# 特徵空間

Eigenspace

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

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from lingeo import random_int_list, random_good_matrix

## Main idea

Let $A$ be a square matrix and $\lambda\in\spec(A)$.
The **eigenspace** of $A$ with respect to $\lambda$ is defined as

$$
E_\lambda = \ker(A - \lambda I).
$$

By definition, we have $\gm(\lambda) = \dim(E_\lambda)$.

Similarly, if $f:V \rightarrow V$ is a linear function and $\lambda\in\spec(f)$,
then the **eigenspace** of $f$ with respect to $\lambda$ is

$$
E_\lambda = \ker(f - \lambda\idmap_V) = \{\bv\in V: f(v) = \lambda \bv\}.
$$

Recall that a set of subspaces $\{V_1,\ldots,V_k\}$ is linearly independent if
the only choice of ${\bf v}_1\in V_1, \ldots, {\bf v}_k\in V_k$ satisfying

$$
{\bf v}_1 + \cdots + {\bf v}_k = {\bf 0}
$$

is ${\bf v}_1 = \cdots = {\bf v}_k = {\bf 0}$.
(See 211 for more details and exercises.)

Let $A$ be an $n\times n$ matrix with distinct eigenvalues $\lambda_1, \ldots, \lambda_q$.
Then $\{E_{\lambda_1}, \ldots, E_{\lambda_q}\}$ is linearly independent.

Therefore, if $\gm(\lambda_i) = \am(\lambda_i)$,
or, equivalently, the sum of geometric multiplicity is $n$,
then one may pick a basis $\beta_{\lambda_i}$ for each $E_{\lambda_i}$ and $\beta = \beta_{\lambda_1} \cup \cdots \cup \beta_{\lambda_q}$ is a basis of $\mathbb{R}^n$.
Therefore, $A$ is diagonalizable.

In a different language, the following are equivalent.

- $A$ is diagonalizable.
- $\mathbb{R}^n = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_q}$.

That is, $\mathbb{R}^n$ can be written as the direct sum of the eigenspaces of $A$.

## Side stories

- simultaneously diagonalizable

## Experiments

##### Exercise 1

執行以下程式碼。
令 $\bu_1$ 為 $A_1$ 的行向量、
$\bu_2,\bu_3$ 為 $A_2$ 的行向量、
$\bu_4,\bu_5$ 為 $A_3$ 的行向量、
而 $R$ 為 $\begin{bmatrix} A_1 A_2 A_3 \end{bmatrix}$ 的最簡階梯型。
（已知 $\ker(A_1) = \ker(A_2) = \ker(A_3) = \{\bzero\}$。）

<!-- eng start -->
Run the code below. Let $\bu_1$ be the column of $A_1$, $\bu_2,\bu_3$ the columns of $A_2$, $\bu_4,\bu_5$ the columns of $A_3$, and $R$ the reduced echelon form of $\begin{bmatrix} A_1 A_2 A_3 \end{bmatrix}$. (Suppose $\ker(A_1) = \ker(A_2) = \ker(A_3) = \{\bzero\}$.)
<!-- eng end -->

### code
set_random_seed(0)
print_ans = False

indep = choice([True, False])

n = 6
A = random_good_matrix(n,5,5)
if not indep:
    while True:
        l = random_int_list(4)
        if 0 not in l:
            break
    A[:,-1] = A[:,[0,1,2,3]] * matrix(4, l)
    
A1 = A[:,[0]]
A2 = A[:,[1,2]]
A3 = A[:,[3,4]]
R = A.rref()

pretty_print(LatexExpr("A_1 ="), A1, LatexExpr(", A_2 ="), A2, LatexExpr("A_3 ="), A3)
pretty_print(LatexExpr("R ="), R)

if print_ans:
    print("Linearly independent?", indep)
    if indep:
        print("If v1 + v2 + v3 = 0,")
        print("then v1 = a1 u1, v2 = a2 u2 + a3 u3, v3 = a4 u4 + a5 u5,")
        print("and a1 u1 + a2 u2 + a3 u3 + a4 u4 + a5 u5 = 0.")
        print("But then {u1, u2, u3, u4, u5} is linearly independent.")
    else:
        v1 = A1 * vector(l[:1])
        v2 = A2 * vector(l[1:3])
        v3 = A3 * vector(l[3:] + [-1])
        print("v1 = (%s)u1"%l[0], v1)
        print("v2 = (%s)u2 + (%s)u3"%(l[1], l[2]), v2)
        print("v3 = (%s)u4 + (%s)u5"%(l[3], -1), v3)

##### Exercise 1(a)

判斷 $\{\Col(A_1), \Col(A_2), \Col(A_3)\}$ 是否線性獨立。

<!-- eng start -->
Determine if $\{\Col(A_1), \Col(A_2), \Col(A_3)\}$ is linearly independent.
<!-- eng end -->

##### Exercise 1(b)

若線性獨立，請說明原因。
若不線性獨立，請找出 $\bv_1\in\Col(A_1)$、$\bv_2\in\Col(A_2)$、及 $\bv_3\in\Col(A_3)$
使得 $\bv_1 + \bv_2 + \bv_3 = \bzero$ 且三向量不全為零向量。

<!-- eng start -->
If yes, provide your reasonse. If no, find nonzero vectors $\bv_1\in\Col(A_1)$, $\bv_2\in\Col(A_2)$, and $\bv_3\in\Col(A_3)$ such that $\bv_1 + \bv_2 + \bv_3 = \bzero$.
<!-- eng end -->

:::info
What do the experiments try to tell you? (open answer)

...
:::

## Exercises

##### Exercise 2

令

$$
A = \begin{bmatrix}
8 & -2 & 3 & -5 & -1 \\
-11 & 7 & -5 & 11 & 1 \\
-34 & 14 & -14 & 33 & 4 \\
0 & 0 & 0 & 3 & 0 \\
-40 & 16 & -19 & 39 & 7
\end{bmatrix}.
$$

已知 $A$ 有三個相異的特徵值 $1,2,3$，
對於 $\lambda = 1,2,3$，
找出特徵空間 $E_\lambda$ 的一組基底 $\beta_\lambda$。
判斷 $\beta = \beta_1 \cup \beta_2 \cup \beta_3$ 是否為 $\mathbb{R}^5$ 的一組基底。

<!-- eng start -->
Let

$$
A = \begin{bmatrix}
8 & -2 & 3 & -5 & -1 \\
-11 & 7 & -5 & 11 & 1 \\
-34 & 14 & -14 & 33 & 4 \\
0 & 0 & 0 & 3 & 0 \\
-40 & 16 & -19 & 39 & 7
\end{bmatrix}.
$$

It is known that $A$ has three distinct eigenvalues $1,2,3$. For each of $\lambda = 1,2,3$, find a basis for the eigenspace $E_\lambda$. Determine if $\beta = \beta_1 \cup \beta_2 \cup \beta_3$ is a basis of $\mathbb{R}^5$.
<!-- eng end -->

##### Exercise 3

令

$$
A = \begin{bmatrix}
-1 & 5 & -6 & 5 & -2 \\
4 & -9 & 11 & -8 & 3 \\
6 & -18 & 19 & -12 & 4 \\
-1 & -2 & 1 & 2 & 0 \\
-6 & 15 & -13 & 9 & 0
\end{bmatrix}.
$$

已知 $A$ 有三個相異的特徵值 $1,2,3$，
對於 $\lambda = 1,2,3$，
找出特徵空間 $E_\lambda$ 的一組基底 $\beta_\lambda$。
判斷 $\beta = \beta_1 \cup \beta_2 \cup \beta_3$ 是否為 $\mathbb{R}^5$ 的一組基底。

<!-- eng start -->
Let

$$
A = \begin{bmatrix}
-1 & 5 & -6 & 5 & -2 \\
4 & -9 & 11 & -8 & 3 \\
6 & -18 & 19 & -12 & 4 \\
-1 & -2 & 1 & 2 & 0 \\
-6 & 15 & -13 & 9 & 0
\end{bmatrix}.
$$

It is known that $A$ has three distinct eigenvalues $1,2,3$. For each of $\lambda = 1,2,3$, find a basis for the eigenspace $E_\lambda$. Determine if $\beta = \beta_1 \cup \beta_2 \cup \beta_3$ is a basis of $\mathbb{R}^5$.
<!-- eng end -->

##### Exercise 4

對以下線性函數 $f$，描述它的每一個特徵空間。

<!-- eng start -->
For the following linear functions, describe their eigenspaces.
<!-- eng end -->

##### Exercise 4(a)

令 $V$ 為 $\mathbb{R}^3$ 中的一個二維空間，
而 $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ 將向量 $\bv\in\mathbb{R}^3$ 投影到 $V$ 上。

<!-- eng start -->
Let $V$ be a $2$-dimensional subspace of $\mathbb{R}^3$ and $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ the projection that sends vectors in $\bv\in\mathbb{R}^3$ onto $V$.
<!-- eng end -->

##### Exercise 4(b)

令 $V$ 為 $\mathbb{R}^3$ 中的一個二維空間，
而 $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ 將向量 $\bv\in\mathbb{R}^3$ 鏡射到 $V$ 的對面。

<!-- eng start -->
Let $V$ be a $2$-dimensional subspace of $\mathbb{R}^3$ and $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ the reflection that sends vectors in $\bv\in\mathbb{R}^3$ the other side of $V$.
<!-- eng end -->

##### Exercise 5

令 $V_1,\ldots, V_k$ 為同一向量空間中的子空間，
且 $\{V_1, \ldots, V_k\}$ 線性獨立。
若對於 $i = 1,\ldots, k$ 分別有 $\beta_i$ 為 $V_i$ 的基底。
證明 $\beta = \beta_1 \cup \cdots \cup \beta_k$ 為 $V_1 \oplus \cdots \oplus V_k$ 的基底。

<!-- eng start -->
Let $V_1,\ldots, V_k$ be subspaces of the same universal space. Suppose $\{V_1, \ldots, V_k\}$ is linearlyl independent. For each $i = 1,\ldots, k$, let $\beta_i$ be a basis of $V_i$. Show that $\beta = \beta_1 \cup \cdots \cup \beta_k$ is a basis of $V_1 \oplus \cdots \oplus V_k$.
<!-- eng end -->

##### Exercise 6

令 $A$ 為一矩陣且其相異的特徵值為 $\lambda_1,\ldots,\lambda_q$。
證明其所有特徵空間 $\{E_{\lambda_1}, \ldots, E_{\lambda_q}\}$ 線性獨立。

（參考 311-5。）

<!-- eng start -->
Let $A$ be a matrix with distinct eigenvalues $\lambda_1,\ldots,\lambda_q$. Show that the collection of its eigenspaces $\{E_{\lambda_1}, \ldots, E_{\lambda_q}\}$ is linearly independent.

(See 311-5.)
<!-- eng end -->

##### Exercise 7

給定兩大小相同的方陣 $A$ 和 $B$，
若存在一個可逆矩陣 $Q$
使得 $Q^{-1}AQ$ 和 $Q^{-1}BQ$ 同時是對角矩陣，
則我們稱 $A$ 和 $B$ 可 **同時對角化（simultaneously diagonalizable）**。

依照以下步驟說明：
若 $A$ 和 $B$ 皆可對角化，且 $AB = BA$，
則 $A$ 和 $B$ 可同時對角化。

<!-- eng start -->
Let $A$ and $B$ be square matrices of the same size. If there is an invertible matrix $Q$ such that both $Q^{-1}AQ$ and $Q^{-1}BQ$ are diagonal matrices, then we say $A$ and $B$ are **simultaneously diagonalizable** .

Use the given instructions to show:
If $A$ and $B$ are diagonalizable with $AB = BA$, then $A$ and $B$ are simultaneously diagonalizable.
<!-- eng end -->

##### Exercise 7(a)

令 $A$ 和 $B$ 為大小相同的方陣。
令 $\lambda$ 為 $A$ 的一個特徵值，而 $E_\lambda$ 為其特徵空間。
證明若 $AB = BA$，則 $E_\lambda$ 同時為 $A$-不變子空間、也是 $B$-不變子空間。

<!-- eng start -->
Let $A$ and $B$ be square matrices of the same size. Let $\lambda$ be an eigenvalue of $A$ and $E_\lambda$ its eigenspace. Show that if $AB = BA$, then $E_\lambda$ is both $A$-invariant and $B$-invariant.
<!-- eng end -->

##### Exercise 7(b)

令 $A_1$ 和 $A_2$ 分別為 $n_1\times n_1$ 及 $n_2\times n_2$ 的可對角化的方陣。
若一 $n_1\times n_2$ 矩陣 $X$ 滿足 $A_1X = XA_2$，
且 $A_1$ 和 $A_2$ 沒有共同的特徵值，
說明 $X$ 必為零矩陣。

提示：先考慮 $A_1$ 和 $A_2$ 都是對角矩陣的情況。

<!-- eng start -->
Let $A_1$ and $A_2$ be $n_1\times n_1$ and $n_2\times n_2$ diagonalizable matrices, respectively. Suppose $X$ is an $n_1\times n_2$ matrix with $A_1X = XA_2$ and $A$ and $B$ share no common eigenvalues. Then $X$ must be the zero matrix.

Hint: Consider the case when both $A_1$ and $A_2$ are diagonal.
<!-- eng end -->

##### Exercise 7(c)

令 $A$ 和 $B$ 為大小相同的方陣且 $AB = BA$。
假設 $A$ 的相異特徵值為 $\lambda_1, \ldots, \lambda_q$、$q \geq 2$、
且 $E_{\lambda_1},\ldots,E_{\lambda_q}$ 分別為其特徵空間。
令 $V_1 = E_{\lambda_1}$ 而 $V_2 = E_{\lambda_2}\oplus \cdots \oplus E_{\lambda_q}$、
且 $\beta_1$ 和 $\beta_2$ 分別為 $V_1$ 和 $V_2$ 的基底。
取 $\beta = \beta_1 \cup \beta_2$，
證明有以下型式：

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
A_1 & O \\
O & A_2
\end{bmatrix}
\text{ and }
[f_B]_\beta^\beta =
\begin{bmatrix}
B_1 & O \\
O & B_2
\end{bmatrix},
$$

其中 $A_1$ 和 $B_1$ 的大小為 $V_1$ 的維度、
而 $A_2$ 和 $B_2$ 的大小為 $V_2$ 的維度。

<!-- eng start -->
Let $A$ and $B$ be square matrices of the same size such that $AB = BA$. Let $\lambda_1, \ldots, \lambda_q$ be the distinct eigenvalues of $A$ and $E_{\lambda_1},\ldots,E_{\lambda_q}$ their eigenspaces, respectively. Suppose $q \geq 2$. Let $V_1 = E_{\lambda_1}$ and $V_2 = E_{\lambda_2}\oplus \cdots \oplus E_{\lambda_q}$. Let $\beta_1$ and $\beta_2$ be the bases of $V_1$ and $V_2$, respectively. Let $\beta = \beta_1 \cup \beta_2$. Show that the matrices have the following form:

$$
[f_A]_\beta^\beta =
\begin{bmatrix}
A_1 & O \\
O & A_2
\end{bmatrix}
\text{ and }
[f_B]_\beta^\beta =
\begin{bmatrix}
B_1 & O \\
O & B_2
\end{bmatrix},
$$

Here the sizes of $A_1$ and $B_2$ are equal to the dimension of $V_1$, and the sizes of $A_2$ and $B_2$ are equal to the dimension of $V_2$.
<!-- eng end -->

##### Exercise 7(d)

證明將前一題 $q = 1$ 的情況，
並用數學歸納法證明：
若 $A$ 和 $B$ 皆可對角化，且 $AB = BA$，
則 $A$ 和 $B$ 可同時對角化。

<!-- eng start -->
Finish the case when $q = 1$. Then prove the statement by induction:
If $A$ and $B$ are diagonalizable with $AB = BA$, then $A$ and $B$ are simultaneously diagonalizable.
<!-- eng end -->