Powers of Matrices | 矩阵的幂

📚 Powers of Matrices | 矩阵的幂

Matrix powers appear everywhere in advanced mathematics, from modelling population dynamics and Markov chains to solving systems of linear recurrence relations. In the IB Mathematics curriculum, particularly at HL, understanding how to compute Aⁿ efficiently is a vital skill that bridges algebra, eigenvalues, and diagonalization. This article builds the theory from first principles, works through fully solved examples, and explores applications as well as limitations.

矩阵的幂在高等数学中无处不在，从种群动态建模、马尔可夫链到线性递推方程组的求解。在 IB 数学课程（尤其是 HL）中，理解如何高效计算 Aⁿ 是连接代数、特征值与对角化的重要技能。本文从基本原理出发构建理论，给出详尽求解实例，并探讨其应用与局限性。

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1. What Are Matrix Powers? | 什么是矩阵的幂？

For a square matrix A of order n, the k-th power A^k (k ∈ ℕ) is defined as repeated multiplication: A^k = A · A · ... · A (k factors). By convention, A⁰ the identity matrix of the same size. Powers follow index laws similar to real numbers: A^mAⁿ = A^m+n and (A^m)ⁿ = A^mn. However, because matrix multiplication is not commutative, (AB)ⁿ ≠ AⁿBⁿ in general, and we must be careful with the order of factors.

对于 n 阶方阵 A，k 次幂 A^k（k ∈ ℕ）定义为重复自乘：A^k = A · A · ... · A（k 个因子）。按惯例，A⁰ = I（同阶单位矩阵）。矩阵幂满足与实数类似的指数律：A^mAⁿ = A^m+n 且 (A^m)ⁿ = A^mn。然而，由于矩阵乘法不满足交换律，一般情况下 (AB)ⁿ ≠ AⁿBⁿ，我们必须格外注意因子的顺序。

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2. Direct Computation of Low Powers | 低次幂的直接计算

exponent is small, we can simply multiply the matrix by itself. Consider the 2×2 matrix A:

2	1
−1	3

For small exponents we multiply stepwise: A² = A×A, A³ = A²×A, and so on.

指数较小时，我们可直接将矩阵自乘。对于上述 2×2 矩阵，逐步计算：A² = A×A，A³ = A²×A，以此类推。

Calculating A²:

2	1
−1	3

×

2	1
−1	3

=

3	5
−5	8

A³ = A²A yields:

1	18
−18	19

Already the entries grow and the arithmetic becomes tedious. For A¹⁰⁰, direct multiplication is impractical. We need a smarter approach – diagonalization.

此时矩阵元已经变大且运算变得繁琐。若要计算 A¹⁰⁰，直接乘法不切实际。我们需要更聪明的方法 —— 对角化。

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3. Diagonalization: The Key to High Powers | 对角化：计算高次幂的关键

If a square matrix A can be written as A = P D P⁻¹, where D is a diagonal matrix and P is invertible, then raising A to the power n becomes straightforward:

Aⁿ = (P D P⁻¹)ⁿ = P D P⁻¹ P D P⁻¹ ... P D P⁻¹ = P Dⁿ P⁻¹

The diagonal matrix D has eigenvalues λ₁, λ₂, ... on its main diagonal, and Dⁿ simply has λ₁ⁿ, λ₂ⁿ, ... on its diagonal. Thus the hard work reduces to finding P and D, after which any power can be written down explicitly.

若方阵 A 可写为 A = P D P⁻¹，其中 D 为对角矩阵，P 可逆，则求 A 的 n 次幂变得极其简单：Aⁿ = (P D P⁻¹)ⁿ = P Dⁿ P⁻¹。对角矩阵 D 的主对角线元素为特征值 λ₁, λ₂, …，而 Dⁿ 只是将这些元素变为 λ₁ⁿ, λ₂ⁿ, …。因此，繁重的工作转化为寻找 P 和 D，之后任何次幂均可直接写出通项。

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4. Diagonalization Process | 对角化过程

For a 2×2 matrix, the steps are:

1. Find eigenvalues by solving det(A − λI) = 0.
2. For each eigenvalue, find a corresponding eigenvector.
3. Form P with eigenvectors as columns, and D with eigenvalues on the diagonal in the same order.
4. Compute P⁻¹.
5. Write Aⁿ = P Dⁿ P⁻¹.

对 2×2 矩阵，步骤如下：1) 解 det(A − λI) = 0 求特征值；2) 对每个特征值求相应特征向量；3) 以特征向量为列构成 P，按相同顺序将特征值置于 D 的对角线上；4) 计算 P⁻¹；5) 写出 Aⁿ = P Dⁿ P⁻¹。

Let us diagonalize the matrix A =

2	1
1	2

.

Characteristic equation: det(A − λI) = (2−λ)² − 1 = λ² − 4λ + 3 = 0 → λ = 3, 1.

特征方程：(2−λ)² − 1 = λ² − 4λ + 3 = 0，得特征值 3 与 1。

For λ=3: solve (A−3I)v = 0 →

−1	1
1	−1

v = 0 ⇒ v₁ =

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