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矩阵特征值与特征向量：$\begin{bmatrix}2

Eigenvalues and Eigenvectors

专题
Algorithmic Programming / 算法编程
难度
L4

题目详情

对矩阵

A=[2112],A=\begin{bmatrix}2&1\\1&2\end{bmatrix},

求特征值与特征向量。

The determinant of a 2×22\times 2 matrix: >det⁡ ⁣[abcd]=ad−bc,>> \det\!\begin{bmatrix}a & b\\ c & d\end{bmatrix}=ad-bc, > and so on. Properties: det⁡(AB)=det⁡(A)det⁡(B),\det(AB)=\det(A)\det(B), det⁡(AT)=det⁡(A),\det(A^T)=\det(A), etc.

Eigenvalues: If Ax=λx,A x=\lambda x, λ\lambda is an eigenvalue, xx is an eigenvector.

  • Characteristic eqn: det⁡(A−λI)=0.\det(A-\lambda I)=0.
  • ∏λi=det⁡(A),∑λi=trace(A).\prod \lambda_i = \det(A),\quad \sum \lambda_i=\mathrm{trace}(A).

Diagonalizable: if AA has nn linearly independent eigenvectors.

Question: For
A=[2112],A = \begin{bmatrix}2 & 1\\ 1 & 2\end{bmatrix},
find its eigenvalues and eigenvectors.

解析

特征方程：

det⁡(A−λI)=(2−λ)2−1=λ2−4λ+3=0,\det(A-\lambda I)=(2-\lambda)^2-1=\lambda^2-4\lambda+3=0,

所以 λ∈{1,3}\lambda\in\{1,3\}。

  • λ=3\lambda=3 时，特征向量可取 (1,1)(1,1)；
  • λ=1\lambda=1 时，特征向量可取 (1,−1)(1,-1)。

Original Explanation

  1. Direct. Solve Ax=λx.Ax=\lambda x. That is [2112][x1x2]=λ[x1x2].\begin{bmatrix}2 & 1\\1 & 2\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} = \lambda \begin{bmatrix}x_1\\x_2\end{bmatrix}. Summing the two equations suggests λ=3\lambda=3 or 11. The corresponding eigenvectors: (1,1)(1,1) and (1,−1).(1,-1).

  2. Characteristic: det⁡ ⁣(A−λI)=det⁡ ⁣[2−λ112−λ]=(2−λ)2−1=λ2−4λ+3=0,\det\!\bigl(A-\lambda I\bigr) = \det\!\begin{bmatrix}2-\lambda & 1\\1 & 2-\lambda\end{bmatrix} = (2-\lambda)^2 -1 = \lambda^2 -4\lambda +3=0, so λ=1\lambda=1 or 33. Substituting back yields eigenvectors (1,−1)(1,-1) and (1,1).(1,1).

  3. Trace & determinant:

    • det⁡(A)=4−1=3=λ1λ2.\det(A)=4-1=3=\lambda_1\lambda_2.
    • trace(A)=4=λ1+λ2.\mathrm{trace}(A)=4=\lambda_1+\lambda_2.
      Solve λ1+λ2=4\lambda_1+\lambda_2=4 and λ1λ2=3\lambda_1\lambda_2=3 giving (1,3)(1,3) or (3,1).(3,1).

Hence eigenvalues {1,3}\{1,3\} with vectors (1,−1)(1,-1) and (1,1).(1,1).