Diagonalisation | 矩阵的对角化

📚 Diagonalisation | 矩阵的对角化

Diagonalisation is a powerful technique in linear algebra that simplifies a square matrix by expressing it in terms of its eigenvalues and eigenvectors. For an n×n matrix A, diagonalisation finds an invertible matrix P and a diagonal matrix D such that P⁻¹AP = D. This transformation often reveals the essential structure of A and makes computations such as powers and exponentials far more manageable.

对角化是线性代数中一项强大的技术，它通过特征值和特征向量来简化一个方阵。对于 n×n 矩阵 A，对角化旨在找到一个可逆矩阵 P 和一个对角矩阵 D，使得 P⁻¹AP = D。这种变换常常揭示矩阵 A 的本质结构，并使得矩阵幂和指数等计算变得更加容易处理。

1. Eigenvalues and Eigenvectors Refresher | 特征值与特征向量回顾

Before discussing diagonalisation, we must recall the definitions of eigenvalues and eigenvectors. A non-zero vector v is called an eigenvector of a square matrix A if multiplying A by v produces a scalar multiple of v. The scalar λ is called the corresponding eigenvalue.

在讨论对角化之前，我们必须回顾特征值和特征向量的定义。非零向量 v 称为方阵 A 的特征向量，如果 A 乘以 v 的结果是 v 的标量倍数。这个标量 λ 称为对应的特征值。

Av = λv, where v ≠ 0

To find eigenvalues, we solve the characteristic equation det(A − λI) = 0. The eigenvectors are then found by solving (A − λI)v = 0 for each eigenvalue λ.

为了求特征值，我们解特征方程 det(A − λI) = 0。对于每个特征值 λ，再通过求解 (A − λI)v = 0 得到对应的特征向量。

2. The Diagonalisation Condition | 可对角化的条件

An n×n matrix A is diagonalisable if and only if it has n linearly independent eigenvectors. Equivalently, the sum of the geometric multiplicities of all distinct eigenvalues must equal n. The geometric multiplicity of an eigenvalue is the dimension of its eigenspace, which equals n − rank(A − λI).

一个 n×n 矩阵 A 可对角化的充分必要条件是它具有 n 个线性无关的特征向量。等价地说，所有不同特征值的几何重数之和必须等于 n。特征值的几何重数是指其对应特征子空间的维数，等于 n − rank(A − λI)。

A useful practical test is the following: if A has n distinct eigenvalues, then A is automatically diagonalisable. However, repeated eigenvalues may still allow diagonalisation provided each eigenvalue’s geometric multiplicity equals its algebraic multiplicity.

一个实用的判断方法是：如果 A 有 n 个互不相同的特征值，那么 A 必定可对角化。然而，即使有重复的特征值，只要每个特征值的几何重数等于其代数重数，矩阵仍然可能可对角化。

3. Constructing the Matrices P and D | 构造矩阵 P 和 D

Suppose we have n linearly independent eigenvectors v₁, v₂, …, vₙ of A, with corresponding eigenvalues λ₁, λ₂, …, λₙ (not necessarily distinct). To diagonalise A, we place the eigenvectors as columns of P:

假设我们拥有 A 的 n 个线性无关特征向量 v₁, v₂, …, vₙ，对应的特征值为 λ₁, λ₂, …, λₙ（不必然互不相同）。要对 A 进行对角化，我们将这些特征向量按列放入矩阵 P：

P = [v₁ v₂ … vₙ]

The diagonal matrix D is formed by placing the eigenvalues on the main diagonal in the same order as the corresponding eigenvectors in P:

对角矩阵 D 由特征值按与 P 中特征向量相同的顺序排列在主对角线上构成：

D = diag(λ₁, λ₂, …, λₙ)

Then we have the key relation AP = PD, or equivalently P⁻¹AP = D. In many applications, we prefer the equivalent form A = PDP⁻¹.

于是我们得到关键关系式 AP = PD，等价地有 P⁻¹AP = D。在许多应用中，我们更常使用等价形式 A = PDP⁻¹。

4. Powers of Matrices via Diagonalisation | 用对角化计算矩阵幂

One of the main reasons to diagonalise a matrix is to compute its powers with ease. From A = PDP⁻¹, we observe that A² = (PDP⁻¹)(PDP⁻¹) = PD²P⁻¹, and by induction, for any positive integer m:

对角化的主要原因之一是能够轻松计算矩阵的幂。由 A = PDP⁻¹，我们可以得到 A² = (PDP⁻¹)(PDP⁻¹) = PD²P⁻¹，通过归纳法，对于任意正整数 m：

Aᵐ = PDᵐP⁻¹

Because D is diagonal, Dᵐ is simply the diagonal matrix obtained by raising each diagonal entry to the power m. This avoids repeatedly multiplying the original matrix by itself, which is especially efficient for large m.

因为 D 是对角矩阵，Dᵐ 只需将每个对角元素分别取 m 次幂即可得到。这种方法避免了反复将原矩阵自乘，对于较大的 m 特别高效。

For example, if D = diag(2, −1), then D⁵ = diag(32, −1). Applying this to a Markov process or a linear recurrence often yields closed-form solutions.

例如，若 D = diag(2, −1)，则 D⁵ = diag(32, −1)。将此应用于马尔可夫过程或线性递推，通常可以得到闭式解。

5. Worked Example: 2×2 Matrix | 2×2 矩阵实例

Consider the matrix A = [[2, 1], [1, 2]]. We will diagonalise A step by step.

考虑矩阵 A = [[2, 1], [1, 2]]，我们逐步对其进行对角化。

First find the characteristic polynomial:

首先求特征多项式：

det(A − λI) = det([2−λ, 1], [1, 2−λ]) = (2−λ)² − 1 = λ² − 4λ + 3 = 0

Thus λ₁ = 3 and λ₂ = 1. For λ = 3, solve (A − 3I)v = 0:

因此 λ₁ = 3，λ₂ = 1。对于 λ = 3，解 (A − 3I)v = 0：

(−1)x + y = 0 → choose v₁ = (1, 1)

For λ = 1, solve (A − I)v = 0:

对于 λ = 1，解 (A − I)v = 0：

x + y = 0 → choose v₂ = (1, −1)

These eigenvectors are linearly independent, so we set P and D as:

这两个特征向量线性无关，因此我们构造 P 和 D：

P = [[1, 1], [1, −1]], D = [[3, 0], [0, 1]]

Then P⁻¹ = (1/2)[[1, 1], [1, −1]] and A = PDP⁻¹. Notice that the order of eigenvalues in D matches the order of eigenvectors in P.

此时 P⁻¹ = (1/2)[[1, 1], [1, −1]]，且 A = PDP⁻¹。注意 D 中特征值的顺序与 P 中特征向量的顺序一致。

6. Computing Aᵐ for the Example | 计算实例中的 Aᵐ

Using the diagonalisation from the previous section, we can compute Aᵐ for any positive integer m. Since Aᵐ = PDᵐP⁻¹, we obtain:

利用上一节的对角化结果，我们可以计算任意正整数 m 的 Aᵐ。由于 Aᵐ = PDᵐP⁻¹，我们得到：

Aᵐ = (1/2) [[3ᵐ + 1, 3ᵐ − 1], [3ᵐ − 1, 3ᵐ + 1]]

For example, if m = 3, then A³ = (1/2) [[27 + 1, 27 − 1], [27 − 1, 27 + 1]] = [[14, 13], [13, 14]]. This matches direct multiplication of A three times.

例如，当 m = 3 时，A³ = (1/2) [[27 + 1, 27 − 1], [27 − 1, 27 + 1]] = [[14, 13], [13, 14]]。这与直接连续相乘三次的结果一致。

This method is particularly useful when m is large. Instead of performing many matrix multiplications, we only compute 3ᵐ and 1ᵐ, which is computationally far cheaper.

这种方法在 m 很大时尤其有用。我们不需要进行大量矩阵乘法，只需计算 3ᵐ 和 1ᵐ，计算成本大大降低。

7. Diagonalisation of 3×3 Matrices | 3×3 矩阵的对角化

For a 3×3 matrix, the same procedure applies. We find three eigenvalues and three linearly independent eigenvectors. If the eigenvalues are all distinct, the process is straightforward. If there are repeated eigenvalues, we must check whether each eigenvalue provides enough eigenvectors.

对于 3×3 矩阵，同样的步骤依然适用。我们求三个特征值和三个线性无关的特征向量。如果特征值全部互异，过程是直截了当的。如果存在重特征值，我们必须检查每个特征值是否提供了足够的特征向量。

Consider the matrix B = [[2, 0, 0], [0, 3, 1], [0, 0, 3]]. Its eigenvalues are 2, 3, 3. The eigenvalue 3 has algebraic multiplicity 2, but solving (B − 3I)v = 0 gives equations y + z = 0, which yields only one independent eigenvector, say (0, 1, −1). Thus B has only two independent eigenvectors in total and is not diagonalisable.

考虑矩阵 B = [[2, 0, 0], [0, 3, 1], [0, 0, 3]]。其特征值为 2, 3, 3。特征值 3 的代数重数为 2，但解 (B − 3I)v = 0 得到方程 y + z = 0，只产生一个独立特征向量，例如 (0, 1, −1)。因此 B 总共只有两个独立特征向量，不可对角化。

In contrast, if the repeated eigenvalue has a full eigenspace, diagonalisation is possible. For example, the identity matrix has eigenvalue 1 repeated three times, and it has three independent eigenvectors, so it is trivially diagonalisable.

相反，如果重特征值的特征子空间维数足够，则可以对角化。例如，单位矩阵的特征值 1 重复三次，但它有三个独立特征向量，因此显然可对角化。

8. Applications to Recurrence Relations | 在递推关系中的应用

Diagonalisation is frequently used to solve linear recurrence relations, especially systems of first-order recurrences. Suppose we have a vector sequence defined by xₖ₊₁ = Axₖ, where xₖ is a column vector. Then xₖ = Aᵏx₀. If A is diagonalisable, we can write xₖ = PDᵏP⁻¹x₀, which gives an explicit closed form.

对角化常用于求解线性递推关系，尤其是一阶递推方程组。假设向量序列由 xₖ₊₁ = Axₖ 定义，其中 xₖ 是列向量。那么 xₖ = Aᵏx₀。如果 A 可对角化，我们可以写成 xₖ = PDᵏP⁻¹x₀，从而得到显式的闭式解。

For a single scalar recurrence such as uₙ₊₂ = uₙ₊₁ + uₙ (Fibonacci-like), one can convert it to a 2×2 matrix system and then diagonalise to derive Binet’s formula. This demonstrates the power of diagonalisation beyond pure matrix theory.

对于单个标量递推关系，如 uₙ₊₂ = uₙ₊₁ + uₙ（类似斐波那契数列），可以将其化为 2×2 矩阵系统，然后通过对角化推导出比内公式。这展示了对角化在纯矩阵理论之外的力量。

9. Limitations and Non-Diagonalisable Matrices | 局限性与非对角化矩阵

Not every matrix can be diagonalised. The fundamental obstruction is a lack of enough linearly independent eigenvectors. This occurs when some eigenvalue has geometric multiplicity strictly less than its algebraic multiplicity. Such matrices are called defective or non-diagonalisable.

并非每个矩阵都可以对角化。根本障碍在于缺乏足够数量的线性无关特征向量。当某个特征值的几何重数严格小于其代数重数时，就会出现这种情况。这类矩阵被称为亏损矩阵或不可对角化矩阵。

For defective matrices, we may instead use the Jordan normal form, which is a nearly diagonal form containing Jordan blocks on the diagonal. While the Jordan form is beyond the scope of A-Level Further Mathematics, it is helpful to know that diagonalisation is part of a broader theory.

对于亏损矩阵，我们可以改用若尔当标准形，这是一种近似对角的形态，其对角线上包含若尔当块。虽然若尔当标准形超出 A-Level 进阶数学的范围，但了解对角化是更广泛理论的一部分是有益的。

In an AQA A-Level context, you will generally work with matrices that have distinct eigenvalues, so diagonalisation is guaranteed. Always check for repeated eigenvalues and verify the number of independent eigenvectors before using the diagonalisation formula.

在 AQA A-Level 的语境中，你通常会遇到具有互异特征值的矩阵，因此对角化是有保证的。在使用对角化公式之前，务必检查是否存在重特征值，并验证独立特征向量的个数。

10. Summary | 总结

Diagonalisation expresses a square matrix A as A = PDP⁻¹, where D is diagonal and P contains linearly independent eigenvectors. The key steps are: find eigenvalues, find corresponding eigenvectors, check linear independence, and construct P and D.

对角化将方阵 A 表示为 A = PDP⁻¹，其中 D 是对角矩阵，P 由线性无关的特征向量组成。关键步骤包括：求特征值、求对应的特征向量、检验线性无关性，以及构造 P 和 D。

Once diagonalised, powers of A become easy to compute via Aᵐ = PDᵐP⁻¹. This technique has wide applications in solving recurrence relations, Markov chains, and systems of differential equations. Mastery of diagonalisation is an essential skill for A-Level Further Mathematics and beyond.

一旦完成对角化，矩阵幂可通过 Aᵐ = PDᵐP⁻¹ 方便计算。该技术在求解递推关系、马尔可夫链和微分方程组中有广泛应用。掌握对角化是 A-Level 进阶数学及更高阶段学习的一项重要技能。

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Diagonalisation | 矩阵的对角化