特征值/特征向量数量

对实系数的 $n\times n$ 矩阵 $A$：

• 在复数域上，特征多项式次数为 $n$，因此有 $\boxed{n}$ 个特征值（按代数重数计）。在实数域上，实特征值个数不一定是 $n$（可能出现共轭复根成对）。

• 对任意特征值 $\lambda$，其对应特征向量集合是一个非零线性子空间（特征子空间），因此“特征向量”按比例缩放有无穷多个。

• 线性无关的特征向量最多 $\boxed{n}$ 个，最少至少 1 个（每个特征值至少有 1 个非零特征向量）；是否能凑满 $n$ 个取决于是否可对角化。

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英文解析

Any $n \times n$ matrix with real entries has $n$ eigenvalues, counted with their multiplicities; some of the eigenvalues may be complex numbers. Any $n \times n$ matrix has at most $n$ eigenvectors.

Let $A$ be an $n \times n$ matrix. Let $\lambda$ be an eigenvalue of $A$ with corresponding eigenvector $\nu \neq 0$ , and let $P_A(x) = \operatorname *{det}(xI_n - A)$ be the characteristic polynomial of $A$ , where $I_n$ is the $n \times n$ identity matrix. Note that

$$\begin{array}{r l}{A\nu = \lambda \nu ,\nu \neq 0\Longleftrightarrow} & {(\lambda I_{n} - A)\nu = 0,\nu \neq 0}\\ {\Longleftrightarrow} & {\lambda I_{n} - A\mathrm{~singular~matrix}}\\ {\Longleftrightarrow} & {\operatorname *{det}(\lambda I_{n} - A) = 0}\\ {\Longleftrightarrow} & {P_{A}(\lambda) = 0.} \end{array}$$

In other words, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is a root of the corresponding characteristic polynomial $P_A(x)$ . Since $P_A(x)$ is a polynomial of degree $n$ , it follows from the Fundamental Theorem of Algebra that $P_A(x)$ has exactly $n$ (complex) roots when counted with their multiplicities. We conclude that any $n \times n$ matrix has $n$ eigenvalues, counted with their multiplicities.

An eigenvalue of multiplicity $m$ has at least one eigenvector and at most $m$ linearly independent corresponding eigenvectors, but it may have less than $m$ linearly independent eigenvectors. Thus, an $n \times n$ matrix has at most $n$ eigenvectors, and at least

as many eigenvectors as the number of distinct eigenvalues of the matrix.