📚 Eigenvalues of Matrices: Definition and Computation | 矩阵特征值的定义与计算方法
Eigenvalues are among the most important concepts in linear algebra, appearing throughout the IB Mathematics Analysis and Approaches HL syllabus as part of the Matrices topic. They reveal the fundamental structure of a matrix and enable powerful applications in diagonalisation, systems of differential equations, Markov chains, and graph theory. In this article, we will define eigenvalues precisely, derive the characteristic equation, and practise effective computational methods for 2×2 and 3×3 matrices.
特征值是线性代数中最核心的概念之一，贯穿 IB 数学分析与方法 HL 课程中矩阵部分的内容。特征值揭示了矩阵的基本结构，并为对角化、微分方程组、马尔可夫链和图论等强大应用奠定基础。本文将精确定义特征值，推导特征方程，并系统练习 2×2 与 3×3 矩阵的特征值计算方法。
1. What is an Eigenvalue? | 什么是特征值？
When a matrix A multiplies a vector v, the result Av is generally a new vector with a different direction and length. However, for some special non-zero vectors, the product Av is simply a scalar multiple of v: the direction stays the same, and only the length changes. These special vectors are called eigenvectors, and the corresponding scalar factor is called an eigenvalue.
当矩阵 A 乘以向量 v 时，结果 Av 通常是一个方向和长度都不同的新向量。然而，对于某些特殊的非零向量，乘积 Av 恰好是 v 的标量倍数：方向不变，只有长度改变。这些特殊向量称为特征向量，对应的标量因子称为特征值。
For example, let A be the matrix [[2, 0], [0, 3]]. If we take v = [1, 0], then Av = [2, 0] = 2v, so v is an eigenvector with eigenvalue 2. This geometric intuition is the starting point for the formal definition.
例如，设矩阵 A = [[2, 0], [0, 3]]。取 v = [1, 0]，则 Av = [2, 0] = 2v，因此 v 是特征向量，对应特征值为 2。这种几何直观是正式定义的出发点。
2. Formal Definition and the Eigenvalue Equation | 正式定义与特征值方程
Let A be an n×n matrix. A scalar λ is an eigenvalue of A if there exists a non-zero vector v in ℝⁿ such that the eigenvalue equation holds:
设 A 是一个 n×n 矩阵。若存在非零向量 v ∈ ℝⁿ 使得以下特征值方程成立，则标量 λ 称为 A 的特征值：
Av = λv, v ≠ 0
The non-zero vector v is called an eigenvector corresponding to λ. The condition v ≠ 0 is essential: if v were zero, the equation 0 = 0 would hold for every λ, which carries no information.
非零向量 v 称为对应于 λ 的特征向量。条件 v ≠ 0 至关重要：若 v 为零向量，则等式 0 = 0 对任意 λ 都成立，不包含任何信息。
Rearranging the eigenvalue equation gives (A − λI)v = 0, where I is the identity matrix. For this homogeneous system to have a non-trivial solution, the matrix A − λI must be singular, which means its determinant must be zero:
将特征值方程移项可得 (A − λI)v = 0，其中 I 是单位矩阵。为使该齐次方程组存在非零解，矩阵 A − λI 必须是奇异的，即其行列式必为零：
det(A − λI) = 0
3. The Characteristic Equation | 特征方程
The equation det(A − λI) = 0 is called the characteristic equation of A. For an n×n matrix, the left-hand side is a polynomial of degree n in λ, known as the characteristic polynomial. The eigenvalues of A are exactly the roots of this polynomial.
方程 det(A − λI) = 0 称为 A 的特征方程。对于 n×n 矩阵，其左侧是关于 λ 的 n 次多项式，称为特征多项式。A 的特征值恰好是该多项式的根。
For a general 2×2 matrix A = [[a, b], [c, d]], we compute:
对于一般的 2×2 矩阵 A = [[a, b], [c, d]]，我们计算：
A − λI = [[a−λ, b], [c, d−λ]]
det(A − λI) = (a−λ)(d−λ) − bc = λ² − (a+d)λ + (ad − bc)
Thus the characteristic equation of a 2×2 matrix takes the elegant form:
因此，2×2 矩阵的特征方程可写成简洁的形式：
λ² − tr(A)λ + det(A) = 0
where tr(A) = a + d is the trace (sum of diagonal entries) and det(A) = ad − bc is the determinant. This shortcut saves time and is strongly recommended in exams.
其中 tr(A) = a + d 是矩阵的迹（对角线元素之和），det(A) = ad − bc 是矩阵的行列式。这个捷径可以节省时间，是考试中强烈推荐的方法。
4. Computing Eigenvalues of 2×2 Matrices | 2×2 矩阵的特征值计算
The procedure for finding eigenvalues of a 2×2 matrix can be summarised in three steps. We will illustrate each step with a worked example.
求 2×2 矩阵特征值的过程可以归结为三个步骤。我们将通过一个例题逐步说明。
Step 1: Form the matrix A − λI. Subtract λ from each diagonal entry, leaving the off-diagonal entries unchanged.
步骤 1：构造矩阵 A − λI。从每个对角线元素中减去 λ，非对角线元素保持不变。
Step 2: Compute its determinant. For a 2×2 matrix, multiply the two diagonal entries and subtract the product of the two off-diagonal entries.
步骤 2：计算其行列式。对于 2×2 矩阵，将两个对角线元素相乘，再减去两个非对角线元素的乘积。
Step 3: Set the determinant equal to zero and solve. The resulting quadratic equation is solved either by factorisation or by the quadratic formula.
步骤 3：令行列式等于零并求解。所得二次方程可通过因式分解或求根公式求解。
Worked Example. Find the eigenvalues of A = [[4, 1], [2, 3]].
例题。求 A = [[4, 1], [2, 3]] 的特征值。
Following Step 1, we write A − λI = [[4−λ, 1], [2, 3−λ]]. By Step 2, the determinant is:
按照步骤 1，写出 A − λI = [[4−λ, 1], [2, 3−λ]]。由步骤 2，行列式为：
det(A − λI) = (4−λ)(3−λ) − 1×2 = 12 − 7λ + λ² − 2 = λ² − 7λ + 10
Setting det = 0 gives λ² − 7λ + 10 = 0. Factorising: (λ − 5)(λ − 2) = 0, so the eigenvalues are λ = 5 and λ = 2. Note that we could have used the shortcut directly: tr(A) = 7 and det(A) = 10, giving exactly the same quadratic equation.
令行列式等于零，得 λ² − 7λ + 10 = 0。因式分解：(λ − 5)(λ − 2) = 0，因此特征值为 λ = 5 与 λ = 2。注意我们也可以直接使用捷径：tr(A) = 7，det(A) = 10，得到完全相同的二次方程。
5. Computing Eigenvalues of 3×3 Matrices | 3×3 矩阵的特征值计算
For 3×3 matrices, the same idea applies, but the determinant expansion requires more care. We use cofactor expansion along a convenient row or column.
对于 3×3 矩阵，基本思路相同，但行列式展开需要更加小心。我们沿选定的行或列进行余子式展开。
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