1}AQ)^n[\bv]_\beta = Q^{-1}A^nQ[\bv]_\beta,\\
((Q^{-1}AQ)^n - Q^{-1}A^nQ)[\bv]_\beta = \bzero.
$$
Since this equation is true for all $\bv$, $(Q^{-1}AQ)^n = Q^{-1}A^nQ$.

##### Exercise 5(e)

說明若 $A$ 可逆，則

$$
(Q^{-1}AQ)^{-1} = Q^{-1}A^{-1}Q.
$$

<!-- eng start -->
Show that

$$
(Q^{-1}AQ)^{-1} = Q^{-1}A^{-1}Q
$$

if $A$ is an invertible matrix.
<!-- eng end -->

**[由孫心提供]**
##### Exercise 5(e)-- answer here:
Can be directly calculated

$$
Q^{-1}AQ(Q^{-1}A^{-1}Q) = Q^{-1}AA^{-1}Q = Q^{-1}Q = I
$$

and

$$
(Q^{-1}A^{-1}Q)Q^{-1}AQ = Q^{-1}A^{-1}AQ = Q^{-1}Q = I.
$$

In this way, it can be verified that the inverse matrix of $Q^{-1}AQ$ is $Q^{-1}A^{-1}Q$.

##### Exercise 6

對以下矩陣 $A$，利用 5(c) 的結果計算 $A^n$。

<!-- eng start -->
For the following matrices $A$, use the result in 5(c) to find $A^n$.
<!-- eng end -->

##### Exercise 6(a)

$$
A = \begin{bmatrix}
0 & 1 \\
-6 & 5
\end{bmatrix}.
$$

（參考 2(d)。）

<!-- eng start -->
$$
A = \begin{bmatrix}
0 & 1 \\
-6 & 5
\end{bmatrix}.
$$

See 2(d).
<!-- eng end -->

**[由孫心提供]**
##### Exercise 6(a)-- answer here:
There exists $\bv_1=\begin{bmatrix}1\\3\end{bmatrix}, \bv_2=\begin{bmatrix}1\\2\end{bmatrix}$
Make $A\bv_1=\begin{bmatrix}3\\9\end{bmatrix}=3\bv_1, A\bv_2=\begin{bmatrix}2\\4\end{bmatrix}=2\bv_2.$
Therefore, $Q=\begin{bmatrix}1&1\\3&2\end{bmatrix}$ can be set so that $Q^{-1}AQ=\begin{bmatrix}3&0\\0&2\end{bmatrix}=D,$
$A^n=Q^{-1}D^n Q=\begin{bmatrix}-2&1\\3&-1\end{bmatrix}\begin{bmatrix}3^n &0\\0&2^n\end{bmatrix}\begin{bmatrix}1&1\\3&2\end{bmatrix}=\begin{bmatrix}3(2^n)-2(3^n)&2(2^n)-2(3^n)\\-3(2^n)+3(3^n)&-2(2^n)+3(3^n)\end{bmatrix}.$

##### Exercise 6(b)

$$
A = \begin{bmatrix}
0 & 1 \\
-4 & 4
\end{bmatrix}.
$$

（參考 3(b)。）

<!-- eng start -->
$$
A = \begin{bmatrix}
0 & 1 \\
-4 & 4
\end{bmatrix}.
$$

See 3(b).
<!-- eng end -->

**[由孫心提供]**
##### Exercise 6(b)-- answer here:

Exists $\bv_1=\begin{bmatrix}2\\4\end{bmatrix},
\bv_2=\begin{bmatrix}-1\\0\end{bmatrix}$
Make $A\bv_1=\begin{bmatrix}4\\8\end{bmatrix}=2\bv_1,
A\bv_2=\begin{bmatrix}0\\4\end{bmatrix}=\bv_1+2\bv_2.$
Therefore, $Q=\begin{bmatrix}2&-1\\4&0\end{bmatrix}$ can be set so that $Q^{-1}AQ=\begin{bmatrix}2&1\\0&2\end{bmatrix}=D$
$A^n
=Q^{-1}D^n Q
=\begin{bmatrix}0&1/4\\-1&1/2\end{bmatrix}
\begin{bmatrix}2^{n}&(2^{n}n)/2\\0&2^{n}\end{bmatrix}
\begin{bmatrix}2&-1\\4&0\end{bmatrix}.$

##### Exercise 7

令 $A$ 及 $B$ 為兩 $n\times n$ 矩陣，而 $Q$ 為一 $n\times n$ 可逆矩陣。
若 $Q^{-1}AQ$ 和 $Q^{-1}BQ$ 都是對角矩陣，證明 $AB = BA$。

<!-- eng start -->
Let $A$ and $B$ be $n\times n$ matrices and $Q$ an $n\times n$ invertible matrix. Suppose both $Q^{-1}AQ$ and $Q^{-1}BQ$ are diagonal matrices. Show that $AB = BA$.
Answer:

Let $D_1$ and $D_2$ be two diagonal matrices. We may assume

$$
D_1 = \begin{bmatrix}
\ a_i & ~ & ~ \\
~ & \ddots & ~ \\
~ & ~ & \ a_n
\end{bmatrix} \text{ and } D_2 = \begin{bmatrix}
\ b_i & ~ & ~ \\
~ & \ddots & ~ \\
~ & ~ & \ b_n
\end{bmatrix}.
$$

From the two diagonal matrices above, we may check

$$
D_1D_2 = \begin{bmatrix}
\ a_ib_i & ~ & ~ \\
~ & \ddots & ~ \\
~ & ~ & \ a_nb_n
\end{bmatrix} = \begin{bmatrix}
\ b_ia_i & ~ & ~ \\
~ & \ddots & ~ \\
~ & ~ & \ b_na_n
\end{bmatrix} = D_2D_1.
$$

Since $Q^{-1}AQ$ and $Q^{-1}BQ$ are diagonal matrices, we have $(Q^{-1}AQ)(Q^{-1}BQ) = (Q^{-1}BQ)(Q^{-1}AQ)$, which means $Q^{-1}ABQ = Q^{-1}BAQ$ since $QQ^{-1} = I$.
After that, we __pre-multiply $Q$__ on the both sides and get $ABQ = BAQ$.
Similarly, we __post-multiply $Q^{-1}$__ on the both sides and get $AB = BA$.

Therefore, $AB = BA$.
<!-- eng end -->

:::info

collaboration: 1
4 problems: 4

- 2abcd

extra: 4.5

- 2e, 3abc, 4abc, 5a, 7

moderator: 1
qc: 1
:::