矩阵的特征分解

实对称矩阵的谱分解·

实对称矩阵的特征值分解成为谱分解。给定一个 $n\times n$ 的实对称矩阵 $A$ ，其特征值分解可以写成

A = V \Lambda V^{-1} = V \Lambda V^{\top}

其中 $V$ 为正交矩阵，每一列对应一个特征向量， $\Lambda$ 为对角矩阵，对角元素为特征值。

将矩阵 $A$ 作用于二维向量 $x$ ，对应运算

Ax = y

将其拆解为三个连续变换

Ax = V \Lambda V^{\top}x = y

用参考资料中的例子， $A = \begin{bmatrix} \dfrac{5}{4} & -\dfrac{3}{4}\\-\dfrac{3}{4} & \dfrac{5}{4}\end{bmatrix}$

\begin{aligned} A &= \begin{bmatrix} \dfrac{5}{4} & -\dfrac{3}{4}\\-\dfrac{3}{4} & \dfrac{5}{4}\end{bmatrix}\\ V \Lambda V^{\top}&= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} 2 & 0\\ 0 & \dfrac{1}{2}\\ \end{bmatrix} \begin{bmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2}\\ \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \end{aligned}

对两个标准正交基做几何变换

\begin{aligned} Ae_1 &= \begin{bmatrix} \dfrac{5}{4} & -\dfrac{3}{4}\\-\dfrac{3}{4} & \dfrac{5}{4}\end{bmatrix} \begin{bmatrix} 1\\0 \end{bmatrix} = \begin{bmatrix} \dfrac{5}{4}\\-\dfrac{3}{4} \end{bmatrix}\\ Ae_1 &= \begin{bmatrix} \dfrac{5}{4} & -\dfrac{3}{4}\\-\dfrac{3}{4} & \dfrac{5}{4}\end{bmatrix} \begin{bmatrix} 0\\1 \end{bmatrix} = \begin{bmatrix} -\dfrac{3}{4}\\\dfrac{5}{4} \end{bmatrix} \end{aligned}

首先 $V^{\top}$ 先做旋转变换，

\begin{aligned} V^{\top}e_1 &= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2}\\ \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} 1\\0 \end{bmatrix} = \begin{bmatrix} \dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2} \end{bmatrix}\\ V^{\top}e_2 &= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2}\\ \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} 0\\1 \end{bmatrix} = \begin{bmatrix} -\dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2} \end{bmatrix}\\ \end{aligned}

$\Lambda$ 做缩放，

\begin{aligned} \Lambda(V^{\top}e_1) &= \begin{bmatrix} 2 & 0\\ 0 & \dfrac{1}{2}\\ \end{bmatrix} \begin{bmatrix} \dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2} \end{bmatrix} = \begin{bmatrix} \dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{4} \end{bmatrix}\\ \Lambda(V^{\top}e_2) &= \begin{bmatrix} 2 & 0\\ 0 & \dfrac{1}{2}\\ \end{bmatrix} \begin{bmatrix} -\dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2} \end{bmatrix} = \begin{bmatrix} -\dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{4} \end{bmatrix}\\ \end{aligned}

最后 $V$ 做旋转变换，

\begin{aligned} V(\Lambda V^{\top}e_1) &= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} \dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{4} \end{bmatrix} = \begin{bmatrix} \dfrac{5}{4}\\-\dfrac{3}{4} \end{bmatrix}\\ V(\Lambda V^{\top}e_2) &= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} -\dfrac{\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{4} \end{bmatrix} = \begin{bmatrix} -\dfrac{3}{4}\\\dfrac{5}{4} \end{bmatrix}\\ \end{aligned}

矩阵 $A$ 的特征向量为

v_1 = \begin{bmatrix} \dfrac{\sqrt{2}}{2}\\ -\dfrac{\sqrt{2}}{2}\\ \end{bmatrix},\space v_2 = \begin{bmatrix} \dfrac{\sqrt{2}}{2}\\ \dfrac{\sqrt{2}}{2}\\ \end{bmatrix}

变换 $V\Lambda V^{\top}$ 对特征向量 $v_1, v_2$ 的作用为

\begin{aligned} V^{\top} \begin{bmatrix} v_1 & v_2 \end{bmatrix} &= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & -\dfrac{\sqrt{2}}{2}\\ \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}\\ \Lambda V^{\top} \begin{bmatrix} v_1 & v_2 \end{bmatrix} &= \begin{bmatrix} 2 & 0\\ 0 & \dfrac{1}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix} = \begin{bmatrix} 2 & 0\\ 0 & \dfrac{1}{2}\\ \end{bmatrix}\\ V\Lambda V^{\top} &= \begin{bmatrix} \dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ -\dfrac{\sqrt{2}}{2} & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} 2 & 0\\ 0 & \dfrac{1}{2}\\ \end{bmatrix} = \begin{bmatrix} 2v_1 & \dfrac{1}{2}v_2\\ \end{bmatrix} \end{aligned}

可视化的 Python 程序见附录。

瑞利商·

给定实数矩阵 $A$ ，定义其瑞利商为

R(x) = \dfrac{x^{\top}Ax}{x^{\top}x}

当向量 $x$ 为矩阵特征向量时， $R(x)$ 为对应的特征值

v_1^{\top}Av_1 = v_1^{\top}\lambda v_1\Rightarrow \lambda = \dfrac{v_1^{\top}Av_1}{v_1^{\top} v_1}

对于实对称矩阵 $A$ ，瑞利商的最大值为最大特征值，最小值为最小特征值，驻点对应所有特征向量。

圆锥曲线——椭圆·

设椭圆半长轴的长度为 $\lambda_1$ ，半短轴的长度为 $\lambda_2$ ，椭圆的解析式为

\dfrac{x_1^2}{\lambda_1^2} + \dfrac{x_2^2}{\lambda_2^2} = 1

写成矩阵乘法的形式

\begin{bmatrix} x_1 & x_2\\ \end{bmatrix} \begin{bmatrix} \dfrac{1}{\lambda_1^2} & 0\\ 0 & \dfrac{1}{\lambda_2^2}\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\\ \end{bmatrix} = 1

设单位向量 $z = \begin{bmatrix} z_1 \\ z_2 \end{bmatrix}$ ，满足 $z^{\top}z = z_1^2 = z_2^2 = 1$ ，

矩阵 $A$ 作用与 $z$ 后得到椭圆上的向量 $x$ ， $Az=x$ ，则 $z=A^{-1}x$

则

\begin{aligned} (A^{-1}x)^{\top}A^{-1}x &= 1\\ x^{\top}(A^{-1})^{\top}A^{-1}x &= 1\\ x^{\top}(V\Lambda^{-1}V^{\top})^{\top}V\Lambda^{-1}V^{\top}x &= 1\\ x^{\top}V\Lambda^{-1}V^{\top}V\Lambda^{-1}V^{\top}x &= 1\\ (V^{\top}x)^{\top}\Lambda^{-2}(V^{\top}x) &= 1\\ \end{aligned}

令 $y = V^{\top}x$ ，则

y^{\top} \Lambda^{-2}y = 1

即

\begin{aligned} \begin{bmatrix} y_1 & y_2\\ \end{bmatrix} \begin{bmatrix} \dfrac{1}{\lambda_1^2} & 0\\ 0 & \dfrac{1}{\lambda_2^2}\\ \end{bmatrix} \begin{bmatrix} y_1 \\ y_2\\ \end{bmatrix} &= 1\\ \dfrac{y_1^2}{\lambda_1^2} + \dfrac{y_2^2}{\lambda_2^2} &= 1 \end{aligned}

非对称方阵的特征分解·

Gram 矩阵·

设矩阵 $A$ 为 $m\times p$ 的矩阵， $A$ 的 Gram 矩阵为

G = A^{\top}A = \begin{bmatrix} \langle a_0, a_0 \rangle & \langle a_0, a_1 \rangle & \cdots & \langle a_0, a_{p-1} \rangle\\ \langle a_1, a_0 \rangle & \langle a_1, a_1 \rangle & \cdots & \langle a_1, a_{p-1} \rangle\\ \vdots & \vdots & \ddots & \vdots\\ \langle a_{m-1}, a_0 \rangle & \langle a_{m-1}, a_1 \rangle & \cdots & \langle a_{m-1}, a_{p-1} \rangle\\ \end{bmatrix}

形状为 $p\times p$ ，是实对称矩阵，并且 Gram 矩阵是半正定矩阵，对于任意 $p$ 维非零列向量 $x$ 有

x^{\top}Gx = x^{\top}A^{\top}Ax = (Ax)^{\top}(Ax) = ||Ax||^2 \geq 0

非对称方阵的 EVD·

设矩阵 $A=\begin{bmatrix} 2 & 1\\0 & 1\\ \end{bmatrix}$ ，分解为

A = V\Lambda V^{-1}= \begin{bmatrix} 1 & -\dfrac{\sqrt{2}}{2}\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} 2 & 0\\ 0 & 1\\ \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & \sqrt{2}\\ \end{bmatrix}

下面考查非对称矩阵 $A$ 对标准正交基的作用，与对称矩阵不同的是 $V$ 和 $V^{-1}$ 的几何变换

将 $V$ 拆成两个矩阵相乘的形式

V = \begin{bmatrix} 1 & -\dfrac{\sqrt{2}}{2}\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} = \begin{bmatrix} 1 & -1\\ 0 & 1\\ \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} = SK

分别对应了缩放和剪切的几何变换，同理

V^{-1} = \begin{bmatrix} 1 & 1\\ 0 & \sqrt{2}\\ \end{bmatrix} = \begin{bmatrix} 1 & 0\\ 0 & \sqrt{2}\\ \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix} = K^{-1}S^{-1}

因为 $\det{(V)}\det{(V^{-1})}=1$ ，因此 $V, V^{-1}$ 对于面积的影响可以相互抵消。因此矩阵 $A$ 对于面积的影响全来自于 $\Lambda$
$\det{(A)} = \det{(\Lambda)}$

总得来说，非对称矩阵 $A$ 表示为

A = V\Lambda V^{-1}= = \underbrace{SK}_V \Lambda \underbrace{K^{-1}S^{-1}}_{V^{-1}}

对应的几何变换为 剪切 → 缩放 → 缩放 → 缩放 → 剪切。

A = V\Lambda V^{-1}= \begin{bmatrix} 1 & -\dfrac{\sqrt{2}}{2}\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} \begin{bmatrix} 2 & 0\\ 0 & 1\\ \end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & \sqrt{2}\\ \end{bmatrix}

矩阵 $A$ 的特征向量为

[v_1, v_2] = \begin{bmatrix} 1 & -\dfrac{\sqrt{2}}{2}\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix}

矩阵 $A$ 对特征向量 $[v_1, v_2]$ 的作用为

A [v_1, v_2] = \begin{bmatrix} 2 & 1\\0 & 1\\ \end{bmatrix} \begin{bmatrix} 1 & -\dfrac{\sqrt{2}}{2}\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} = \begin{bmatrix} 2 & -\dfrac{\sqrt{2}}{2}\\ 0 & \dfrac{\sqrt{2}}{2}\\ \end{bmatrix} = [2v_1, v_2]

参考资料·

Visualize-ML/Linear-Algebra-Made-Easy—Learn-with-Python-and-Visualization

附录·

可视化：分别从标准正交基和特征向量的角度看谱分解（Python 代码由 GPT-5.1 参考正文内容生成，仅供参考）

import numpy as np
import matplotlib.pyplot as plt


# Standard basis
e1 = np.array([1, 0])
e2 = np.array([0, 1])
# Specify eigenvectors and eigenvalues
v1 = np.array([np.sqrt(2)/2, -np.sqrt(2)/2])   # eigenvector for eigenvalue 2
v2 = np.array([np.sqrt(2)/2,  np.sqrt(2)/2])   # eigenvector for eigenvalue 1/2

V = np.column_stack([v1, v2])
Lambda = np.diag([2, 0.5])
VT = V.T

# Construct A = V Λ V^T
A = V @ Lambda @ VT

# A acting on standard basis
Ae1 = A.dot(e1)
Ae2 = A.dot(e2)

# Unit circle
theta = np.linspace(0, 2*np.pi, 400)
circle = np.vstack([np.cos(theta), np.sin(theta)])
ellipse = A.dot(circle)  # image of unit circle → ellipse

# A acting on eigenvectors (pure scaling)
Av1 = A.dot(v1)
Av2 = A.dot(v2)

# ------------------------------------------------------
# Visualization
# ------------------------------------------------------
fig, axes = plt.subplots(1, 2, figsize=(12, 6))

# ------------------------------------------------------
# Figure 1: Standard basis view
# ------------------------------------------------------
ax = axes[0]
ax.set_title("Spectral Decomposition from Standard Basis")
ax.set_xlim(-2, 2)
ax.set_ylim(-2, 2)
ax.axhline(0, color='gray')
ax.axvline(0, color='gray')
ax.set_aspect('equal')

# Unit circle
ax.plot(circle[0], circle[1], 'k--', label="Unit Circle")

# Ellipse
ax.plot(ellipse[0], ellipse[1], 'r', label="Ellipse = A(Unit Circle)")

# Standard basis
ax.arrow(0, 0, e1[0], e1[1], width=0.01, color='black', label='e1')
ax.arrow(0, 0, e2[0], e2[1], width=0.01, color='black', label='e2')

# A acting on basis vectors
ax.arrow(0, 0, Ae1[0], Ae1[1], width=0.01, color='red', label='A e1')
ax.arrow(0, 0, Ae2[0], Ae2[1], width=0.01, color='blue', label='A e2')

ax.legend(loc="upper left")


# ------------------------------------------------------
# Figure 2: Eigenvector basis view
# ------------------------------------------------------
ax = axes[1]
ax.set_title("Spectral Decomposition from Eigenvector Basis")
ax.set_xlim(-2, 2)
ax.set_ylim(-2, 2)
ax.axhline(0, color='gray')
ax.axvline(0, color='gray')
ax.set_aspect('equal')

# Unit circle
ax.plot(circle[0], circle[1], 'k--', label="Unit Circle")

# Ellipse
ax.plot(ellipse[0], ellipse[1], 'r', label="Ellipse = A(Unit Circle)")

# Eigenvectors
ax.arrow(0, 0, v1[0], v1[1], width=0.01, color='purple', label='v1')
ax.arrow(0, 0, v2[0], v2[1], width=0.01, color='brown', label='v2')

# Eigenvalues scaling the eigenvectors → ellipse axes
ax.arrow(0, 0, Av1[0], Av1[1], width=0.01, color='purple', alpha=0.6, label='lambda1 * v1')
ax.arrow(0, 0, Av2[0], Av2[1], width=0.01, color='brown', alpha=0.6, label='lambda2 * v2')

ax.legend(loc="upper left")

plt.tight_layout()
plt.show()

可视化：圆锥曲线——椭圆（Python 代码由 GPT-5.1 参考正文内容生成，仅供参考）

import numpy as np
import matplotlib.pyplot as plt

# Ellipse parameters: semi-major axis λ1, semi-minor axis λ2
lambda_1 = 3.0
lambda_2 = 1.5

# Construct the diagonal matrix Λ
Lambda = np.diag([lambda_1, lambda_2])

# A = V Λ V^T, with V taken as a rotation matrix (angle is arbitrary)
theta = np.deg2rad(30)  # ellipse rotation angle: 30°
V = np.array([[np.cos(theta), -np.sin(theta)],
              [np.sin(theta),  np.cos(theta)]])

A = V @ Lambda @ V.T

# Parametric form of the ellipse (unit circle → ellipse)
t = np.linspace(0, 2*np.pi, 400)
z = np.vstack([np.cos(t), np.sin(t)])  # points on the unit circle

# Apply A to z to get the ellipse: x = A z
x = A @ z

# Coordinate transformation y = V^T x (ellipse → axis-aligned ellipse)
y = V.T @ x

# ==== ==== ==== ==== ==== ====
#       Visualization
# ==== ==== ==== ==== ==== ====
plt.figure(figsize=(7,7))
plt.plot(x[0, :], x[1, :], label='Ellipse x = A z', linewidth=2)
plt.plot(y[0, :], y[1, :], label='y = V^T x (Axis-aligned ellipse)', linestyle='--')

# Unit circle
plt.plot(z[0, :], z[1, :], label='Unit Circle z', linestyle=':')

# Draw axes
plt.axhline(0, color='black', linewidth=0.5)
plt.axvline(0, color='black', linewidth=0.5)

plt.gca().set_aspect('equal', 'box')
plt.legend()
plt.title("Ellipse Visualization by Matrix Transformations")
plt.xlabel("x1 / y1")
plt.ylabel("x2 / y2")
plt.grid(True)

plt.show()

矩阵的特征分解

实对称矩阵的谱分解·

瑞利商·

圆锥曲线——椭圆·

非对称方阵的特征分解·

Gram 矩阵·

非对称方阵的 EVD·

参考资料·

附录·

评论

搜索文章

实对称矩阵的谱分解·

瑞利商·

圆锥曲线——椭圆·

非对称方阵的特征分解·

Gram 矩阵·

非对称方阵的 EVD·

参考资料·

附录·

相关文章

从复数到四元数

信息论基础

泊松分布

评论