Chapter 5 (Eigenvalues and Eigenvectors): Eigenvectors and eigenvalues (特征向量和特征值)

本文为《Linear algebra and its applications》的读书笔记

Eigenvectors and Eigenvalues

Eigenvectors and Eigenvalues

• Although a transformation $\mathbf{x}\mapsto A\mathbf{x}$ may move vectors in a variety of directions, it often happens that there are special vectors which are transformed by $A$ into scalar multiples of themselves.

• Note that an eigenvector must be nonzero, by definition, but an eigenvalue may be zero.
• Let $\lambda$ be an eigenvalue of an invertible matrix $A$. Then ${\lambda }^{-1}$ is an eigenvalue of $A^{-1}$.

Eigenspace

特征空间

• $\lambda$ is an eigenvalue of an $n\times n$ matrix $A$ if and only if the equation
  $$(A-\lambda I)\mathbf{x}=0(3)$$has a nontrivial solution. The set of all solutions of (3) is just the null space of the matrix $A-\lambda I$. So this set is a subspace of ${\mathbb{R}}^n$ and is called the eigenspace (特征空间) of $A$ corresponding to $\lambda$.
  • The eigenspace consists of the zero vector and all the eigenvectors corresponding to $\lambda$.
  • Note that the linear combination of the eigenvectors corresponding to $\lambda$ is still an eigenvector corresponding to $\lambda$.

  
  

EXAMPLE 3

Show that 7 is an eigenvalue of matrix $A=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]$, and find the corresponding eigenvectors.

SOLUTION

• The scalar 7 is an eigenvalue of $A$ if and only if the equation
  $$A\mathbf{x}=7\mathbf{x}$$has a nontrivial solution, which is equivalent to $(A-7I)\mathbf{x}=0$
  
• To solve this homogeneous equation, form the matrix
  The general solution has the form $x_2\left[\begin{array}{c}1\\ 1\end{array}\right]$. Each vector of this form with $x_2\ne 0$ is an eigenvector corresponding to $\lambda =7$.
  

EXAMPLE 4

• Let $A=\left[\begin{array}{ccc}4& -1& 6\\ 2& 1& 6\\ 2& -1& 8\end{array}\right]$. An eigenvalue of $A$ is 2. It can be shown that the basis for the corresponding eigenspace is { [ 1 2 0 ] , [ − 3 0 1 ] } \{\begin{bmatrix}1\\2\\0\end{bmatrix},\begin{bmatrix}-3\\0\\1\end{bmatrix}\} {
  ⎣⎡​120​⎦⎤​,⎣⎡​−301​⎦⎤​}. So the eigenspace is a two-dimensional subspace of ${\mathbb{R}}^3$.
  
  

推论: $A$ 与 $A^T$ 的特征值相同

• Show that $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $A^T$.

SOLUTION

• For any $\lambda$, $(A–\lambda I{)}^T=A^T–(\lambda I{)}^T=A^T–\lambda I$. Since $(A–\lambda I{)}^T$ is invertible if and only if $(A–\lambda I)$ is invertible, we conclude that $A^T–\lambda I$ is not invertible if and only if $A–\lambda I$ is not invertible. That is, $\lambda$ is an eigenvalue of $A^T$ if and only if $\lambda$ is an eigenvalue of $A$.

EXERCISES 30

Consider an $n\times n$ matrix $A$ with the property that the column sums all equal the same number $s$. Show that $s$ is an eigenvalue of $A$.

SOLUTION

• [Hint: the row sums of $A^T$ all equal the same number $s$]
  
  
• 也可以证明 $det(A-sI)=0$

Eigenvalues of a triangular matrix

• The following theorem describes one of the few special cases in which eigenvalues can be found precisely. Calculation of eigenvalues will also be discussed in Section 5.2.

PROOF

• For simplicity, consider the $3\times 3$ case. If $A$ is upper triangular, then $A-\lambda I$ has the form
  The scalar $\lambda$ is an eigenvalue of $A$ if and only if the equation $(A-\lambda I)\mathbf{x}=0$ has a nontrivial solution, that is, if and only if the equation has a free variable. It is easy to see that $(A-\lambda I)\mathbf{x}=0$ has a free variable if and only if at least one of the entries on the diagonal of $A-\lambda I$ is zero. This happens if and only if $\lambda$ equals one of the entries $a_{11},a_{22},a_{33}$ in $A$.
  

Eigenvalue of 0

• What does it mean for a matrix $A$ to have an eigenvalue of 0? This happens if and only if the equation
  $$A\mathbf{x}=0\mathbf{x}$$has a nontrivial solution. But (4) is equivalent to $A\mathbf{x}=0$, which has a nontrivial solution if and only if $A$ is not invertible.
  
• Thus 0 is an eigenvalue of $A$ if and only if $A$ is not invertible. This fact will be added to the Invertible Matrix Theorem.

不同特征值对应的特征向量之间线性无关

PROOF

• Suppose { v 1 , . . . , v r } \{\boldsymbol v_1,…, \boldsymbol v_r\} {
  v1​,...,vr​} is linearly dependent. Since ${\mathbf{v}}_1$ is nonzero, One of the vectors in the set is a linear combination of the preceding vectors. Let $p$ be the least index such that ${\mathbf{v}}_{p+1}$ is a linear combination of the preceding (linearly independent) vectors. Then there exist scalars $c_1,...,c_p$ such that
  $$c_1{\mathbf{v}}_1+...+c_p{\mathbf{v}}_p={\mathbf{v}}_{p+1}(5)$$Multiplying both sides of (5) by $A$, we obtain
  $$\begin{gathered} c_{1}A{\mathbf{v}}_1+...+c_{p}A{\mathbf{v}}_p=A{\mathbf{v}}_{p+1} \\ {c}_1{\lambda }_1{\mathbf{v}}_1+...+c_p{\lambda }_p{\mathbf{v}}_p={\lambda }_{p+1}{\mathbf{v}}_{p+1}(6) \end{gathered}$$Multiplying both sides of (5) by ${\lambda }_{p+1}$ and subtracting the result from (6), we have
  $$c_1({\lambda }_1-{\lambda }_{p+1}){\mathbf{v}}_1+...+c_p({\lambda }_p-{\lambda }_{p+1}){\mathbf{v}}_p=0(7)$$
  
  
  
• Since { v 1 , . . . , v p } \{\boldsymbol v_1,…, \boldsymbol v_p\} {
  v1​,...,vp​} is linearly independent, the weights in (7) are all zero, which is impossible. Hence { v 1 , . . . , v r } \{\boldsymbol v_1,…, \boldsymbol v_r\} {
  v1​,...,vr​} cannot be linearly dependent and therefore must be linearly independent.

推论: $n\times n$ 矩阵最多只有 $n$ 个不同特征值

• Show that an $n\times n$ matrix can have at most $n$ distinct eigenvalues.

PROOF

• If an $n\times n$ matrix has $p$ distinct eigenvalues, then by Theorem 2 there would be a linearly independent set of $p$ eigenvectors (one for each eigenvalue). Since these vectors belong to an $n$-dimensional vector space, $p$ cannot exceed $n$.

Eigenvectors and Difference Equations

• This section concludes by showing how to construct solutions of the first-order difference equation:
  $${\mathbf{x}}_{k+1}=A{\mathbf{x}}_k(k=0,1,2,...)(8)$$
  
• If $A$ is an $n\times n$ matrix, then (8) is a recursive description of a sequence { x k } \{\boldsymbol x_k\} {
  xk​} in ${\mathbb{R}}^n$. A solution of (8) is an explicit description of { x k } \{\boldsymbol x_k\} {
  xk​} whose formula for each ${\mathbf{x}}_k$ does not depend directly on $A$ or on the preceding terms in the sequence other than the initial term ${\mathbf{x}}_0$.
• The simplest way to build a solution of (8) is to take an eigenvector ${\mathbf{x}}_0$ and its corresponding eigenvalue $\lambda$ and let
  $${\mathbf{x}}_k={\lambda }^k{\mathbf{x}}_0(k=1,2,...)(9)$$This sequence is a solution because
  $$A{\mathbf{x}}_k=A({\lambda }^k{\mathbf{x}}_0)={\lambda }^k(A{\mathbf{x}}_0)={\lambda }^k(\lambda {\mathbf{x}}_0)={\lambda }^{k+1}{\mathbf{x}}_0={\mathbf{x}}_{k+1}$$Linear combinations of solutions in the form of equation (9) are solutions, too!