The Characteristic Equation | 特征方程

📚 The Characteristic Equation | 特征方程

In AQA Further Mathematics, the characteristic equation is the key algebraic tool for finding eigenvalues of a square matrix. It turns the geometric idea of an invariant direction into a polynomial equation that can be solved using standard algebraic techniques.

在 AQA 进阶数学中，特征方程是求方阵特征值的关键代数工具。它将“不变方向”这一几何思想转化为一个可用标准代数方法求解的多项式方程。

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1. What is the Characteristic Equation? | 什么是特征方程？

fined as det(A − λI) = 0, where I is the n × n identity matrix and λ is an unknown scalar.

对于 n × n 方阵 A，特征方程定义为 det(A − λI) = 0，其中 I 是 n × n 单位矩阵，λ 是未知标量。

The left-hand side, det(A − λI), is called the characteristic polynomial. Its roots are the eigenvalues of A.

左边的 det(A − λI) 称为特征多项式。它的根就是 A 的特征值。

det(A − λI) = 0

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2. Eigenvalues and Eigenvectors: The Link | 特征值与特征向量的联系

= λv for some non-zero vector v. The vector v is called an eigenvector corresponding to λ.

矩阵 A 的特征值 λ 满足 Av = λv，其中 v 是非零向量。向量 v 称为对应于 λ 的特征向量。

Geometrically, an eigenvector is a direction that remains unchanged under the transformation represented by A, apart from being stretched or compressed by the factor λ.

从几何上看，特征向量是在矩阵 A 所表示的线性变换下方向保持不变的向量，只被 λ 拉伸或压缩。

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3. Deriving the Characteristic Equation | 特征方程的推导

Starting from Av = λv, we can rearrange using the identity matrix I:

从 Av = λv 出发，利用单位矩阵 I 可以重写为：

Av − λIv = 0, so (A − λI)v = 0

Since v is non-zero, the matrix A − λI must be singular. Therefore its determinant must be zero.

因为 v 非零，所以矩阵 A − λI 必须是奇异的，因此它的行列式必须为零。

characteristic equation det(A − λI) = 0, which is the condition for λ to be an eigenvalue.

这就得到了特征方程 det(A − λI) = 0，它是 λ 成为特征值的条件。

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4. The 2 × 2 Case: Trace and Determinant Form | 2 × 2 情形：迹与行列式形式

s a, b, c, d, the characteristic polynomial can be written directly using the trace and determinant.

对于一般的 2 × 2 矩阵 A，其元素为 a, b, c, d，特征多项式可以直接用迹和行列式写出。

If A is given by A = [[a, b], [c, d]], then:

若 A 为 A = [[a, b], [c, d]]，则：

λ² − (a + d)λ + (ad − bc) = 0

Equivalently, this is:

等价地，可以写成：

λ² − tr(A)λ + det(A) = 0

t. This form is very useful in exams because it avoids expanding det(A − λI) from scratch.

其中 tr(A) = a + d 是迹，det(A) = ad − bc 是行列式。这种形式在考试中非常有用，因为它避免了从头展开 det(A − λI)。

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5. Worked Example for a 2 × 2 Matrix | 2 × 2 矩阵例题

Let A = [[4, 1], [2, 3]]. We find the characteristic equation and the eigenvalues.

设 A = [[4, 1], [2, 3]]。我们来求特征方程和特征值。

First compute the trace and determinant:

首先计算迹和行列式：

tr(A) = 4 + 3 = 7, det(A) = 4 × 3 − 1 × 2 = 10

So the characteristic equation is:

所以特征方程为：

λ² − 7λ + 10 = 0

Factorising gives (λ − 2)(λ − 5) = 0, so the eigenvalues are λ = 2 and λ = 5.

因式分解得 (λ − 2)(λ − 5) = 0，因此特征值为 λ = 2 和 λ = 5。

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6. The 3 × 3 Case and Cubic Characteristic Polynomial | 3 × 3 情形与三次特征多项式

a 3 × 3 matrix A, the characteristic equation det(A − λI) = 0 is a cubic polynomial in λ.

对于 3 × 3 矩阵 A，特征方程 det(A − λI) = 0 是一个关于 λ 的三次多项式。

It can be expanded in the general form:

它可以展开为一般形式：

λ³ − tr(A)λ² + (sum of principal minors)λ − det(A) = 0

below the diagonal, then the characteristic equation is simply (λ − 2)(λ − 3)(λ − 4) = 0.

当 A 是三角矩阵时会有一个更快的特殊情况。例如，若 A 的对角线元素为 2, 3, 4，且对角线以下全为零，则特征方程就是 (λ − 2)(λ − 3)(λ − 4) = 0。

Expanding this gives λ³ − 9λ² + 26λ − 24 = 0, so the eigenvalues are 2, 3, and 4.

展开后得到 λ³ − 9λ² + 26λ − 24 = 0，因此特征值为 2、3 和 4。

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7. Finding Eigenvectors Once Eigenvalues Are Known | 已知特征值求特征向量

as been found, the corresponding eigenvectors are the non-zero solutions of (A − λI)v = 0.

一旦求得特征值 λ，对应的特征向量就是 (A − λI)v = 0 的非零解。

For the earlier example A = [[4, 1], [2, 3]] with λ = 2, we have:

对于前面的例子 A = [[4, 1], [2, 3]]，当 λ = 2 时，有：

A − 2I = [[2, 1], [2, 1]]

Solving (A − 2I)(x, y)ᵀ ives 2x + y = 0, so y = −2x. Thus the eigenvectors are k(1, −2) for any non-zero scalar k.

解 (A − 2I)(x, y)ᵀ = 0 得 2x + y = 0，因此 y = −2x。所以特征向量为 k(1, −2)，其中 k 为任意非零标量。

For λ = 5, solving (A − 5I)v = 0 gives −x + y = 0, so the eigenvectors are k(1, 1).

对于 λ = 5，解 (A − 5I)v = 0 得 −x + y = 0，所以特征向量为 k(1, 1)。

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8. Repeated Eigenvalues and Algebraic Multiplicity | 重复特征值与代数重数

tr(A)² − 4det(A) determines whether the characteristic equation has distinct or repeated roots.

对于 2 × 2 矩阵，判别式 Δ = tr(A)² − 4det(A) 决定了特征方程有相异根还是重复根。

e. Its algebraic multiplicity is the number of times it appears as a root of the characteristic polynomial.

若 Δ = 0，则存在重复特征值。它的代数重数就是它作为特征多项式根出现的次数。

wever, its geometric multiplicity may be smaller, meaning there may not be two independent eigenvectors.

例如，A = [[2, 1], [0, 2]] 的特征方程为 (λ − 2)² = 0，所以 λ = 2 的代数重数为 2。然而它的几何重数可能更小，意味着可能不存在两个线性无关的特征向量。

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9. Complex Eigenvalues and Geometric Interpretation | 复特征值与几何解释

2 matrix, the characteristic equation has two conjugate complex roots of the form α ± iβ.

若实 2 × 2 矩阵满足 Δ < 0，则特征方程有两个共轭复根，形式为 α ± iβ。

For example, A = [[0, −1], [1, 0]] has characteristic equation λ² + 1 = 0, giving λ = ±i.

例如，A = [[0, −1], [1, 0]] 的特征方程为 λ² + 1 = 0，解得 λ = ±i。

Geometrically, complex eigenvalues indicate that the transformation involves a rotation combined with a scaling, rather than simple stretching along fixed directions.

从几何上看，复特征值表明该变换包含旋转与缩放的组合，而不是沿固定方向的简单拉伸。

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10. Common Mistakes and Exam Tips | 常见错误与考试提示

One common mistake is using det(λI − A) instead of det(A − λI). Both have the same roots, but the polynomial may differ by a factor of −1, so signs in the expansion must be handled carefully.

一个常见错误是使用 det(λI − A) 而不是 det(A − λI)。两者有相同的根，但多项式可能相差一个 −1 因子，因此展开时的符号需要小心处理。

−tr(A), not tr(A). Always check that the sum of the eigenvalues equals the trace and the product equals the determinant.

另一个错误是忘记在 2 × 2 情形中 λ 的系数是 −tr(A)，而不是 tr(A)。始终要检查特征值之和等于迹，特征值之积等于行列式。

rise the cubic carefully and check each root by substitution back into the characteristic equation.

求特征向量时，记住任意非零标量倍数都是可以的。在 3 × 3 问题中，要仔细因式分解三次多项式，并将每个根代回特征方程进行检验。

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